Author Topic: Final Decision (Read 6414 times)

Zwerg · « **Reply #15 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%.

MANDO · « **Reply #16 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.

MANDO · « **Reply #17 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

(Right click and save)

BlauK · « **Reply #18 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!

BlauK · « **Reply #19 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!!!

vorticon · « **Reply #20 on:** November 27, 2004, 01:22:25 PM »

assume the 2 duds were loaded first then drop the one that was loaded last...

MANDO · « **Reply #21 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.

ALF · « **Reply #22 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, 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'

BlauK · « **Reply #23 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!

Dawggus · « **Reply #24 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

ccvi · « **Reply #25 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.

BlauK · « **Reply #26 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.

BlauK · « **Reply #27 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?

ALF · « **Reply #28 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

ccvi · « **Reply #29 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.