Final Decision

[MANDO]

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.

[MANDO]

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.

[MANDO]

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.

[lasersailor184]

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.

[dedalos]

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

[MANDO]

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

Dedalos, RB cannot be lost.

[lasersailor184]

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?

[MANDO]

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?

[lasersailor184]

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.

[ccvi]

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?

[Raider179]

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.

[MANDO]

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.

[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.

[BlauK]

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.

[BlauK]

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???"  

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.