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General Forums => Aces High General Discussion => Topic started by: MANDO on November 26, 2004, 07:10:56 PM

Title: Final Decision
Post by: MANDO on November 26, 2004, 07:10:56 PM: The following is an interesting logic problem:

An attack pilot took off with a payload of three bombs under the fuselage. Few minutes later he received a radio call, by mistake, the ground crew loaded 2 training bombs (red) and only 1 real bomb (dark green).

The airplane has a lever in the cockpit with three positions to select and arm each bomb (left, center and right pilons), at the left of each position of the lever is a illuminated red led inditating that the ordenance is present at the corresponding pilon. This lever was at the lower position, so left pilon was selected.

To drop a bomb, the pilot needs to move the lever to the desired position, click a button on the lever to arm the bomb, wait three seconds and then press the trigger at the stick to release the bomb.

As he was aproaching the target, the center pilon led switched off, by some mechanical problem that bomb was released. The pilot inverted the plane quickly and looking at his high six saw a bright red object descending. He was lucky, it was one of the two training bombs.

Minutes later he was just over the target and due the heavy flak he will not have more than a single chance to drop one bomb and run back to home.

What should he do with the bomb selector lever?
Title: Final Decision
Post by: hornet63 on November 26, 2004, 08:10:50 PM: *****punt of course*****:rofl
Title: Final Decision
Post by: MANDO on November 27, 2004, 04:06:15 AM: To clarity the final situtation of the pilot:
He now has only two bombs attached to left and right aircraft underbelly pilons, one of these bombs is real (dark green), the other is a practice bomb (red). Remember that he lost the center pilon bomb and it was a red practice bomb.
The bomb lever selector is still at its original position, marking left pilon.

The pilot is starting the final attack dive, he has time left only to make a final decision: move the lever to select the other pilon or not. He will not have time to do a second attack. What should he do?
Title: Final Decision
Post by: DieAz on November 27, 2004, 04:10:56 AM: do like in MA arm both and be suicide dweeb ;)
Title: Final Decision
Post by: MANDO on November 27, 2004, 04:18:53 AM: Quote
Originally posted by DieAz
do like in MA arm both and be suicide dweeb ;)

The lever only allows you to select and arm one pilon. His final decision will show how dweeb he is (hopefuly he will not select center pilon, he would be dweeb and retard at the same time :p )
Title: Final Decision
Post by: bozon on November 27, 2004, 05:25:14 AM: switch to the other bomb. 66% that this is the live one.
that just goes to show the absurdity of statistical probability.

Bozon
Title: Final Decision
Post by: BlauK on November 27, 2004, 05:38:46 AM: Just drop either one of them before the attack and check the color... then u know if it is worth to do the final attack or to just bug off home :)

Bozon, how did u figure out that 66%?
I would suggest 50% at that point, whereas before the failure it was only 33%... or am I totally missing something?
Title: Final Decision
Post by: bozon on November 27, 2004, 06:24:39 AM: from statistics point of view, this is a problem of partitioning.
1st, the pilot chose one option out of the 3. So it means that he has got 33% to get it right, but 66% to be wrong.

Think of it as dividing his options into to groups, one with 33% to have the live bomb and the other with 66%.

The new information, that the center one was dud, left him with 2 bombs to choose from, but the partitioning to groups is still the same - one group has 33% and the other group (now containing only one bomb) has 66%. So he should switch.

The absurdity is that this assumes a cognitive first selection. Had the center bomb fallen without him looking at the selection lever position, and then he was asked to choose, he'd have 50-50% chance. In order for the new information (center is not live) to change something, a first decision had to be made prior to the info arriving. Alternatively, the random position of the lever before him choosing is also information.

Bozon
Title: Final Decision
Post by: MANDO on November 27, 2004, 07:22:14 AM: If we look at the final situation, the pilot has two bombs, one of them real. So, it would seem that he has same chances selecting any of them, 50%.

But Bozon is right, knowing that the center pilon had a training bomb is the key to change the lever from left pilon to right pilon.

If the pilot does not change the lever, or simply decices randomly which pilon to select (left or right), he will have 50% chances of dropping the real bomb, but changing to the right pilon he has 66%.
Title: Final Decision
Post by: BlauK on November 27, 2004, 07:45:29 AM: Yeah, right.
And if his Grandma had said no to his grandpa, the chance would be 88,9%

Looking at he lever or not does not change the chance % when he knows that one training bomb was lost. Prove me wrong, drop the bomb in similar case 100 times and prove that you get even close to 66 cases of the real bomb :p
Title: Final Decision
Post by: Furball on November 27, 2004, 07:55:33 AM: most boring thread EVER!

(http://www.hill.anorak.org.uk/pics/cbgbig.gif)
Title: Final Decision
Post by: MANDO on November 27, 2004, 08:53:23 AM: BlauK, may be you will understand with a different example:

Lets suppose the plane had 1000 bombs, only one of them real. Then the plane releases 998 bombs (all of them training bombs), and only two remain: the one initially selected by the lever and another. What would you do?

If you need an even more clear example, lets suppose you play Lottery. You have your number, and only one wizard knows the winning number (this wizard does not play Lottery). The wizard ask you to see your number, and then he tells you: "ok, the winning number is yours or this other one (and shows you another number)". You may change your number now by the other. What would you do?
Title: Final Decision
Post by: MANDO on November 27, 2004, 08:57:19 AM: Quote
Originally posted by BlauK
drop the bomb in similar case 100 times and prove that you get even close to 66 cases of the real bomb :p

You would be really surprised ... To reach a stable 66% probably you will need to repeat the case 150 times.
Title: Re: Final Decision
Post by: DREDIOCK on November 27, 2004, 09:06:54 AM: Quote
Originally posted by MANDO

.

What should he do with the bomb selector lever?

Nothing. He returns home dumping the last two along the way in case there were any other malfunction, then beats the crap out of his mechanic
Title: Final Decision
Post by: ALF on November 27, 2004, 10:22:42 AM: I musta missing something in college, if you flip a coin 10 times and it comes out heads all 10 times, what are the odds it will come out heads on try 11....50% because the next flip is a new subset, completely removed from the old set with no direct imperical relation.

You drop 2 of 3 bombs, only one of which is a live good bomb....your odds of dropping a dud are 66%, you now have 2 bombs, the odds are not 50%, because you are now dealing with a NEW probability, that is completely seperate from the initial one. Dropping the 1st bomb has nothing to do with the other two.

I believe that where some are making their mathmatical error is in the assertion that if we have a 33% chance of getting the 'live' one, and the first is a dud, they we can take the 33% chance we had initially and add the new 50% chance we have now, and somehow conclude that its a good bet you will get the live bomb on this second drop. And while if you fly the plane 1000 times, in the end you will see a 66% rate of a live bomb being dropped, this is only because 33% of the time the damned live bomb will drop on the initial malfunction.
You cannot take the information that the initial bomb is a dud and still use the probability of a three bomb subset, because youve made it a two bomb equasion.
Title: Final Decision
Post by: Zwerg on November 27, 2004, 10:34:52 AM: Quote
Originally posted by MANDO
BlauK, may be you will understand with a different example:

Lets suppose the plane had 1000 bombs, only one of them real. Then the plane releases 998 bombs (all of them training bombs), and only two remain: the one initially selected by the lever and another. What would you do?

If you need an even more clear example, lets suppose you play Lottery. You have your number, and only one wizard knows the winning number (this wizard does not play Lottery). The wizard ask you to see your number, and then he tells you: "ok, the winning number is yours or this other one (and shows you another number)". You may change your number now by the other. What would you do?

Yet anothter example:

You have 2 bombs (1 real, 1 training)

When you are near target you imagine that you carry an endless number of bombs. Then you imagine that you drop them all except the 2 bombs under your crate. Now you must switch for sure because the probability for success is 100%.
Title: Final Decision
Post by: MANDO on November 27, 2004, 11:25:31 AM: Quote
Originally posted by ALF
You cannot take the information that the initial bomb is a dud and still use the probability of a three bomb subset, because youve made it a two bomb equasion.

ALF, you are completely wrong. But dont worry, the problem is not easy to "see". I'll develop a very small program to show the effects of chosing the other pilon, once done I'll post it with the source code to check the algorythm.
Title: Final Decision
Post by: MANDO on November 27, 2004, 01:01:30 PM: Ok, here is the program. Is a very small windows program (virus free) with a .cpp source file to check the algorythm.

The program follows exactly the original description of the problem. It allows you to perform individual attacks manually or run automatically trying bombings every 100 ms.

The Final Decision Simulator (http://www.terra.es/personal2/matias.s/TheBomber.zip)

(Right click and save)
Title: Final Decision
Post by: BlauK on November 27, 2004, 01:10:26 PM: I comletely agree with Alf... It will be interesting to see your program, Mando. If it proves your point, it will overturn the field of statistics :)

As far as I have understood it, you are claiming that if you have planned to flip a coin twice and get heads with the first flip, you are somehow more likely to get tails with your second flip...

You are probably mixing the initial situation befere the first toss and the new situation where you have already flipped one coin. They both have different propabilities for this actual outcome. Things that have happened already do not affect the new propability, unless they take away the possibility of the same thing happening again (like numbers coming out of the lottery machine). Flipped coins do not reduce the possibility of heads or tails occuring again in the next flip..... neither does your first dud affect the propabilities between the last two bombs.

There is no possible LOGIC in choosing between 2 options if you dont know which is which... there is just pure luck!
Title: Final Decision
Post by: BlauK on November 27, 2004, 01:22:06 PM: Mando, how come your program shows situations where the real bomb is in the center and the accidentally dropped one in the wing? That is not what you described in the questions :)

... but even with this setup... initially the lever is in whichever position (not the on that drops accidentally though)... you see you lost a dud... you still have the selected bomb and one unselected bomb left when you start to figure out the propability... which is 50:50 anyhow!!!
Title: Final Decision
Post by: vorticon on November 27, 2004, 01:22:25 PM: assume the 2 duds were loaded first then drop the one that was loaded last...
Title: Final Decision
Post by: MANDO on November 27, 2004, 01:36:57 PM: Quote
Originally posted by BlauK
Mando, how come your program shows situations where the real bomb is in the center and the accidentally dropped one in the wing? That is not what you described in the questions :)

Blauk, the program shows the pilons graphically, as well as leds indicating bomb presence and initial and final lever positions. You may decide between three diferent decisions: Never change your lever, change always your lever or change your lever randomly.

For each run, the progam decides randomly which one is the real bomb (left, center or right). Then it decices randomly where is placed initially the lever. Now the program drops one of the training bombs (one that does not coincide with the initial lever position), and it is indicated by a black led in the corresponding pilon. And, finally, depending on your choice, the program dedices what to do with the lever.

Basically, if you decide to keep always the initial lever position, you will have a 33% of success. If you dedices randomly to change the lever position, you will have a 50% of success. And if you always change the lever, you will have a 66% of success.

I know you are still a non-believer ;) Try the program and dont forget to have some hard alchool nearby, you will be really surprised.

The source code of the algorythm is inside the zip file, you can check it.
Title: Final Decision
Post by: ALF on November 27, 2004, 01:51:54 PM: God knows Ive been wrong before, however in this case you must be operating from a flawed pretext, or we are not agreeing on the actual details of the problem.

Let me try and clarify (this should do nothing but obfuscate the issue...but I'll try)

If we take a plane with two bombs, one good one dud, what are the probabilities that the live bomb will be dropped 1st in a randome situation....50% thats easy to understand

If we then add a third bomb and make it a dud, we have 2 duds and one live bomb. The probability that the live bomb will be droped in the 1st TWO attempts is 66%.

HOWEVER

We cannot extrapolate from the second drop and ignore the results of the first to use the 66% figure, because the 66% INCLUDES the 33% chance of the live bomb being dropped on the FIRST drop.

or, to put it another way, there is a 33% chance the live bomb will drop on the 1st pull, a 33% it would be on the seconds pull and 33% on the last pull.......but we cannot take the initial 1st pull 33% and add it to pull # two.

Once you set the predisposition of bomb drop #1 being a DUD, it is no longer a variable (http://www.cogsci.princeton.edu/cgi-bin/webwn?stage=1&word=variable), you have made it a constant. As a constant it is no longer part of probability or the equasion.

or (just to make this a nice long post)

Lets say in your initial assesment that we have three bombs, one is LIVE, one is a DUD, and one is your maint cheif commiting suicide.

If the 1st drop is your beloved maint chief, what would there be a 66% the next drop would be live instead of the dud? Does that explain how the 1st drop is seperate from the last 2 and how telling me that you dropped the main cheif 1st makes him a NON FACTOR? Or better yet....why isnt there also a 66% chance of dropping the dud as the second drop?

I think you are confusing the non scientific 'likelyhood' with the mathmatical 'probability'
Title: Final Decision
Post by: BlauK on November 27, 2004, 01:58:59 PM: Mando,

the problem with your logic is that you are actually just calculating the probabilities of having a dud selected initially before take-off. That is naturally 66,7% because there are 2 duds and one live bomb.

If you then just take one of the duds away and keep the original assumption, you end up with those result you have now. Your equation shoul change when one known dud is taken away from the equation... it is just like loading the plane with only 2 bombs, of which one is real and one is dud... chances are 50:50!

You are now calculating chances of flipping heads with three coins but only giving 2 flips ;)

You are not considering that the dud that is dropped cannot be selected initially!!!! Therefore even the initial chances are 50:50 for kept dud and kept live bomb!
Title: Final Decision
Post by: Dawggus on November 27, 2004, 02:16:18 PM: Yep, Monty Hall made a living off this counterintuitive phenomenon for a long time on "Let's Make a Deal":

http://mathforum.org/dr.math/faq/faq.monty.hall.html

Cya Up!

Dawg
Title: Final Decision
Post by: ccvi on November 27, 2004, 02:52:18 PM: Quote from there: "The contestant picks a door and Monty opens one of the remaining doors, one he knows doesn't hide the car."

And that exactly is the difference between a random mechanical failure and a wizzard helping in playing lottery or monty opening doors.
Title: Final Decision
Post by: BlauK on November 27, 2004, 02:58:32 PM: Ok, I buy the Monty Hall deal.... but would it still work if door A was already selected for the player (instead of allowing him to select any) and the door B was shown as having a goat?

I suppose it works with thinking that the lever has originally a 2/3 chance of being selected for a dud, which would mean 2/3 of a chance of getting the real bomb by switching in the end.....

It looks like I have to admit my defeat :) S!... I suppose it all relies on the fact that your original selection cannot be taken away. If it could be, then selecting between 2 remaining ones would be 50:50.
Title: Final Decision
Post by: BlauK on November 27, 2004, 03:18:52 PM: Mando, can you change your program so that the left bomb is always originally selected and a dud is always dropped from middle? Would that make a difference?
Title: Final Decision
Post by: ALF on November 27, 2004, 03:27:17 PM: Where we are running into a diference is the CONSCIOUS SELECTION of a dud was never implied in the intial question, and that a malfunction does not a selection make. If we interpret a pourposful shoice that the initial bomb was a dud, and that it was made only after a conscious knowlege of the initial lever position, then we have a diferent mathmatical equation.

I refer to the key point :

Quote

He was lucky, it was one of the two training bombs.

Isn't it amazing how just a small little thing can make such a change. A randome act in this case is much diferent than a decision to reveal. It is part of my original point that we must 'eliminate' the possibility of the 1st drop being a live bomb to make it work, thats what Monte did, and thats what a randome event, as described does not do.

Let me know if I need to get my wifes Phychology degree involved...Im out of fields once we pass Mathmatics and Pre Law :eek:
Title: Final Decision
Post by: ccvi on November 27, 2004, 03:32:32 PM: Fixed in the description of the problem:
- A random bomb of the three is real, other two are duds
- Selected bomb is left one
- Bomb dropped by failure is the center one

let's go through all three cases of positions of the real bomb:

monty-doors / lottery-wizzard version
- left bomb real: not changing drops the correct one
- center bomb real: wizzard drops the right bombs, thus switching to the center drops the correct bomb
- right bomb real: wizzard drops the center one, thus switching drops the correct one
-> 66.7% for switching.

mechanical failure version
- left bomb real: not switching drops the correct one
- center bomb real: mechanical failure drops the center bomb. This does not happen in the described problem.
- right bomb real: switching drops the correct bomb
-> 50% for switching.

These results are also true if you start not only with a random position of the real bomb, but also additionally of a random position of the leveler and a mechanical failure on a random pylon - and then remove all cases that do not match the described scenario (= all cases where the mechanical failure drops the real bomb or the selected bomb).

The mechanical failure does not know the position of the real bomb, monty and the wizzard do.
Title: Final Decision
Post by: ccvi on November 27, 2004, 03:46:32 PM: Quote
Originally posted by ALF
I refer to the key point :

MANDO: "He was lucky, it was one of the two training bombs. "

Isn't it amazing how just a small little thing can make such a change. A randome act in this case is much diferent than a decision to reveal. It is part of my original point that we must 'eliminate' the possibility of the 1st drop being a live bomb to make it work, thats what Monte did, and thats what a randome event, as described does not do.

253 for (la_index = 0; la_index != 3; la_index++)
254 {
255 if ((la_index != la_realbomb) && (la_index != la_lever))
256 {
257 la_lost = la_index;
258 }
259 }

This doesn't seem to match "he was lucky".
Title: Final Decision
Post by: mechanic on November 27, 2004, 05:00:07 PM: very strange indeed. not sure i understnd the purpose of this action. is it in any reference to bombing tactics or merely a 16 year olds math problem?
Title: Final Decision
Post by: MANDO on November 27, 2004, 07:36:24 PM: Quote
Originally posted by BlauK
Mando, can you change your program so that the left bomb is always originally selected and a dud is always dropped from middle? Would that make a difference?

Yes, I can, but the difference would be 0%.

Think on the steps:

1 - randomly a bomb is selected as real bomb.
2 - randomly a placement is selected for the lever.
3 - a fake bomb is dropped (BUT BEING SURE THAT THIS TRAINING BOMB IS NOT SELECTED BY THE INITIAL POSITION OF THE LEVER).
4 - Final decision is taken.
Title: Final Decision
Post by: ccvi on November 27, 2004, 07:46:26 PM: Quote
Originally posted by MANDO
3 - a fake bomb is dropped (BUT BEING SURE THAT THIS TRAINING BOMB IS NOT SELECTED BY THE INITIAL POSITION OF THE LEVER).

This is not what happens by pure luck, this is wizzards work. A random mechanical failure could also have dropped the real bomb.

You need to drop a random bomb and ignore the cases where the real one drops.
Title: Final Decision
Post by: MANDO on November 27, 2004, 07:46:54 PM: Quote
Originally posted by ccvi
The mechanical failure does not know the position of the real bomb, monty and the wizzard do.

Certainly, but the case is that the lost bomb is a training one. If you lost the real bomb from the beginning, this would not increase your chances changing the lever or not. You are against a "secure fact" (not sure about the english term), you will miss for sure.
Title: Final Decision
Post by: MANDO on November 27, 2004, 07:51:27 PM: Quote
Originally posted by ccvi
You need to drop a random bomb and ignore the cases where the real one drops.

ccvi, read the initial description. When the pilot noticed the black led, he inverted the pane so see the bomb falling. At this point, the pilot know whether the bomb is red (training) or green (real). Of course, if the pilot sees the green bomb faling, he will abort the mission and return to base.
Title: Final Decision
Post by: ccvi on November 27, 2004, 08:04:00 PM: Of course he will, but it could still have happened. Those cases you're not ignoring, but you're adding them to those where he needs to switch.

a) 1/3: real bomb selected, fake dropped -> keep switch where it is
b) 1/3: fake bomb selected, real dropped -> rtb
c) 1/3: fake bomb selected, fake dropped -> switch to other bomb

Your failure-wizzard is manipulating every case of b) to be a case of c), of course you end up with a) vs. b)+c) then.

Dropping 1 random bomb, keeping the real one by pure luck, can and will result in loss of the real one in the case b), which just isn't the case in the scenario. So you just need to compare a) and c).
Title: Final Decision
Post by: MANDO on November 27, 2004, 08:09:11 PM: ccvi,

case B is rtb, period. You should not consider it to compute the chances of destroy the target once you are diving ready to press the trigger.

The question was not "what are the chances of hitting the target after takeoff", in any case, "what are the chances of hitting the target when diving over it with a bomb selected". Those cases with 0 chances (green bomb is lost and noticed by the pilot) should not be considered.
Title: Final Decision
Post by: ccvi on November 27, 2004, 08:17:38 PM: Quote
Originally posted by MANDO
Those cases with 0 chances (green bomb is lost and noticed by the pilot) should not be considered.

Correct. But you do consider them by manipulating them to match one of the cases that need to be considered. By always dropping the fake, even when in realtiy it would have been a case of early rtb.
Title: Final Decision
Post by: Kweassa on November 27, 2004, 08:55:46 PM: It's gonna be whatever is bad for the pilot.

Common wisdom of tendencies suggest that when something goes wrong, things always tend to happen to rush to the worst consequences of them all.

So, usually, a realistic approach is what Dredidock would suggest - the pilot will abort the bomb run altogether and dump whatever is left during the way, with no regards to statistics at all.

It doesn't make any difference whatever he chooses, as it is highly likely that whatever he does is not gonna work, unless he gives the act up altogether.

There is no 'logic' in this matter at all :D - only empirical 'tendencies'. Times like these granny's advices work better than mathematics. :)
Title: Final Decision
Post by: hawker238 on November 27, 2004, 09:51:56 PM: MANDO,

How come in the program, the good bomb can be on the middle pylon? I thought it was absolute that the middle bomb was a dud and that it would always be the one being dropped? I don't think I understand the problem.

Also, I ran the sim for 1500 drops, with "Never change the lever", and I got 33% success rate (1501 attacks, 497 hits). How does this figure?
Title: Final Decision
Post by: bozon on November 27, 2004, 11:48:50 PM: the difference between the 50% of selecting between the two remaining bombs and the 66% case is whether you "re-shuffle" the deck (take off the 2 bombs and randomly reattach them) after the center one has fallen.

normally we are used to the case in which we shuffle the deck or re-roll the dice before each gamble and therefor all previous information in erased. In this case, the "dice" was rolled only once, before the game started.

Bozon
Title: Final Decision
Post by: MANDO on November 28, 2004, 03:47:45 AM: Quote
Originally posted by hawker238
MANDO,

How come in the program, the good bomb can be on the middle pylon? I thought it was absolute that the middle bomb was a dud and that it would always be the one being dropped? I don't think I understand the problem.

Also, I ran the sim for 1500 drops, with "Never change the lever", and I got 33% success rate (1501 attacks, 497 hits). How does this figure?

Hawker, my first post was a single "case" of the problem. The program selects any pilon (centerline included) for the real bomb. The initial placement of the real bomb and the initial placement of the bomb selector lever can be any one, the problem will not change.

The KEY information is that the pilot will always see a training bomb falling and that bomb is placed in a different pilon than the initially selected by the lever.

If you run no more than 150 or 200 attacks with "Always change" you will have a 66% instead of a 33%, If you run no more than 150 or 200 attacks with "Change the lever randomly" you will have a 50% instead of a 33%, this is the beauty of the problem.
Title: Final Decision
Post by: RTSigma on November 28, 2004, 04:59:37 AM: I'm waiting to see if this will turn into a "you don't bury survivor's" question/joke
Title: Final Decision
Post by: MANDO on November 28, 2004, 06:15:48 AM: Here is a new version of the simulator, now you can also play manually to test your "luck".

If you select "Play manualy" option, you will be able to start manual attacks clicking on "New manual attack" button. The initial bombs will be initially hidden, but you will see the led indicators and the initial lever position and then you need to click one of the buttons to select manually the final position of the lever.

Final Decision Simulator 2 (http://www.terra.es/personal2/matias.s/TheBomber2.zip)
Title: Final Decision
Post by: hawker238 on November 28, 2004, 09:15:57 AM: This is a very interesting phenomena. Thanks, MANDO.
Title: Final Decision
Post by: Zwerg on November 28, 2004, 09:33:22 AM: Make a program of this: :)

Initial switch left = real bomb. Your middle training bomb goes off.
Stay left:win
Change to right:loose

Initial switch left = real bomb. Your right training bomb goes off.
Stay left:win
Change to middle:loose

Initial switch middle = training bomb. Your right training bomb goes off.
Stay middle:loose
Change to left:win

Initial switch right = training bomb. Your middle training bomb goes off.
Stay right:loose
Change to left:win
Title: Final Decision
Post by: LoneStarBuckeye on November 28, 2004, 10:45:15 AM: This is an interesting problem, and I think that Alf, ccvi, and Blauk have it right. The critical difference between the Monte Hall scenario and the one Mandoble proposed is that in the Monte Hall scenario, it is a given when one is partitioning the sample space that Monte will always select a door hiding a goat. That, of course, is not true in Mandoble's scenario--the bomb that was dropped might have been (with 1/3 probability) the live bomb.

The reason that this is important is that in the Monte Hall case, you know that you've got a 1/3 chance of having selected the correct door and a 2/3 chance that you're wrong. Once Monte shows you one of the two wrong choices, which he will always do, there's still just a 1/3 chance that your initial choice is correct. By switching, then, you'll increase your chances of being correct to 2/3.

We can represent the difference mathematically using a priori sample spaces, in which we assign probabilities to all of the permutations in the two cases. Here are the two a priori sample spaces, which show the initially selected choice followed by the revealed choice. The three options are labeled C (correct), I1 (incorrect choice #1), and I2 (incorrect choice #2).

Monte Hall:

C I1 1/6 chance
C I2 1/6 chance
I1 I2 1/3 chance
I2 I1 1/3 chance

Mandoble:

C I1 1/6 chance
C I2 1/6 chance
I1 I2 1/6 chance
I1 C 1/6 chance
I2 I1 1/6 chance
I2 C 1/6 chance

As you can see, the critical difference between the two cases is that in the Monte Hall case, when the initial selection is incorrect, the revealed choice will always be the other incorrect choice. In the Mandoble case, on the other hand, when the initial selection is incorrect, the revealed choice (i.e., the dropped bomb) may be either correct (i.e., live bomb) or incorrect.

If we define event A to be that the initial choice is correct and event B to be that the revealed choice is incorrect, our a priori sample spaces yield the following probabilies:

Monte Hall:

P(AB) = 1/3 (i.e., probability of both A and B happening, or the sum of the probabilities of the first two lines of Monte's chart)
P(B) = 1 (i.e., Monte ALWAYS reveals a door with the incorrect choice)

Mandoble:

P(AB) = 1/3 (same as Monte)
P(B) = 4 * 1/6 = 2/3 (i.e., sum of the rows 1, 2, 3, and 5 of Mandoble's chart)

From basic statistics, we know that the probability of A given B, which is often written P(A|B), can be calculated as follows:

P(A|B) = P(AB) / P(B)

Plugging in our numbers, we see that in Monte's case,

P(A|B) = (1/3) / 1 = 1/3 ==> 1/3 chance that our initial guess was correct, given that Monte revealed an incorrect choice. Thus, there is a 2/3 chance that if we switch, we'll win the car.

In Mandoble's case,

P(A|B) = (1/3) / (2/3) = 1/2 ==> there's a 1/2 chance that the initial selection was the live bomb, given that the revealed bomb was a dud.

Best regards,

JNOV
Title: Final Decision
Post by: MANDO on November 28, 2004, 11:13:55 AM: Quote
Originally posted by LoneStarBuckeye
P(A|B) = (1/3) / (2/3) = 1/2 ==> there's a 1/2 chance that the initial selection was the live bomb, given that the revealed bomb was a dud.

Wrong. Initial selection is fully random between three pilons. Real bomb is fully ramdom between the three pilons. Initial selection has 1/3 always. The dud bomb is revealed later. The funny point is that you cannot change the history of the events without changing the final probability of success.
Title: Final Decision
Post by: LoneStarBuckeye on November 28, 2004, 11:22:25 AM: No, I'm correct, and, what's more, my analysis agrees with you. The initial selection is fully random and has a 1/3 probability of being correct. That's event A, and if you notice, its probability is 1/3 in my analysis. The catch is that the probability of event B (i.e., of the revealed choice being incorrect) is only 2/3, not 1 as in the Monte Hall case.

The charts in my post set out the a priori probabilities. That is, they are the probabilities of events before the "experiment" begins. And the key is that your hypothetical, unlike the Monte Hall scenario, doesn't have an a priori rule that the revealed choice will always be incorrect. Perhaps you think you specified a Monte-Hall-type scenario, but as I read it, you did not.

- JNOV
Title: Final Decision
Post by: MANDO on November 28, 2004, 11:52:08 AM: Quote
Originally posted by LoneStarBuckeye
your hypothetical, unlike the Monte Hall scenario, doesn't have an a priori rule that the revealed choice will always be incorrect.

But if the revealed bomb is green (the real), the pilot will not attack, he will RTB.
Title: Final Decision
Post by: ccvi on November 28, 2004, 11:54:53 AM: Your car has been making strange noises for a few times. Everytime that happend you went to the garage and they replaced a spark plug.

Now your car is making strange noises again. You're learning by immitation and replace a spark plug yourself.

But your car still makes strange noises.

You cannot transfer the solution of one problem to another problem without verifying that the premises are the same.

That's also the problem with todays engineering. Lot's of fomulae are beeing taught, but the premises under which they are valid or even the approximations used to derive them are kept in the dark.
Title: Final Decision
Post by: ccvi on November 28, 2004, 11:56:14 AM: Quote
Originally posted by MANDO
But if the revealed bomb is green (the real), the pilot will not attack, he will RTB.

But this wont prevent a mechanical failure from dropping a real bomb. In the scenario the pilot was lucky that a dud dropped.
Title: Final Decision
Post by: LoneStarBuckeye on November 28, 2004, 11:59:56 AM: Quote
Originally posted by MANDO
But if the revealed bomb is green (the real), the pilot will not attack, he will RTB.
Whether the pilot attacks or RTBs is immaterial; when the dropped bomb is revealed, the experiment is over.
Title: Final Decision
Post by: BlauK on November 28, 2004, 12:10:58 PM: For some reason I tend to meve to Mando's side...

Why would it matter how the door with a goat is selected, or that it just happened to be the dud that dropped by itself. In both cases we are NOW in a situation with 2 choices, of which one is already selected. The question is: "Should you switch the selection now?". It is the same question in both cases because the one possibility is now dropped and identified. The question is about probability of benefit of switching and not switching... is it not? It is not about selecting one or other... it is about switching the selection. That is the difference.... I suppose :)

I tried to search the internet for pages which explain how the Monty Hall case has been proven false... but did not find any.
Title: Final Decision
Post by: MANDO on November 28, 2004, 12:21:07 PM: Quote
Originally posted by BlauK
not? It is not about selecting one or other... it is about switching the selection. That is the difference.... I suppose :)

Exactly :aok
Title: Final Decision
Post by: ccvi on November 28, 2004, 12:28:52 PM: Quote
Originally posted by BlauK
Why would it matter how the door with a goat is selected, or that it just happened to be the dud that dropped by itself. In both cases we are NOW in a situation with 2 choices, of which one is already selected.

Those scenarios are not equal. Think about the large number of bombs version.

The probability of having the leveler initially set to the right bomb is low.
The probability that the real bomb does not drop when all but two bombs drop is also low. In this scenario this is the case because of luck. The situation the pilot is in is rare.
As a result, the probabilities that one or the other bomb is real are equal.

If some greater beeing ensures that the real bomb is never dropped, you're comparing the low probability of the leveler beeing initially set to the right bomb, and an extremely high probability of the other bomb beeing real.

But there is no such greater beeing that influences mechanical failures.
Title: Final Decision
Post by: Arlo on November 28, 2004, 12:37:32 PM: You fly back with both bombs, land, round up Moe, Larry and Curly and show them the difference between the color green and the color red. You then request their transfer to the infantry where color blindness isn't as critical (well it probably is but you request it anyway).
Title: Final Decision
Post by: LoneStarBuckeye on November 28, 2004, 12:40:42 PM: The difficulty in comparing the two situations comes from not looking at the a priori probabilities. ccvi is quite right; the two situations are not equivalent. Indeed, to make Mandoble's hypothetical equivalent to the Monte Hall scenario, one would have to specify, a priori (e.g., before the pilot takes off), that a bomb will fall before the pilot signals its release AND that it will be one of the two incorrect (i.e., dud) bombs. In that case, switching to the other remaining bomb would increase the chances of dropping the correct bomb to 2/3. With that qualifcation, Mandoble's a priori sample space would change to look like this:

C I1 1/6 chance
C I2 1/6 chance
I1 I2 1/3 chance
I2 I1 1/3 chance

In that case, P(AB) = 1/6 + 1/6 = 1/3, and P(B) = 1. Thus, P(A|B) = P(AB) / P(B) = 1/3, just as in the Monte Hall scenario.

Without such a qualification, the two remaining choices have equal (i.e., 1/2) probability of being correct, as my foregoing analysis shows.

- JNOV
Title: Final Decision
Post by: MANDO on November 28, 2004, 02:52:27 PM: LoneStarBuckeye, my example was just that. Read the case at the beginning.

1 - The lost bomb is always training, being real one the pillot will abort the mission and case is over.
2 - THE INITIAL LEVER POSITION DOES NOT COINCIDE WITH THE LOST BOMB <- KEY RULE
3 - The pilot perceives the lost bomb (pylon and color).

The program works that way and demostrates 1/3 for "I wont change never", 1/2 for "may be I'll change" and 2/3 for "I always change".

While the pilot ignores rule 2, the real chances do not vary.
Title: Final Decision
Post by: vorticon on November 28, 2004, 03:07:51 PM: he had a right handed colour blind ground crew (ever meet a left handed colour blind dude?). therefore they loaded left to right. they loaded the training bombs first, and realising there mistake, but not having time to fix it, they put on a real one, intending to warn the pilot before takeoff.

so naturally, he should set the lever to the RIGHT BOMB.
Title: Final Decision
Post by: ccvi on November 28, 2004, 03:15:54 PM: Quote
Originally posted by MANDO
1 - The lost bomb is always training, being real one the pillot will abort the mission and case is over.

The lost bomb can also be the real one. Just not in your simulation. In case of mechanical failure on the pylon of the real bomb a wizzard helps and drops the dud. The pilot in the simulation isn't lucky not to lose the real bomb.

You are modifying the case of early rtb to match the case where switching is required instead of ignoring it.

edit: You're manipulating a "does not happen" to "cannot happen".
Title: Final Decision
Post by: Raider179 on November 28, 2004, 03:23:34 PM: 50%. With no knowledge of in what order the bombs were loaded this pilot has a 50% chance to drop the correct bomb. He can either select to switch or stay with the one he has. That is 2 choices and with no idea of which bomb is next that is a 50% chance of success/failure. You are confused because you keep interjecting the bomb that was jettissioned into the chances that you will get the right bomb. This bomb no longer exists as it relates to the problem therefore it is irrelevent.

1 : 2 thats 50% you either move it or you dont. You stand no better/worse chance because you have no idea where it is.
Title: Final Decision
Post by: BlauK on November 28, 2004, 03:24:23 PM: ccvi,

the case was not about a random mechanical failure... it was: "you lost one training bomb... what now?" .. what are the chances in this case?

.. or should we also consider the possibility that the "mechanical failure" would drop more than one bomb, or that it would concern the engine.. etc. Just disregard it. The dud WAS dropped already ;)
Title: Final Decision
Post by: BlauK on November 28, 2004, 03:26:53 PM: Raider,
read them all. That is what I also thought in the beginning. But the case is not about selecting between 2 bombs... it is about keeping the alredy selected one or switching the selection. That is what makes the unbeliavable difference :\
Title: Final Decision
Post by: ccvi on November 28, 2004, 03:54:11 PM: Quote
Originally posted by BlauK
the case was not about a random mechanical failure.../B]

MANDO in his initial post: "He was lucky, it was one of the two training bombs."

This clearly indicates a random mechanical failure, and the pilot was just lucky.

Otherwise it should have read something like: A bomb was lost. By more fragile design of the pylons for training bombs the pilot knew that only a training bomb could drop.

But it didn't. The pilot was lucky.
Title: Final Decision
Post by: MANDO on November 28, 2004, 04:56:45 PM: Quote
Originally posted by ccvi
MANDO in his initial post: "He was lucky, it was one of the two training bombs."

ccvi, you are missing completely the point. Being a real bomb does not change the % of destroying the target once you start the real attack, because you are not going to perform that attack.

The question was "WHAT TO DO ONCE YOU ARE GOING TO DIVE AND ATTACK?", not just what were the chances of the plane since the very beginning once airborne.
Title: Final Decision
Post by: ccvi on November 28, 2004, 06:18:12 PM: Please replace

253 for (la_index = 0; la_index != 3; la_index++)
254 {
255 if ((la_index != la_realbomb) && (la_index != la_lever))
256 {
257 la_lost = la_index;
258 }
259 }

with

la_lost = rand() % 3;
if ((la_lost == la_realbomb) || (la_lost == la_lever))
return (FALSE);

This IGNORES cases where the real one drops, not changes them to match the scenario described. IOW: The pilot is luck not to lose the real one, he doesn't keep the real one by design.

edit: replaced wrong 2 with 3.
Title: Final Decision
Post by: bozon on November 28, 2004, 06:32:57 PM: JNOV has a good point about the selection of the fallen bomb which I assumed given.

If it is truely random, and by repeating the experiment we allow for any one of the 3 to fall even if it is the live (or even the selected one!), than the probability relevant is the one for an entirly random process P(A|B) = 1/2. In this case the initial selection is meaningless since it does not affect the next possible options.

If only a dud can fall, then the probability of the groups is conserved (1/3 for one group and 2/3 for the other group) since the 1st selection restricts the following options. So it's a question whether you think the one that fell was due to divine intervention...

Bozon
Title: Final Decision
Post by: eskimo2 on November 28, 2004, 06:55:22 PM: At this point we all would have augered; too much of a brain teaser to comprehend in so little time. If the mission really is this dangerous, then this was the best choice:

Quote
Originally posted by BlauK
Just drop either one of them before the attack and check the color... then u know if it is worth to do the final attack or to just bug off home :)

eskimo
Title: Final Decision
Post by: Blue Mako on November 28, 2004, 08:09:52 PM: Man this is the dumbest thread ever but I'll bite.

There is only one choice that the pilot needs to make. Choose between one of two bombs. Each have equal probability of being correct.

The fact that a third bomb fell off the aircraft earlier is absolutely irrelevant to the final choice between two bombs. Your set of probability has been reduced to one chance in two.

50:50

Equal chance.

Any other ways to say it?

In this scenario he is just going to guess left or right. Moving the lever does not change the odds at all.
Title: Final Decision
Post by: Purzel on November 29, 2004, 02:05:15 AM: I think we cann agree that the described problem suggests being different from what it really is.

Arriving at the target, the pilot has to decide which bomb he drops. There are only two hanging from the plane, and he knows one is live, the other dud. What he does with the lever (how he flips the coin) is not important.

How he got there is also unimportant. If he had a plane with only 2 pylons an one was loaded with a training bomb he would end up at the same situation. With the same probabilities.

You can write a program doing that too, and you will see, the chance of selecting the right one (or wrong one) will be 50%.

The interesting part about what happened here is the the initial descriptioin seemed to be what I said last (the 2-bomb-problem). But it isnt. Thats why it is so surprising.
Title: Final Decision
Post by: MANDO on November 29, 2004, 02:22:23 AM: ccvi, by doing so you are going to "ruin" the "good luck" of the pilot for the described problem. That luck is mainly based on the selection of the lost bomb based on two other random selections. If we look for a pure random combinations where the lost bomb is red, the real bomb is green and the lever marks a present bomb, then there is no more "good luck" initially present.

The original case indicates that the pilot see a red dot falling. If you repeat that case 1000 times, he will see the red dot 1000 times because it is part of the description. He may ask himself, ok, what if the ground crew painted the real bomb red by mistake? But what he see is red, not green.
Title: Final Decision
Post by: dedalos on November 29, 2004, 09:29:40 AM: Quote
Originally posted by ALF
I musta missing something in college, if you flip a coin 10 times and it comes out heads all 10 times, what are the odds it will come out heads on try 11....50% because the next flip is a new subset, completely removed from the old set with no direct imperical relation.
[/B]

You missed the three door example in the probability class :D
Title: Final Decision
Post by: midnight Target on November 29, 2004, 03:25:19 PM: The left one is already selected... drop it now, then select and pickle off the right one. Duh!
Title: Final Decision
Post by: dedalos on November 29, 2004, 03:32:25 PM: Quote
Originally posted by midnight Target
The left one is already selected... drop it now, then select and pickle off the right one. Duh!
:aok
Title: Final Decision
Post by: ccvi on November 29, 2004, 05:31:46 PM: Quote
Originally posted by MANDO
ccvi, by doing so you are going to "ruin" the "good luck" of the pilot for the described problem. That luck is mainly based on the selection of the lost bomb based on two other random selections. If we look for a pure random combinations where the lost bomb is red, the real bomb is green and the lever marks a present bomb, then there is no more "good luck" initially present.

The original case indicates that the pilot see a red dot falling. If you repeat that case 1000 times, he will see the red dot 1000 times because it is part of the description. He may ask himself, ok, what if the ground crew painted the real bomb red by mistake? But what he see is red, not green.

The selection in you solution attempt is not "by luck" but "by design", whereas in the described scenario the selected bombs drops by "luck", not "design".
Title: Final Decision
Post by: MANDO on November 29, 2004, 05:53:46 PM: Quote
Originally posted by ccvi
The selection in you solution attempt is not "by luck" but "by design", whereas in the described scenario the selected bombs drops by "luck", not "design".

How do you know that? You know that the lost bomb is red, you dont know that the lost bomb may be green. You may consider that only training bombs can be lost by its engeenering or design. The case describes a single mechanical faillure that affected a training bomb.

If you want to look at the scenario considering that there is no "pre-designed good luck" (the lost bomb is red and always will be red), then you should consider also the following:

1 - The ground crew commit mistakes. In fact, in the described case mistook 2 of 3 (2 reds and 1 green instead of 3 greens). You may suspect that the ground crew doesnt know to differentiate between real and training bombs. So, they may repeat mistakes loading bombs (all training, for example, 1 training plus two real, all real, may be they even forget to load some pilons ...).

2 - The plane may have a mechanical faillure in any bomb and/or pilon, so, more than 1 mechanical faillure are also possible. So, due ground crew mistakes, even losing 2 bombs does not guarantee that the third is green.

3 - The ground crew may load all real bombs also.

4 - The plane may have also NO mechanical failures.
Title: Final Decision
Post by: ccvi on November 30, 2004, 02:32:24 PM: You can add all those four cases to your simulation. If you're doing it correctly, the result will not change.
All cases of "might have happend but did not happen" (possible, but do not match the story) need to be treated in the same way: Return() before evaluation.
This is true for those four cases as well as for an unlucky drop of the real bomb or a drop of the bomb that was selected.
Title: Final Decision
Post by: MANDO on November 30, 2004, 03:14:59 PM: ccvi, I can add those factors, but they are not related to the initial problem.

You saw a red bomb falling, and you conclude that green bombs may fall also. This is too much of an asumption before having a single case where the green bomb is lost. A real (green)bomb is a different object than a training (red) bomb. Is like saying, I saw a red bomb falling once, so, next time a wing may fall also or even the pilot. Losing a green bomb is not a "possible case" without a single real intance, else, losing anything (not only real bombs) should be possible also.

In the entire history of the events, 1 case only, only one red bomb was lost from a pilon. All the pilons are the same type, so you should assume that the problem was in the red bombs. The only factor that we may add without breaking the iniital case is considering that the second red bomb may also be lost. For these cases the pilot will be success all the times.
Title: Final Decision
Post by: BlauK on November 30, 2004, 04:09:49 PM: ccvi,
howabout considering it this way:

The possibility of having the real one originally selected is 1/3 and either of the training bombs (=tb) selected is 2/3.

Then one tb is lost from unselected position.....

If real bomb is selected, losing tb1 is 1/6 and losing tb2 is also 1/6 of the chances.

If tb1 is selected, the lost one was tb2, 1/3 possibility.

If tb2 was selected, the lost one was tb1, 1/3 possibility.

So we end up in a situation with only 2 bombs left and the chances are 1/3 that the real one is selected, 2/3 that a tb is selected..... should you switch? ;)

Where is the fault in the above logic with these given conditions?

It is all about do you switch or not!!! It is not about which of the 2 should you select, since one of them is already selected.

-------------

What is interesting is that one can actually still make 3 choices: decide to switch, decide not to switch or not to decide at all :) .... in 2/3 of these cases the selection is not switched. It looks like the odds are against going for better chances (=switching) :D
Title: Final Decision
Post by: ccvi on November 30, 2004, 05:09:21 PM: Quote
Originally posted by MANDO
In the entire history of the events, 1 case only, only one red bomb was lost from a pilon.

You cannot do statistics with a single event.

Quote
Originally posted by BlauK
Where is the fault in the above logic with these given conditions?

The fault is that the solution does not match the initial description of this scenario. "He was lucky, it was one of the two training bombs." If only duds could drop, he wouldn't have been lucky, because no matter how often he was in that situation always duds would have dropped.

Re-check the original post closely:
- "ground crew loaded 2 training bombs (red) and only 1 real bomb (dark green)" - no order given, therefore assumed to be random
- "This lever was at the lower position, so left pilon was selected."
- "the center pilon led switched off, by some mechanical problem that bomb was released"

These are fixed. With a random order of the bombs there's a probability of 1/3 that the real one was dropped and he could have rtb'ed.

But "He was lucky, it was one of the two training bombs."

There's a probability of 1/3 that the real one is on the left wing, and a probability of 1/3 that the real one is on the right wing. With the center bomb gone it's equally likely that the real one is on the left or right wing.

To get 2/3 for the right wing you either need a special the aircraft that whenever a real bomb is loaded to th
e center position moves it to the right wing, or a wizzard that whenever the real bomb was at the center and about to drop catches it and drops the dud on the right wing instead. Neither of these exists.
Title: Final Decision
Post by: MANDO on November 30, 2004, 05:55:50 PM: Quote
Originally posted by ccvi
There's a probability of 1/3 that the real one is on the left wing, and a probability of 1/3 that the real one is on the right wing. With the center bomb gone it's equally likely that the real one is on the left or right wing.

Wrong asumption. The real bomb was as safe as any other part of the plane that didnt fall, the problem was on a training one. You only can assume that losing both training bombs is also possible.

You decided to put the "blame" on the pilon instead on the red bomb. There are three pilons, all are the same, there are two training bombs, a combination of a pilon/training bomb failed. Suspecting on the training bombs: 1 failed, 1 worked. Suspecting on the pilons: 1 failed, 2 worked. It is more probable that the problem is on the training bomb, not the pilon. In any case, you cannot suspect on the real bomb itself, it remained and was only one: 0 failed, 1 worked.

If you want the a definitive excuse for the expresion "he was lucky" then consider the following:

When the pilot saw the "red" led (indicator of bomb presence) switching off, he may suspect also on the led itself. When the pilot saw the red bomb falling, he confirmed that the led was OK. That gave him the chance of getting 66% of success. Being the faillure on the led instead of on the bomb, the success chance would be 33%.
Title: Final Decision
Post by: BlauK on December 01, 2004, 02:53:40 AM: ccvi,

it is not statistics... it is probabilities ;)
Title: Final Decision
Post by: ccvi on December 01, 2004, 04:54:15 PM: The 2/3 for switching would be true if no red bomb and no pylon a red bomb is attached to and no screws that hold a pylon that red bomb is attached to ... had ever failed and wont in future.

Apart from that beeing pretty far fetched because such failure free systems do not exist in reality, which point in your initial story describes that kind of warped reality? Please provide appropriate qutotes from the initial post.
Title: Final Decision
Post by: BlauK on December 02, 2004, 03:29:08 AM: Sorry ccvi, I missed what was your point.

Are you trying to say that The Monty Hall example would not work if this was the very first show and he had never before opened a door with a goat for anyone? What does it matter if the door could get stuck, or the handle would fall off when opening, etc? The case states that the door IS already open... or that one bomb is already lost.

It is about probability of switching the selection in this given situation.... "someone had three doors of which one was selected and saw that one of the unselected doors had a goat" ..... or "you had three bombs of which one was selected and you saw that une unselected pylon had a training bomb".

Is it not irrelevant for the actual question of switching, how you get the information of outruling one unselected option??? A man opened the door (that door is missing form your further options) or you saw one bomb drop by itself (it also now misses).

The point is:
You are in this situation now... should you switch the selection? What are the probabilities for getting a live bomb by switching and by not switching? Forget the screws and pylons.
Title: Final Decision
Post by: ccvi on December 02, 2004, 04:03:44 PM: Quote
Originally posted by BlauK
Is it not irrelevant for the actual question of switching, how you get the information of outruling one unselected option???

It is relevant. Only if both the probability of losing a real bomb and the probability of losing a bomb that is selected are 0 you'll get the real one by switching in 2/3 of the attempts. Both these conditions are not described in the initial story.

The way to get into such a situation is important. Imagine a slight modifed version of mandos warped reality. Keep the probability of 0 for failures of bombs that are selected. Because of cheap mass-production during war time real bombs tend to fail a lot more than old training bombs from pre-war times. Now imagine a very very lucky pilot that lost a training bomb in exactly the situation mando described. Do you think he has to switch, just because he's in that situation? The only thing that saves his bomb is that the selected one never drops, even if it is real.
Title: Final Decision
Post by: dedalos on December 02, 2004, 04:24:12 PM: http://astro.uchicago.edu/rranch/vkashyap/Misc/mh.html
Title: Final Decision
Post by: MANDO on December 02, 2004, 04:44:52 PM: ccvi, you cannot make asumptions, you should work only with the data presented in the case, and there is a single case. Based on the described case, you know that red bombs fail, green bombs dont. Probability of losing a red bomb that is selected by the lever is irrelevant. If you lost the selected red bomb, you will need to switch the lever also. Switching the lever will increase always your chances from 33% to 66% or from 0% to 50%.
Title: Final Decision
Post by: BlauK on December 03, 2004, 05:56:31 AM: ccvi,
it is exactly like you described. This is that exact case. You are there.. you HAVE LOST that one red bomb. What will you do NOW?

Mando,
If the selected bomb is lost, then it is a whole different story and situation. It is not about switching anymore since nothing is selcted. It is then about selection one of the 2 remaining bombs and the chances are then 50:50 ;)
Title: Final Decision
Post by: MANDO on December 03, 2004, 09:22:15 AM: Quote
Originally posted by BlauK
If the selected bomb is lost, then

then you also NEED to change the lever. The question was, what should do the pilot?

CHANGE THE LEVER ALWAYS.
Title: Final Decision
Post by: dedalos on December 03, 2004, 09:49:51 AM: It is 50/50. The formula used to calculate the probability that gives the 2/3 is the other bomb/door, is based on knowing that the bomb is red before droping it.

http://astro.uchicago.edu/rranch/vkashyap/Misc/mh.html

__________________
Title: Final Decision
Post by: nazgulAX on December 03, 2004, 09:56:40 AM: select the left pilon
Title: Final Decision
Post by: BlauK on December 03, 2004, 01:12:48 PM: Mando,
If he does not switch in such case when the selected bomb is lost, there is no point to attack at all! I agree with your earlier case, but now you are mixing things. The Monty Hall scenarion does not include a case where the selected door (with a goat) is opened and then a new selection is made. Then the question would not be "to switch or not?" anymore, but "where to switch?"
Title: Final Decision
Post by: BlauK on December 03, 2004, 01:19:46 PM: dedalos,
where in the formulas is that "knowing before opening" indicated?

IMO, the condition is that "an unselected door with a goat IS/WAS opened". It is exactly the same as "an unselected red bomb was lost"

That is the situation you are in now. What are the success probabilities of keeping and switching?
Title: Final Decision
Post by: TEShaw on December 03, 2004, 02:54:10 PM: I'm enjoying this debate and the articulate way you guys are putting forth your arguments.

However, if Monte Hall is God, then MANDO's argument is irrefutable.

regards,

Airman T. E. Shaw
Title: Final Decision
Post by: lasersailor184 on December 03, 2004, 03:58:58 PM: Umm guys, you are making this way way too difficult. It doesn't take a member of Mensa to figure it out (but don't worry, I'm here to solve it for you).

On the way to the target the guy finds a clear open space and drops a random bomb. Doesn't matter which one.

One of two things can happen:

1.) The bomb will impact with the ground, but not explode. Hence the pilot just dropped the practice bomb. If this happens, he continues with the mission.

2.) The bomb will impact with the ground and detonate. Hence the pilot just dropped the real bomb. If this happens, there is no use risking his life and the plane to drop a practice bomb that won't do anything anyway. If the pilot dropped the real bomb, then he just flies home and chalks the day up as a loss.

HOTDAMN! I'm good.
Title: Final Decision
Post by: dedalos on December 03, 2004, 04:12:15 PM: Quote
Originally posted by BlauK
dedalos,
where in the formulas is that "knowing before opening" indicated?

IMO, the condition is that "an unselected door with a goat IS/WAS opened". It is exactly the same as "an unselected red bomb was lost"

That is the situation you are in now. What are the success probabilities of keeping and switching?

I bolded the part that to me means knowing before opening (He is not going to open the one witht he price). This is how I understand it atlist.

The a priori probability that the prize is behind door X, P(X) = 1/3

The probability that Monty Hall opens door B if the prize were behind A,
P(Monty opens B|A) = 1/2

The probability that Monty Hall opens door B if the prize were behind B,
P(Monty opens B|B) = 0

The probability that Monty Hall opens door B if the prize were behind C,
P(Monty opens B|C) = 1

The probability that Month Hall opens door B is then
p(Monty opens B) = p(A)*p(M.o. B|A) + p(B)*p(M.o. B|B) + p(C)*p(M.o. B|C)
       = 1/6 + 0 + 1/3 = 1/2

Then, by Bayes' Theorem,

P(A|Monty opens B) = p(A)*p(Monty opens B|A)/p(Monty opens B)
       = (1/6)/(1/2)
       = 1/3
and
P(C|Monty opens B) = p(C)*p(Monty opens B|C)/p(Monty opens B)
       = (1/3)/(1/2)
       = 2/3

So, since th ebomb is droped accidently this does not apply and the prob now is 50/50
Title: Final Decision
Post by: lasersailor184 on December 04, 2004, 10:52:31 AM: I'm sorry, did I ruin it for everyone? :D
Title: Final Decision
Post by: BlauK on December 04, 2004, 10:57:44 AM: dedalos,
ok... show me the formula all the way from the beginning with an accidentally lost training bomb. If you lose a real bomb, you can disregard those cases, because they do not fulfill the stated case.

How will you end up with 50:50 for switching case?

What is your bases for the statement that this does not apply for an accidentally lost bomb (which WAS already lost and therefore happens in 100% of the possible relevant cases!!!)
Title: Final Decision
Post by: yuto on December 04, 2004, 02:26:36 PM: So you play paper/rock/scissors with a robot which you are told is completely random.

First try:
you - rock
robot - paper

Second try:
you - rock
robot - paper

Third try:
you - rock
robot - paper

. . . .(continues for 10 minutes)

On the 101st try,
- a mathematician/statistician would stick with his rock
- a guy with some sense would go with scissors
Title: Final Decision
Post by: DieAz on December 04, 2004, 09:15:53 PM: Quote
Originally posted by yuto
So you play paper/rock/scissors with a robot which you are told is completely random.

First try:
you - rock
robot - paper

Second try:
you - rock
robot - paper

Third try:
you - rock
robot - paper

. . . .(continues for 10 minutes)

On the 101st try,
- a mathematician/statistician would stick with his rock
- a guy with some sense would go with scissors

and get pounded by a rock :rofl
Title: Final Decision
Post by: MANDO on December 05, 2004, 05:13:50 AM: Quote
Originally posted by BlauK
Mando,
If he does not switch in such case when the selected bomb is lost, there is no point to attack at all!

BlauK, in this case he needs to switch (is "forced" to switch by the circumstances). We can consider that, for this case in particular, "to switch" is not a decission, the decission for this case is "where to switch" with only 1/2 of success.

So, when the pilot may switch, he "should" switch. When the pilot is forced to switch, he will switch.
Title: Final Decision
Post by: lasersailor184 on December 05, 2004, 11:51:59 AM: Mando, can you confirm that I'm right?
Title: Final Decision
Post by: vorticon on December 05, 2004, 11:58:17 AM: hmm, couldnt the pilot do a roll to see wich side is "heavier" (im assuming training bombs are lighter...not sure if thats correct) then switch to that side?
Title: Final Decision
Post by: MANDO on December 05, 2004, 12:05:51 PM: lasersailor, no, I cant confirm that, you are wrong ;)

He should not drop a "random" bomb. But you have an interesting idea.

Just after losing the "training" bomb, and with the determined decission to change the lever before the attack, he can just drop the bomb CURRENTLY SELECTED (not a random one), he has 33% chances that this is a real bomb. If it is real, he can RTB without risking into the enemy ack.

You have, indirectly, optimized the solution, WTG :aok
Title: Final Decision
Post by: ccvi on December 05, 2004, 01:35:42 PM: Quote
Originally posted by BlauK
If you lose a real bomb, you can disregard those cases, because they do not fulfill the stated case.

That's correct. But this is not what mandos solution (a solution to a different problem) is doing.
Title: Final Decision
Post by: MANDO on December 05, 2004, 01:40:05 PM: ccvi, you are describing a different problem, not me. Knowing that single case, losing the real bomb has the very same probabilities than the pilot losing his left leg.
Title: Final Decision
Post by: ccvi on December 05, 2004, 01:50:41 PM: And the same probability as losing an unselected dud.
Title: Final Decision
Post by: MANDO on December 05, 2004, 02:05:23 PM: Quote
Originally posted by ccvi
And the same probability as losing an unselected dud.

You dont know whether or not the remaining dud was unselected.
Title: Final Decision
Post by: ccvi on December 05, 2004, 02:25:17 PM: At the moment of drop it's also unknown whether the lost bomb was real or a training one.

Using quotes from the initial post, please explain why you think that
- the probability of losing a selected bomb is 0.
- the probability of losing a real bomb is 0.
Title: Final Decision
Post by: lasersailor184 on December 05, 2004, 03:24:00 PM: Vorticon, I thought the exact same thing initially. But training bombs are made out of all metal to weigh the exact same.

However mando, I'm still right. It's the only solution there is with decreasing risk for nothing.
Title: Final Decision
Post by: vorticon on December 05, 2004, 03:34:34 PM: Quote
Originally posted by lasersailor184
Vorticon, I thought the exact same thing initially. But training bombs are made out of all metal to weigh the exact same.

However mando, I'm still right. It's the only solution there is with decreasing risk for nothing.

ah...in which case, you response is the only acceptable course of action...unless he has a wingman.
Title: Final Decision
Post by: MANDO on December 05, 2004, 04:48:31 PM: Quote
Originally posted by lasersailor184
However mando, I'm still right. It's the only solution there is with decreasing risk for nothing.

But not droping a bomb randomly, just drop the selected bomb.
Title: Final Decision
Post by: MANDO on December 05, 2004, 04:54:02 PM: Quote
Originally posted by ccvi
At the moment of drop it's also unknown whether the lost bomb was real or a training one.

Using quotes from the initial post, please explain why you think that
- the probability of losing a selected bomb is 0.
- the probability of losing a real bomb is 0.

It is unknown by the pilot, but the lost bomb was RED, and the pilot was aware of that few seconds later.

If selected bomb is lost, the pilot is forced to move the lever. So, the solution does not change "move the lever".

The probability of losing real bombs is 0, because the only lost one is RED. Remember, for TWO red, you lost ONE, for one green, you lost CERO, for one pilot head, you lost 0, etc. Only 1 RED bomb failed, you cannot from a single case assume the green bomb will fail also (or the left leg or the pilot).
Title: Final Decision
Post by: lasersailor184 on December 05, 2004, 05:28:52 PM: It is unknown which bomb is on which rack. Therefor, it doesn't matter if you switch the lever or not. You drop a bomb. If it explodes you RTB. If it doesn't you go on to target.
Title: Final Decision
Post by: BlauK on December 05, 2004, 05:31:29 PM: Originally posted by ccvi
Using quotes from the initial post, please explain why you think that
- the probability of losing a selected bomb is 0.
- the probability of losing a real bomb is 0.

---

Quoting the original text:
"This lever was at the lower position, so left pilon was selected."
" As he was aproaching the target, the center pilon led switched off, by some mechanical problem that bomb was released."

Left one was selected, center one was lost. Therefore this question does not deal with losing a selected bomb!

Quote:
"The pilot inverted the plane quickly and looking at his high six saw a bright red object descending. He was lucky, it was one of the two training bombs."

Like you see again, it was the training bomb hat was lost.

Only now comes the question about switching and its probabilities in tis particular case.....

Mando's program takes a bit wider view though. It allows also other initial selections and training bomb losses also from other positions than center. The described case falls under the same category though, since it is just describing one of the cases.

ccvi,
to me it seems that you want to change the question as: "What is the probability that a pilot with these three bombs in unknown positions would lose any one of the bombs, recognize it, and then think of switching the selection or not?"

That is not the original question though.
Title: Final Decision
Post by: MANDO on December 05, 2004, 06:15:19 PM: Quote
Originally posted by lasersailor184
It is unknown which bomb is on which rack. Therefor, it doesn't matter if you switch the lever or not. You drop a bomb. If it explodes you RTB. If it doesn't you go on to target.

lasersailor184, to make it simpler, the chances to have the lever where the real bomb is from the very beginning are 1/3. So, there are 2/3 that the real bomb is on the other group of racks. One of the other racks is empty now, you lost one training bomb, so, you still have 2/3 chances of having the real bomb on the other rack. If you want to drop a bomb and see what happens, drop the one currently selected, but ONLY if you have the determined decission of changing the lever at the moment of the attack.
Title: Final Decision
Post by: lasersailor184 on December 05, 2004, 09:45:39 PM: You are overlooking the simplicity of this problem, as is everyone who is trying to quote difficult statistics.

The real bomb is either on the left pilon, or the right pilon. You do not know, nor can any statician justifiy that there is more than a 50% chance whether or not you have the real bomb selected.

Next, factor in that the pilot will be flying through a heavily defended area. Does he want to go in there with a chance of him being killed to drop a practice bomb and do nothing? **** no!

So my solution still stands except for another which I'll touch upon in a minute. He picks one pylon **DOESN'T MATTER WHICH BECAUSE THE CHANCES ARE THE SAME.** He drops it over open ground and watches the effect. If he has a real bomb left, then it's worth the chance to fly into the defended area. If he doesn't he flies home because the result isn't worth the risk.
Title: Final Decision
Post by: BlauK on December 06, 2004, 04:39:13 AM: laser,

read the links to Monty Hall scenario or search the web for it. Then tell us whu the mathematicians and statisticians cannot prove it wrong!

You yourself are not considering the fact that the whole case is not the same as making a selection between 2 options.. it is about switching th eoriginal selection or not switching. And that makes the difference, believe it or not!

The question here is simply about to switch or not to switch. Solutions like.. fly over friendly base to identify the bombs.. are very nice work-arounds, but they do not tackle the actual question of probabilities.
Title: Final Decision
Post by: MANDO on December 06, 2004, 06:41:37 AM: lasersailor,
now imagine your plane has 1000 bombs, you lost 998 training bombs, what is the probability that the lever was selecting the real bomb from the very beginning? 1/1000? But what is the probability for the other rack? 999/1000?
Title: Final Decision
Post by: dedalos on December 06, 2004, 09:41:38 AM: Quote
Originally posted by BlauK
dedalos,
ok... show me the formula all the way from the beginning with an accidentally lost training bomb. If you lose a real bomb, you can disregard those cases, because they do not fulfill the stated case.

How will you end up with 50:50 for switching case?

What is your bases for the statement that this does not apply for an accidentally lost bomb (which WAS already lost and therefore happens in 100% of the possible relevant cases!!!)

I don't think I can because of this line:
The probability that Monty Hall opens door B if the prize were behind C,
P(Monty opens B|C) = 1

If the location of the real bomb/prize is not known, then P(Monty opens B|C) = .5 and therefore the result of the the formula will not be 2/3.

The base of this prob problem is the fact the who ever opens the door or drops the bomb, knows where the real one/prize is. If not, then it is just random acts and at the end you have 2 choices to make a random decison.

Lets try it anyway, without the knowlege of where the prize is:
The probability that Monty Hall opens door B if the prize were behind A,
P(Monty opens B|A) = 1/3

The probability that Monty Hall opens door B if the prize were behind B,
P(Monty opens B|B) = 1/3

The probability that Monty Hall opens door B if the prize were behind C,
P(Monty opens B|C) = 1/3

The probability that Month Hall opens door B is then
p(Monty opens B) = p(A)*p(M.o. B|A) + p(B)*p(M.o. B|B) + p(C)*p(M.o. B|C)
= 1/9 + 1/9 + 1/9 = 3/9 = 1/3

Then, by Bayes' Theorem,

P(A|Monty opens B) = p(A)*p(Monty opens B|A)/p(Monty opens B)
= (1/3)/(1/2)
= 2/3
and
P(C|Monty opens B) = p(C)*p(Monty opens B|C)/p(Monty opens B)
= (1/3)/(1/2)
= 2/3

All choices are equal and therefore the selction will be random.
Title: Final Decision
Post by: BlauK on December 06, 2004, 11:31:46 AM: Dedalos,
why are you loking at the situation BEFORE opening the door or losing the bomb. Try to look at the situation when the final decision has to be made. Th edoor is already opened P(goat was shown)=1 just like P(training bomb was lost)=1

Actually one does not even have to know the LOCATION of the real bomb or the price to be able to open an unselected door with a goat to lose an unselected training bomb. One only need to know that a goat was shown or that a training bomb was lost. That knowledge is what counts in the final P(switch selection)=2/3 and P(dont switch)=1/3.

The accident of losing the real bomb is not included in the question or the case of probability. If you are asked about how many bananas you can eat, why would you start calculating also some bananas?

Consider the following case: You had 2 coins, lost one accidentally or someone took one from you intentionally. Now you flip all your coins once. Is the P(heads) at this particular moment somehow different in these 2 cases only because in one case the the prior happening imcluded intention and the other was accident???
Title: Final Decision
Post by: dedalos on December 06, 2004, 11:45:54 AM: Blauk

The reason I look at the situation before opening the door (look at the formula above) is that for it to work and give you the desired 2/3 result, the person openning the door or dropping the bomb, must have knowlege of where the prize or the real bomb is.

If not, and the door/bomb is opened/droped randomly then your probability is the following:
3 bombs --> prob = 1/3,1/3,1/3 for any of the bombs to be droped.

Once one bomb is droped, the prob is 1/2 for the one on the left and 1/2 for the one on the right. Real or fake does not make a difference at this point. Because you did not select the droped bomb but it was droped randomly, you have a 50/50 situation.

Appologies if I cannot explain it any better (Einglish not my first language), but the key to the problem is knowing that the dropped bomb was selected to not be the real one for a reason. If it was an accident then it is a random event resulting in the 50/50 situation.
Title: Final Decision
Post by: BlauK on December 06, 2004, 12:13:56 PM: Same here dedalos, English is not my 1st language. Still I wonder why you insist that there is any probability included in the case of losing the first training bomb. The description of the problem already states that IT WAS LOST!!!! Therefore P(training bomb was lost)=1... it is a pre-condition for the whole case. Just as it is for the Monty Hall scenario. When the competitor is asked to switch or not, a goat is already shown.

This question is about switching after training bomb is lost, not about probability of losing a training bomb. If you want to include the latter, you have to find out somewhere the probability for accidentally losing a bomb.

ps. I just find this phenomenon very captivating and always enjoy good logical arguments :) That is why I have sticked with this thread. Nothing personal is meant towards anyone here :)
Title: Final Decision
Post by: MANDO on December 06, 2004, 01:02:53 PM: Quote
Originally posted by dedalos
3 bombs --> prob = 1/3,1/3,1/3 for any of the bombs to be droped.

Sorry, but wrong: one of them has 0% to be lost. You know that training bombs fail, real bombs dont.
Title: Final Decision
Post by: lasersailor184 on December 06, 2004, 01:22:57 PM: If you lost 998 training bombs, you suddenly have 1 real bomb and 1 training bomb.

Thus you have a 50% chance of selecting the right bomb. It doesn't matter how many bombs you have before you have to make the decision. It only matters how many bombs you have to make the decision. You have 2, one real one training bomb.

So it doesn't matter whether you flip the lever or not, it's a 50% chance you have the correct bomb selected.

So my solution still stands.
Title: Final Decision
Post by: dedalos on December 06, 2004, 01:24:20 PM: Quote
Originally posted by BlauK
Just as it is for the Monty Hall scenario. When the competitor is asked to switch or not, a goat is already shown.

I think this is where we disagree. It is not about the goat/bomb being shown. The difference is that in the Monty example, his door was not selected randomly. He is not going to open the door with the prize. If door A had the prize then the prob of opening door A=0. This is not the case with the bombs. The probability of a good bomb droping by accident is 1/3. Once a bomb is droped then there is a 50/50 that one of the remaining is good (since you know a training bomb was lost)

The only way the Monty example would apply to the airplane would be if tyhe story changed to:

The grownd crew knows where the good bomb is and they remotly made a traning bomb drop. Then, the pilot should switch since there is a 2/3

Quote
Still I wonder why you insist that there is any probability included in the case of losing the first training bomb. The description of the problem already states that IT WAS LOST!!!! Therefore P(training bomb was lost)=1... it is a pre-condition for the whole case. Just as it is for the Monty Hall scenario. When the competitor is asked to switch or not, a goat is already shown.

Not the same. What makes the difference is not that the door/bomb was opened/droped. The difference is on the why the door/bomb was selected. In any case, try it using prob formulas and you will see. The only way to get a 2/3 is by knowing that the who ever made the bomb drop, knew that it was not going to be a real one.
Title: Final Decision
Post by: dedalos on December 06, 2004, 01:26:03 PM: Quote
Originally posted by MANDO
Sorry, but wrong: one of them has 0% to be lost. You know that training bombs fail, real bombs dont.

Right, did not think of that :D
Title: Final Decision
Post by: dedalos on December 06, 2004, 01:27:48 PM: Quote
Originally posted by lasersailor184
If you lost 998 training bombs, you suddenly have 1 real bomb and 1 training bomb.

Thus you have a 50% chance of selecting the right bomb. It doesn't matter how many bombs you have before you have to make the decision. It only matters how many bombs you have to make the decision. You have 2, one real one training bomb.

So it doesn't matter whether you flip the lever or not, it's a 50% chance you have the correct bomb selected.

So my solution still stands.

If those 998 were lost completly randomly/accidentaly then you are right.
Title: Final Decision
Post by: BlauK on December 06, 2004, 01:33:59 PM: lasersailor,
sorry, but you are wrong there :) This thing has been already explained it tnis thread and in some of th elinks provided. I wont repeat it. Go find it out yourself :)

I repeat once more.. it is not about selecting one or another. It is about keeping the original selection or switching to a new one. If you cannot see the difference between those, it is useless to continue :(

dedalos,
The competitor in Monty Hall scenario does not really know how Monty picked the door, does he? :) And why would it matter since it already happened. Not statistics, but probabilities in this situation for switching :)
Title: Final Decision
Post by: BlauK on December 06, 2004, 01:38:16 PM: for those 1000 bombs there is an initial chance or 1/1000 that the real bomb is selected. In 999 of 1000 cases it is not.

Since the condition is that the real bomb cannot be dropped by accident, in 999/1000 cases the only remaining (not dropped) unselected bomb is the real one and in 1/1000 cases it is the selected one...... naturally everyone switch the lever, right? ;)

----

If nothing was selected initially, all those dropped bombs dont mean a thing!!! One just selects between 2 bombs and the chance of getting the real bomb is 50:50.

---

Having something selected and switching or not is a completely different case from just choosing between two previously unselected options. Funny, interesting and almost unbeliavable... but true :eek:
Title: Final Decision
Post by: MANDO on December 06, 2004, 01:56:03 PM: lasersailor, on the first or second page of this thread there is a link to my "game" where you can try your "luck". Download it and test it by yourself, it is very small.
Title: Final Decision
Post by: lasersailor184 on December 06, 2004, 02:37:44 PM: I have seen and read everything on this post. (Just to make sure someone didn't post my idea first).

This isn't some complex statistics problem. It's something a second grader could solve easily.

There is a 50-50 chance that the person would select the real bomb.

Say you have a bag with 3 MM's in it. 2 red, one blue. Just by dumb luck, a red falls out of a hole in the bottom of the bag. You reach in and grab 1 MM. What are the chances it's going to be blue? What are the chances it's going to be red?

The problem is no different then the one I just described.
Title: Final Decision
Post by: dedalos on December 06, 2004, 02:39:54 PM: Quote
Originally posted by BlauK

dedalos,
The competitor in Monty Hall scenario does not really know how Monty picked the door, does he? :) And why would it matter since it already happened. Not statistics, but probabilities in this situation for switching :)

So, we are in agreement that (Monty) knowing makes the difference?

The competitor knows unless it is the first time that game is run. He does not really need to know. What matters is how Monty is selecting the door. The contestant needs to know only so that he can realize what is going on and switch the door. His knowledge has nothing to do with what he should do. He may not do the right thing if he does now, but his knowlege does not effect the results of the problem. Clear as mud?

This has nothing to do with statistics. It is one of the first probability problems you will find in probability classes. I've seen it 4 times, 2 dif Universities, 3 diff teachers.

Any teachers here? Heeeelp, we are stuck
Title: Final Decision
Post by: MANDO on December 06, 2004, 02:58:28 PM: Quote
Originally posted by lasersailor184
Say you have a bag with 3 MM's in it. 2 red, one blue. Just by dumb luck, a red falls out of a hole in the bottom of the bag. You reach in and grab 1 MM. What are the chances it's going to be blue? What are the chances it's going to be red?

Believe it or not, the original problem has nothing in common with your example.

1 - In your example any ball can fall, not only red ones (dumb luck).
2 - In your example you are not selecting a ball from the begining. You make a single selection at the end.
Title: Final Decision
Post by: ccvi on December 06, 2004, 03:46:42 PM: Quote
Originally posted by BlauK
Originally posted by ccvi
Using quotes from the initial post, please explain why you think that
- the probability of losing a selected bomb is 0.
- the probability of losing a real bomb is 0.

Quoting the original text:
"This lever was at the lower position, so left pilon was selected."
" As he was aproaching the target, the center pilon led switched off, by some mechanical problem that bomb was released."
"The pilot inverted the plane quickly and looking at his high six saw a bright red object descending. He was lucky, it was one of the two training bombs."

What you're trying to say is, that the single special case the pilot is in says anything about probabilities? Try again.
Title: Final Decision
Post by: lasersailor184 on December 06, 2004, 05:20:18 PM: It's the exact same. The dude reaches in and feels 2 MM's. He has to make a decision. He wants a blue one. Which one does he pick?

It doesn't matter which he picks because it's going to be a half and half chance whether or not he gets a blue MM.
Title: Final Decision
Post by: BlauK on December 06, 2004, 05:20:32 PM: ccvi,
You do not have to (and prolly cannot) do statistics to calculate the probability of getting tails with a coin for the very first time it was ever flipped in the known universe.
This single case asks about a probablility in this single case! Is that so hard to understand? :) It does not tell the answer, it asks a question. Then someone just has to figure out what is relevant for calculating the result.

dedalos,
no we are not in an agreement. The competitor knowing that an unselected goat was revealed makes the difference! How it ended up to become revealed is irrelevant. IT IS ALREADY REVEALED :) Dont calculate probabilities for it becoming revealed. It is out of the equation already.

laser...,
I give up :) Either you cannot read, I cannot write or you just stubbornly refuse to understand the difference between apples and oranges. They should teach that on 2nd grade ;)
Title: Final Decision
Post by: BlauK on December 06, 2004, 05:26:34 PM: laser,
do you REALLY think that there is no difference between A and B?

A: you buy the very first lottery ticket (1/1000), then all but one ticket are thrown away. Someone looks at that last ticket and then your ticket and says that one of them is the winner. Now you can switch tickets if you want to.

B: There are only 2 tickets (1 winner, 1 loser). Now you can select which one you want to buy.

...no difference at all?

----------

just to make your MM example equal to Mando's case it should go like...

"You pick one MM out of the bag and put it in your pocket without looking at it. Then a red one drops out from the bag. Should you now pick the last MM from the bag or the one in your pocket to get a blue one?"

The answer is that the one in your pocket is 1/3 likely to be blue, the one in the bag is 2/3 likely to be blue.

----

To get the 50:50 probability you just drop a red MM from the bag and then pick on of the 2 remaining ones from the bag. No pre-selections before losing the red one.
Title: Final Decision
Post by: ccvi on December 06, 2004, 06:21:46 PM: Quote
Originally posted by BlauK
"You pick one MM out of the bag and put it in your pocket without looking at it. Then a red one drops out from the bag. Should you now pick the last MM from the bag or the one in your pocket to get a blue one?"

Do me a favor and try it.

Take 3 equal things, one secretly marked as winning-thing.
Select one without looking at it.
Shake the other two till one drops.
If it's the winning-thing, start from scratch.
If the winning-thing is in your pocket note down one point for pocket.
If the winning-thing is in your hand note down one point for hand.
Title: Final Decision
Post by: MANDO on December 06, 2004, 06:28:01 PM: Quote
Originally posted by ccvi
Take 3 equal things ...

Wrong example.
Title: Final Decision
Post by: MANDO on December 06, 2004, 06:29:31 PM: Quote
Originally posted by lasersailor184
It's the exact same.

Sorry, but I give up too.
Title: Final Decision
Post by: ccvi on December 06, 2004, 06:48:40 PM: Quote
Originally posted by MANDO
Wrong example.

It's up to you to show where in your inital post something is hidden that indicates a 0-failure probability of selected bombs and a 0-failure probability of real bombs. Unless you can show that, it's your bomb example that is wrong for the monte-case, or your solution that is wrong for the bombs.
Title: Final Decision
Post by: lasersailor184 on December 06, 2004, 11:25:01 PM: Don't give me that bull****. Just because I don't buy into is not a reason for you to stop.

The situation is as follows. You have the left bomb selected. Either the left bomb is real, or the right bomb is real. What do you do with the switch?

Quote
A: you buy the very first lottery ticket (1/1000), then all but one ticket are thrown away. Someone looks at that last ticket and then your ticket and says that one of them is the winner. Now you can switch tickets if you want to.

B: There are only 2 tickets (1 winner, 1 loser). Now you can select which one you want to buy.

Let me get this straight. In situation A. You have already bought a ticket but can switch it with the other one. You do not know which one won. Do you switch it or not?

In Situation B you are told that there are two tickets for sale. One is a winner, one is a loser. Which one do you pick?

Each draws a 50-50 chance of winning. You cannot change this!
Title: Final Decision
Post by: BlauK on December 07, 2004, 01:17:30 AM: Ok laser,
Lets make a 2nd grade example :)

Three objects all have same original probability. Lets call these things men. Lets call the probability money. These 3 men live in same house, therefore this house has 100% (or 1) money.
Then one man moves away to another house and takes his money with him. Now this other house has 1/3 of the money while the original house has 2/3.
What happens next is that one of the two men who stayed dies! He cannot take his money with him when he dies.
So we end up with a situation with the original house (1 living man, 1 dead) still having 2/3 of the money and the new house (1 man who moved away) still having 1/3 of the money.
Even though the number of men reduced, the amount of money (=probability) stays the same.
Since you selected originally the man/object that was moved away, which one would you rather have now?

Did this make it any clearer? :)
Title: Final Decision
Post by: BlauK on December 07, 2004, 01:23:38 AM: Quote
Originally posted by ccvi
Do me a favor and try it.

Take 3 equal things, one secretly marked as winning-thing.
Select one without looking at it.
Shake the other two till one drops.
If it's the winning-thing, start from scratch.
If the winning-thing is in your pocket note down one point for pocket.
If the winning-thing is in your hand note down one point for hand.

Mando,
I dont think this example is wrong. Only not exact.

-Select one and put it in your pocket without looking at it.
-Shaking is ok.
-Dont start from scratch, but disregard this case (or is that what you meant? i.a. Count only those cases where one non-winner falls from the bag!!!)
-pocket ok
-in your hand... well, only after you pick it from the bag

ccvi,
I counter the challenge. I know that it works. If I do it, why would you still believe me? PLEASE try it yourslef and show me that it does not work :) .... or try Mando's program which does it for you :)
Title: Final Decision
Post by: BlauK on December 07, 2004, 01:41:28 AM: Lets put out one final explanation:

The pre-condition is that the selected bomb was not lost and the real bomb was not lost (to verify this, simply look at the original story). We are dealing only with this once in a lifetime case that "just happened"

So now we have 1 bomb selected, 1 bomb unselected and 1 (identified as training bomb) as lost.

So what has happened? These are ALL the possible scenarios that could have happened so far for this particular case:

RB = real bomb
T1 = training bomb1
T2 = training bomb2

A: RB is selected, T1 was lost, T2 is unselected

B: RB is selected, T2 was lost, T1 is unselected

C: T1 is selected, T2 was lost, RB is unselected

D: T2 is selected, T1 was lost, RB is unselected

Do you agree so far?????? SInce we saw that T-bomb WAS lost, there is no other way it caould have happened.

So in 2 of these listed cases RB is selected and in 2 cases T-bomb is selected (T1 or T2).

Now we figure out the probabilities according to the chance of any of these 3 bombs being selected originally:

RB = 1/3
T1 = 1/3
T2 = 1/3

Since RB is selected in cases A and B, the chance for each of those cases is 1/6

Chance for case C is 1/3 and chance for case D is also 1/3.

Now we need to decide about switching and can add it all up:

Chance for having the RB selected right now is (A+B) = 1/3

Chance for having T1 or T2 selected right now is (C+D) = 2/3

Do you switch?

Where is the fault in this explanation and calculation? Please show it and prove it.... and not just by saying .."because I think so.. or my mother said so" ;)

If you dont agree with the pre-conditions, then you dont believe your own eyes havng seen the training bomb fall off!
Title: Final Decision
Post by: ccvi on December 07, 2004, 02:56:44 AM: Quote
Originally posted by BlauK
-Dont start from scratch, but disregard this case (or is that what you meant? i.a. Count only those cases where one non-winner falls from the bag!!!)

ccvi,
I counter the challenge. I know that it works. If I do it, why would you still believe me? PLEASE try it yourslef and show me that it does not work :) .... or try Mando's program which does it for you :)

Correct, count only those cases where a non-winner falls from the bag.
You cannot try this using mandos program, because it includes a little wizzard that changes something small but important. I posted a modification of a few lines further above. Ask mando to compile the program with those.

This is what mandos program does:
-Select one and put it in your pocket without looking at it.
-Shaking is ok.
-In case the winner drops from the bag, drop the other one from the bag and put the winner back in. <-- mandos magic work
-pocket ok
-in your hand... well, only after you pick it from the bag

blauk, you're trying to argue that it doesn't matter how the pilot got into this situation, or how it does not matter whether it just happens by random chance or by wizzards work that all losing lottery tickets are dropped. The difference between disregarding non-matching cases and modifying them to make them match affects the probabilities of whether you have to switch or not.

mando's trying to argue that there's a zero-failure probability of bombs that are real or selected. The problem is that there's no conclusive evidence why this should be the case - neither in the initial story nor something that could be derived from the hardware properties of the things involved in it.
Title: Final Decision
Post by: BlauK on December 07, 2004, 03:47:56 AM: Originally posted by ccvi

-In case the winner drops from the bag, drop the other one from the bag and put the winner back in. <-- mandos magic work

ccvi,
how does this magic work differ from dirregerding the cases where the winner is dropped?

e.g.
results by accident:

winner (disregard)
loser
winner (disregard)
loser
winner (disregard)
loser

result: 3/3 cases the loser comes out

The difference between disregarding non-matching cases and modifying them to make them match affects the probabilities of whether you have to switch or not.

How does it matter when the result is the same... those cases where a winner is dropped do not apply to this specific problem. Why do you have to disregard them one by one, why dont you just take them all out since they are not counted?

mando's trying to argue that there's a zero-failure probability of bombs that are real or selected. The problem is that there's no conclusive evidence why this should be the case - neither in the initial story nor something that could be derived from the hardware properties of the things involved in it.

For this particular case the selected bomb was not lost and a real bomb was not lost. There IS zero probability for those. You are trying to argue that for calculating the probability of you flipping tails with a coin has something to do with a probability for you being born even though you have already been born!

The question is not about probability of losing a bomb! That is not asked.
Title: Final Decision
Post by: MANDO on December 07, 2004, 03:51:21 AM: Quote
Originally posted by BlauK
Mando,
I dont think this example is wrong.

BlauK, ccvi example has nothing to do with the original case, and ccvi knows it pretty well.

In ccvi example, marked thing may drop (the real bomb). In the original example we have 2 equal things and lots of different things (real bomb, pilots head, lever, bomb presence leds, plane engine, pilots left leg, etc..). ccvi assumes that losing a training bomb justifies that real bomb may be lost also (or plane engine, or whatever else). Not only that, ccvi assumes that any thing has same chances to fail.

ccvi example will change all the probabilities, even dropping the cases where the lost thing is the marked one. ccvi is looking for pure raw random events where the marked thing remains. Believe it or not, that changes everything, the case is a very different one that the original one.
Title: Final Decision
Post by: MANDO on December 07, 2004, 03:53:53 AM: Quote
Originally posted by ccvi
mando's trying to argue that there's a zero-failure probability of bombs that are real or selected.

ccvi, your only evidence is that a training bomb failed.
Title: Final Decision
Post by: MANDO on December 07, 2004, 05:19:03 AM: Quote
Originally posted by lasersailor184
Don't give me that bull****. Just because I don't buy into is not a reason for you to stop.

Ok, imagine you do not have a plane with racks, but three doors. Behind two of them there are goats (instead training bombs), and the other door hides a golden doblon instead of a real bomb.

To select a door, you need to be in front of it.
Currently you are in front of LEFT door.

Suddenly, you ear a goat bleating (is that the english word?) at your door. Should you switch to other door? Yes (0% to 50%).

But, what if the goat bleats at the CENTRAL door? Should you switch to the RIGHT door? Yes also (33% to 66%). (<- This is the equivalent to the original case).

What if both goats bleat? now you know where the golden doblon is, you know where to switch (if needed). 100%

What if both goats keep in silent? Doesnt matter what you choice, 33%.

TIP: Golden doblon does not bleat.
Title: Final Decision
Post by: lasersailor184 on December 07, 2004, 11:07:11 AM: It does not change the fact that it's a 50-50 chance when you go to drop the bomb. You ignore all other situations because they don't matter!

Ignore the whole problem you wrote out mando, because it really doesn't matter. It's called eroneous information. It has no bearing on the final solution.

Here is the **ACTUAL** problem.

You have 2 bombs, each weighing the same. One is on the left rack, one is on the right rack. One is a training bomb, one is the real bomb. What is the chance that you'll drop a real bomb?

Here are the possibilities of what could happen.

1.) Real bomb is on the left rack. Training bomb is on the right rack.
2.) Training bomb is on the left rack. Real bomb is on the right rack.

There is no way in hell that you could reason that the bomb you currently have selected is the real bomb with 66% chance.

The same thing applies to the goats and golden coins. Once 1 goat makes a sound he is stricken from the equation because he is not wanted. Hence you now have 2 doors, one which has some golden coins behind it. Ignore the 3rd door because it is now erroneous information. There is no way you can determine which door has a bigger chance of having the golden coins behind it because you have no further information.
Title: Final Decision
Post by: dedalos on December 07, 2004, 11:19:25 AM: Quote
Originally posted by BlauK
Lets put out one final explanation:

The pre-condition is that the selected bomb was not lost and the real bomb was not lost (to verify this, simply look at the original story). We are dealing only with this once in a lifetime case that "just happened"

So now we have 1 bomb selected, 1 bomb unselected and 1 (identified as training bomb) as lost.

So what has happened? These are ALL the possible scenarios that could have happened so far for this particular case:

RB = real bomb
T1 = training bomb1
T2 = training bomb2

A: RB is selected, T1 was lost, T2 is unselected

B: RB is selected, T2 was lost, T1 is unselected

C: T1 is selected, T2 was lost, RB is unselected

D: T2 is selected, T1 was lost, RB is unselected

Do you agree so far?????? SInce we saw that T-bomb WAS lost, there is no other way it caould have happened.

So in 2 of these listed cases RB is selected and in 2 cases T-bomb is selected (T1 or T2).

Now we figure out the probabilities according to the chance of any of these 3 bombs being selected originally:

RB = 1/3
T1 = 1/3
T2 = 1/3

Since RB is selected in cases A and B, the chance for each of those cases is 1/6

Chance for case C is 1/3 and chance for case D is also 1/3.

Now we need to decide about switching and can add it all up:

Chance for having the RB selected right now is (A+B) = 1/3

Chance for having T1 or T2 selected right now is (C+D) = 2/3

Do you switch?

Where is the fault in this explanation and calculation? Please show it and prove it.... and not just by saying .."because I think so.. or my mother said so" ;)

If you dont agree with the pre-conditions, then you dont believe your own eyes havng seen the training bomb fall off!

Thats one of the things that is wrong. Too, sick to play today, lol

A: RB is selected, T1 was lost, T2 is unselected
B: RB is selected, T2 was lost, T1 is unselected
C: T1 is selected, T2 was lost, RB is unselected
D: T2 is selected, T1 was lost, RB is unselected
E: T1 is selected, RB was lost, T2 is unselected
F: T2 is selected, RB was lost, T1 is unselected
Title: Final Decision
Post by: MANDO on December 07, 2004, 11:50:36 AM: lasersailor, obviously, it is not a so elementary problem as you describe.

Download the program and play manualy. You may also select a "politic of action" and play in auto. Dont forget to reset the counters when you change the way the program will take decissions.

Bomber simulator (http://www.terra.es/personal2/matias.s/TheBomber2.zip)

Dedalos, RB cannot be lost.
Title: Final Decision
Post by: lasersailor184 on December 07, 2004, 12:15:50 PM: You obviously don't get it. It is as simple as a coin flip.

I have played with your program. I've gotten results all over the board, nothing that can draw any kind of conclusions assuming that you could draw conclusions from them! Each time you do it is a seperate and special case. You can't compare them together because the next time has an equal opportunity of happening every time.

Ok Mando, solve this problem.

I have a quarter in one of my hands. I have nothin in the other. I hold both out. What are the chances you will select the correct one?
Title: Final Decision
Post by: MANDO on December 07, 2004, 12:39:27 PM: Quote
Originally posted by lasersailor184
I have played with your program. I've gotten results all over the board, nothing that can draw any kind of conclusions

No way, if you change always the lever placement you will reach a 66%. Just do test it. Play a minimum 100 times, you can do it in no more than 5 mins.

Is that everyone is getting 66% that way except you?
Title: Final Decision
Post by: lasersailor184 on December 07, 2004, 02:25:28 PM: I ran it a bunch of times. I got anything from 30-80 with it landing twice on 55. It doesn't mean anything.

It's like running a coin flipping program and flipping out when it stops at 90% heads 10% tails.

The base chance of something happens has no bearing on what will happen. What will happen is a shot in the dark every single time. You cannot change this. Nor can you expect something else to happen because "SoandSo" happened 200 times before.

Each time you flip a coin, you have a 50-50 chance of it landing on heads or tails. Now, that doesn't mean out of 100 tries, it'll land on heads 50 times. It means that every single flip individually analyzed will have an equal chance of hitting heads or tails and having no bearing on what happened before or what will happen in the future.

The same applies to this airplane. It does not matter what your program says, it does not matter what some statistician with a ruler up his bellybutton says, there is an equal chance of you dropping the real bomb as there is for dropping a fake bomb whether you switch pilons or not. There's only 2 bombs, not 3.
Title: Final Decision
Post by: ccvi on December 07, 2004, 03:13:13 PM: Quote
Originally posted by MANDO
ccvi, your only evidence is that a training bomb failed.

So what you're saying is that because there's one recorded event of a failure of an unselected training bomb, real bombs and selected bombs can never fail?

Only once in my life I stared at the ground for 60 seconds. Only once because it's so boring. The one time I did I found a coin on the ground. Therefore, according to your logic, if I ever tried again, I would find another one?
Title: Final Decision
Post by: Raider179 on December 07, 2004, 04:34:41 PM: I can not believe this post is still going. The first bomb after falling is now irrelevant unless the pilot knows something we dont.

Laser you are 100% correct.

everyone agree flipping a coin is 50:50?

ok now what are the chances after flipping 2 heads in a row the next coin will be heads?

50:50

It does not matter about preceding events. Only the actual odds that the coin is placed under matter. You could flip 1000 heads in a row and your odds of another head is 50:50. It does not increase or decrease because of the precdeding flip.

The problem seems to be whether to switch the selector and drop the bomb or not to switch it and drop it as is. Are there any other choices I am missing? Because those are the only two I see. What changes the odds to 66%? because a bomb fell? lmao you guys have this question fundamentally screwed up.

Does the pilot know anything about the bombs on his wings? Does he know they always load certain bombs onto the hardpoints in a certain order? no no no

he has no friggin clue therefore his odds are 50 50 no matter what.

The monte hall thing is a sham. He just used human nature and thats what produced those kinds of results. Take your kid and a piece of candy. Ask do you want this or whats in a box (have something they cant tell what it is). see what they choose.

seems to me its just confusion. The fact that you had 3 bombs to begin with has no bearing on the final part with 2 bombs.
Title: Final Decision
Post by: MANDO on December 07, 2004, 06:27:10 PM: lasersailor, you cannot get anything near 80% running the program with 100 cases or more. If you run the program and select the option "never change the lever" in auto, you will get 33% after some seconds. If you select "Change always" you will get 66% after some seconds. If you select "change randomly" you will get 50%. If you got different results, then you ran a different program than mine.
Title: Final Decision
Post by: lasersailor184 on December 07, 2004, 09:42:34 PM: No, I definately used yours. And since you claimed that the pilot should change the lever because of a 66% chance, I ran it on auto for roughly 100 runs a bunch of times with always change lever.
Title: Final Decision
Post by: BlauK on December 08, 2004, 02:04:16 AM: Quote
Originally posted by dedalos
Thats one of the things that is wrong. Too, sick to play today, lol

A: RB is selected, T1 was lost, T2 is unselected
B: RB is selected, T2 was lost, T1 is unselected
C: T1 is selected, T2 was lost, RB is unselected
D: T2 is selected, T1 was lost, RB is unselected
E: T1 is selected, RB was lost, T2 is unselected
F: T2 is selected, RB was lost, T1 is unselected

Heeelloooo dedalos :) :)

The pilot ALREADY SAW that a RED TRAINING BOMB WAS LOST! Cases E and F are not possible in this case. If you start adding things to possible cases, you can go on adding everything that is possible in the world.

If that is the only thing we find worth arguing, it is better to let it be :) I cannot show any better evidence than what was said about the situation. If you believe that TB was not lost, we are talking about different cases.
Title: Final Decision
Post by: BlauK on December 08, 2004, 02:14:38 AM: Quote
Originally posted by ccvi
So what you're saying is that because there's one recorded event of a failure of an unselected training bomb, real bombs and selected bombs can never fail?

Only once in my life I stared at the ground for 60 seconds. Only once because it's so boring. The one time I did I found a coin on the ground. Therefore, according to your logic, if I ever tried again, I would find another one?

ccvi, laser, raider,

The question is NOT about probability of losing aany kind of bomb. The bomb WAS lost already. If you cannot grasp that, it is your own loss :)

Consider the following case:
You have accidentally taken a wrong train and have arrived to Atlanta. Then you flip a coin. What is the probability of getting heaads?

ccvi: "Well first we have to calculate the probability of taking a wrong train... are you saying that I can never take a right train???? How can the probability be calculated based on this single case???" :D :D

I apologize the ridicule, but THIS is exactly what you are doing now :)

PLEASE.... do not try to calculate the probability of this kind of situation ever occuring. You cannot do it. You dont know failure probabilities or loading mistake probabilities.

The question is that if you ever happened to be in this kind of situation, what would switching or not switching do for you!!!! Put yourself in this highly improbable situation.
Title: Final Decision
Post by: BlauK on December 08, 2004, 02:25:39 AM: laser,

The your probability of winning in a lottery with an already bought ticket does not get any better if non-winners are revealed to you before finding the winner

Consider this:
-1000 lots, only 1 winner.
-you buy one
-Someone else buys 999 and asks his secretary to open all "no win" lots. Should she find the winner, it should be closed again. If all are "no win" lots. then one lot should still be kept closed.

Now you and the other buyer both have only one lot each. Neither of you know which one is the winner.

What do you think... what are your own chances now? 50:50 ???? or 1:999 ????

With your reasoning so far you should claim 50:50. Do you?
Title: Final Decision
Post by: MANDO on December 08, 2004, 03:43:18 AM: Quote
Originally posted by lasersailor184
No, I definately used yours. And since you claimed that the pilot should change the lever because of a 66% chance, I ran it on auto for roughly 100 runs a bunch of times with always change lever.

Ok, your result with 100 should be near 66%, if not, let it running 2 minutes with each option. And please, post your results.

My results for 200 attacks:

"Never change the lever" 31%
"Change the lever always" 69%
"Change the lever randomly" 54%

If you let the program to run even more attacks, 1000 for example, you will have more accurately 33, 66 and 50%.
Title: Final Decision
Post by: BlauK on December 08, 2004, 04:27:09 AM: This is surely interesting and it is also interesting to see my own position switching... :)

It starts to look like this whole case is purely mathematical and dependent on how the pre-conditions are defined.

mando,
What if we consider th epylos instead of bombs, since they are actually selected instead the bombs, which can be on which ever pylon:

RB = pylon with real bomb
L-TB = pylon with lost training bomb
K-TB = pylon with kept training bomb

Possible cases:
A: RB selected, L-TB lost, K-TB unselected (1/2)
B: K-TB selected, L-TB lost, RB unselected (1/2)

:) :)

So is the problem actually based on interpretations. When we are counting probabilities for this one (once in a lifetime) case where the TB was lost from center position, can we say that it could also have been the right pylon if there was a training bomb?

-TB was lost from an unselected pylon.
or
-TB was lost from center pylon, which was unselected.

Does it affect building the equation, that we know the location (pylon) of the lost bomb? Also we dont know if it was TB1 or TB2 that was lost... just a TB... so how can we differentiate between them? If TB1 was red and TB2 yellow, it would be a completely different case, since we would know which TB was lost.

A: RB selected, TB lost from middle, TB unselected (1/2)
B: TB selected, TB lost from middle, RB unselected (1/2)

So... thinking while writing here :)

What if the bomb selector was kind of automatic (semi-intelligent) in such way that it would not tell the positions, only that one could switch with it to a next available bomb. Then we would not know the position of the lost bomb or the selected bomb.

Mando's program is actually performing like above. But if we are originally told that in this particular case the left was selected and TB was lost from middle, we seem to get too much information to apply it to Monty Hall scenario.

Should I now position myself somewhere in between?? :D

If we had this automatic bomb selector, the question would go: "one unknown bomb is selected, a TB is lost from unknown position, should we switch from originally selected unknown position to the other unknown position?" ..... Yes, with 2/3 probability.

Since Mando originally stated that left was selected and center bomb was lost, it seems that we cannot use Monty Hall scenario in this case.. or can we? Do we calculate the pylons or the bombs

More comments anyone?? :)

Accident for single case against intelligence (with options) for repetitive cases, which kind of make the location info irrelevant seems to make the difference..... like someone stated already quite early in this thread.

OK... made my mind :) Without a selector I just described higher above (lets call it Monty Hall bomb selector) the Monty scenario does not apply to the original case.

Sorry Mando...
I seem to be a weather vane, but I still have enjoyed this argument :) .. and thanks to everyone for your own arguments.. sorry for possible harsh words.
Title: Final Decision
Post by: MANDO on December 08, 2004, 11:22:15 AM: BlauK,
considering that the initially selected bomb may be training and lost, if we change always the lever, we'll get 50%, if we dont, 33%. Even in this case the answer is the same: TO CHANGE THE LEVER.
Title: Final Decision
Post by: dedalos on December 08, 2004, 12:16:00 PM: Quote
Originally posted by BlauK
Heeelloooo dedalos :) :)

The pilot ALREADY SAW that a RED TRAINING BOMB WAS LOST! Cases E and F are not possible in this case. If you start adding things to possible cases, you can go on adding everything that is possible in the world.

If that is the only thing we find worth arguing, it is better to let it be :) I cannot show any better evidence than what was said about the situation. If you believe that TB was not lost, we are talking about different cases.

Ha, thats exactly the point :) Without those cases, the pilot knows that he has two bombs. One good one bad and he flips a coin.

But you are right, we will never agree. That does not mean we cannot argue the point :aok
Title: Final Decision
Post by: lasersailor184 on December 08, 2004, 01:21:02 PM: From what I can understand is that the Monte Hall situation calculates the center pylon into the equation, even though it was lost.

You shouldn't do this. It's almost exactly like the pilot only upped the airplane with 2 bombs, one real and one fake. Hence it is only a coin flip.
Title: Final Decision
Post by: dedalos on December 08, 2004, 02:33:52 PM: Quote
Originally posted by lasersailor184
From what I can understand is that the Monte Hall situation calculates the center pylon into the equation, even though it was lost.

You shouldn't do this. It's almost exactly like the pilot only upped the airplane with 2 bombs, one real and one fake. Hence it is only a coin flip.

What he said
Title: Final Decision
Post by: Raider179 on December 08, 2004, 04:18:10 PM: blauk I understand the question just tell me if this is right.

He has two bombs left..
A lever will switch to either bomb.
He can choose A. To switch

B. Not to swtich

There are two outcomes
A. Drops training bomb
B. Drops real bomb

Is this right?
These are the only choices and only outcomes correct?
Is there something I am missing.

its 50:50 unless he knows exactly where the ordance is on his wings. Then he could just select the right ord and his chances are 100%
Title: Final Decision
Post by: Raider179 on December 08, 2004, 04:24:30 PM: Quote
Originally posted by MANDO
Ok, imagine you do not have a plane with racks, but three doors. Behind two of them there are goats (instead training bombs), and the other door hides a golden doblon instead of a real bomb.

To select a door, you need to be in front of it.
Currently you are in front of LEFT door.

Suddenly, you ear a goat bleating (is that the english word?) at your door. Should you switch to other door? Yes (0% to 50%).

But, what if the goat bleats at the CENTRAL door? Should you switch to the RIGHT door? Yes also (33% to 66%). (<- This is the equivalent to the original case).

This makes no sense. On what are you basing switching is more probable. You have no idea where they put the goat. none The odds dont favor you either way. YRevealing the location of one goat just makes the odds go from 1:3 to 1:2. you have 2 curtains now instead of 3. I have to be misunderstanding this somewhere.
Title: Final Decision
Post by: MANDO on December 08, 2004, 05:27:32 PM: Raider179, the idea is that first time you were in front of a door, you have 2/3 chances of missing the correct door. So, it is more probable that the correct door is any of the other two.

If you dont agree this from the beginning, do not continue.

If a goat sounds in any other door but yours, then you take out that door from the group with 2/3 chances of having the correct door. Now you have your original door and another one. Your door has the same chances as the beginning 1/3. But the group with 2/3 now has a single door left (with 2/3). So, it is better for you to change your door.
Title: Final Decision
Post by: lasersailor184 on December 08, 2004, 05:30:57 PM: No mando, that's wrong.

If you're standing in front of the middle door and suddenly it bleats, then you know a goat is behind that door. Or if it bleats behind another door (but not both).

So you now have *******************TWO******************* doors to choose from.

Without any further information I.E. Another goat bleating, you have 50-50 chances of selecting the correct door.

Switching which door you are standing in front of has *****NO***** bearing on the final problem, as does which door the goat bleated behind.
Title: Final Decision
Post by: Raider179 on December 08, 2004, 06:08:42 PM: Quote
Originally posted by MANDO
Raider179, the idea is that first time you were in front of a door, you have 2/3 chances of missing the correct door. So, it is more probable that the correct door is any of the other two.

yeah that is correct 2:3 but this is because each individual door has a 33% likelihood of being correct. 2 x 33 =66%

If you dont agree this from the beginning, do not continue.

If a goat sounds in any other door but yours, then you take out that door from the group with 2/3 chances of having the correct door. Now you have your original door and another one. Your door has the same chances as the beginning 1/3. But the group with 2/3 now has a single door left (with 2/3). So, it is better for you to change your door.

yeah that is correct 2:3 but this is because each individual door has a 33% likelihood of being correct. 2 x 33 =66%

Its a fundmental problem with your logic here. You take away one of the doors but you are still saying it is part of the group. It is not. each of the three has 33% chance. take one away you have to remove it from the odds.
Title: Final Decision
Post by: lasersailor184 on December 08, 2004, 06:14:18 PM: Yup, just like taking marbles out of a bag.

You have 50 marbles in a bag, 10 red, 10 blue, 10 green, 10 clear, 10 black.

What are the chances you'd pick a black one? 10/50. You keep the marble out of the bag.

Now, what are the chances you'd pick a red one. It's not 10/50 because you took a marble out, it's 10/49.
Title: Final Decision
Post by: MANDO on December 08, 2004, 07:00:36 PM: Ok Raider179,
now imagine you have 1000000 doors with only one golden coin, the rest are goats. Initially you select door number 710567.

Suddenly, all the goats but one piss their doors, only your door and other remain clean. Would you change the door? If you dont change your door, and you win, you are really, really, really lucky.

And yes, at the end you have two doors, one good, one bad, 50%?
Title: Final Decision
Post by: Raider179 on December 09, 2004, 12:28:07 AM: yeah its a 50 50 situation you just described.
Title: Final Decision
Post by: ccvi on December 09, 2004, 01:35:50 AM: Quote
Originally posted by MANDO
Ok Raider179,
now imagine you have 1000000 doors with only one golden coin, the rest are goats. Initially you select door number 710567.

Suddenly, all the goats but one piss their doors, only your door and other remain clean. Would you change the door? If you dont change your door, and you win, you are really, really, really lucky.

And yes, at the end you have two doors, one good, one bad, 50%?

50%. Because there's nothing that preven'ts the possible goat behind the selected door to also releveal itself.
probability for selecting the correct one is 1/1000000
probability that, if 999999-1 random goats releval themseleves (because all goats are equal, and the one behind the selected door does not get any important parts cut off etc - that's not in the story) and the one that does not reveal itself is the one you selected is if it was a goat 999999/1000000 * 1/999999 = 1/1000000.

Both probabilities are equally low.

You need an external all-knowing intervention like a game-show master or a wizzard to get different results. Mechanical failures and goats both lack this part.
Title: Final Decision
Post by: Tilt on December 09, 2004, 09:19:30 AM: Quote
Originally posted by MANDO
BlauK,
considering that the initially selected bomb may be training and lost, if we change always the lever, we'll get 50%, if we dont, 33%. Even in this case the answer is the same: TO CHANGE THE LEVER.

You seem to assume by changing the lever the odds are changed...........

Actually the odds were changed when he saw that the first bomb was a training bomb........

If he did not know if the first bomb was real or not (even after dropping) then the chance was still 33%.

According to your arguement changing the lever changes the odds...........but supposing he changed it and then changed it back again............. your logic would assume that the odds revert to 33%? if your logic assumes that it remains at 50% then it was 50% before the lever was moved also.
Title: Final Decision
Post by: dedalos on December 09, 2004, 10:37:20 AM: Quote
Originally posted by MANDO
Raider179, the idea is that first time you were in front of a door, you have 2/3 chances of missing the correct door. So, it is more probable that the correct door is any of the other two.

If you dont agree this from the beginning, do not continue.

If a goat sounds in any other door but yours, then you take out that door from the group with 2/3 chances of having the correct door. Now you have your original door and another one. Your door has the same chances as the beginning 1/3. But the group with 2/3 now has a single door left (with 2/3). So, it is better for you to change your door.

Mando, with this logic it is bretter not to switch
Title: Final Decision
Post by: MANDO on December 09, 2004, 01:34:12 PM: Quote
Originally posted by ccvi
You need an external all-knowing intervention like a game-show master or a wizzard to get different results. Mechanical failures and goats both lack this part.

Why do you need an external all-knowing intervention? He is just going to show you all except 1 bad and 1 good (the only good). You may force that kind of situations without external all knowing intervention.

Lets get back to the 2 goats and a golden doblon example, and lets change the preparation a bit.

1 - you force the goats to drink a lot of water and place them randomly, also place the doblon.
2 - the "player" selects initially a door and wait in front of the doors.
3 - As soon as a single goat pisses (clearly noticeable below the door because the floor behind the door is inclined) the player will take a final decision.

Should the player change his door when a goat in a different door pisses? Of course, the golden doblon is not going to piss.
Title: Final Decision
Post by: lasersailor184 on December 09, 2004, 01:37:36 PM: Exactly mando. That changes the odds of finding the right door. There is one less door to worry about.
Title: Final Decision
Post by: MANDO on December 09, 2004, 02:21:52 PM: Quote
Originally posted by lasersailor184
Exactly mando. That changes the odds of finding the right door. There is one less door to worry about.

So, he should change or not?
Title: Final Decision
Post by: ccvi on December 09, 2004, 03:13:04 PM: Quote
Originally posted by MANDO
Why do you need an external all-knowing intervention? He is just going to show you all except 1 bad and 1 good (the only good). You may force that kind of situations without external all knowing intervention.

He, who opens the doors, knows what's hidden where.

Quote
Originally posted by MANDO
Lets get back to the 2 goats and a golden doblon example, and lets change the preparation a bit.

1 - you force the goats to drink a lot of water and place them randomly, also place the doblon.
2 - the "player" selects initially a door and wait in front of the doors.
3 - As soon as a single goat pisses (clearly noticeable below the door because the floor behind the door is inclined) the player will take a final decision.

Should the player change his door when a goat in a different door pisses? Of course, the golden doblon is not going to piss. [/B]

a) 1/3: initial selection correct
b) 1/3: initial selection wrong, goat reveals behind the non-selected door
c) 1/3: initial selection wrong, goat reveals itself behind the selected door

In cases a and b (they look the same from the point of view of the player) switching will not improve the probability of winning. It's 1/2.
In case c switching will improve the probability of winning from 0 to 1/2.
Always switching results in the same probability of winning as only switching when required to (when the initial selection is revealed as false).

The pilot with the lost bomb was in case a or b. Switching does not help him.