Final Decision

[MANDO]

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.

[ccvi]

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.

[lasersailor184]

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.

[BlauK]

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

[BlauK]

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.

[ccvi]

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.

[MANDO]

Originally posted by ccvi
Take 3 equal things ...

Wrong example.

[MANDO]

Originally posted by lasersailor184
It's the exact same.

Sorry, but I give up too.

[ccvi]

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.

[lasersailor184]

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?

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!

[BlauK]

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?

[BlauK]

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

[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!

[ccvi]

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.

[BlauK]

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.