Death Probability! Really Hard One!

[roolstar]

Alfred, Bob and Charlie (A, B, C) are three prisoners.

One day, one of the guards came and told them: "Tomorrow, 2 among you will be executed, and 1 will be set free. I know who will die myself but I don't want to tell you now, I prefer to tease you during your last night here."

That night while the other prisoners were sleeping, Alfred was unable to: he was calculating the probaility of being executed the next day. According to his calculations, he had a 2/3 (66.67%) probability of being executed tomorrow.

But then "EUREKA"!

He called the guard and said: "I don't want you to tell me if I'm gonna die tomorrow, but can you tell me which among the other two will die?"

The guard answered him: "Mmmm, ok. Charlie will be one of the two prisoners who will get executed tomorrow. But I will not tell you if the second one will be you or Bob!"

Alfred with a smile: "Thank you so much man!'

Guard: "But I don't understand how this will make you feel any better."

Alfred replied with a huge smile: "Well, now I have a 1/2 (or 50%) chance of being executed instead of the 2/3 (66.67%) I had before your answer: either me or Bob will die tomorrow so 1 out of 2!"

Is Alfred right in his reasoning?

[Guest]

One million people, including yourself, purchase lottery tickets of which there will be three winners. If you could get someone with inside info to rule out 999,994 losers, other than yourself if you should be one, is your probability of being a winner 1/2?

[Guest]

Alfred replied with a huge smile: "Well, now I have a 1/2 (or 50%) chance of being executed instead of the 2/3 (66.67%) I had before your answer: either me or Bob will die tomorrow so 1 out of 2!"

Is Alfred right in his reasoning?

The guard's proclamation of Charlie's impending doom has given Alfred no new information. Alfred already knew that one of his cohorts would die, so being told that Charlie was going to die doesn't really tell him anything new.

On the contrary, had he asked the guard, "Will Charlie die?" and the guard had responded "Yes", then Alfred could infer that he has only a 50% chance of being executed.

[bonanova]

Who's that standing over there in the corner, smiling at us?

Why, it's ... Monty Hall!

Let's see: Alfred is the door #1 that I chose,

Bob is the door #2 that Monty opened to show a losing choice,

and Charlie is the door #3 that I should trade my choice for.

If only I had that choice!

But I don't.

Ooops, poor Al still has only a 1/3 chance of surviving.

Charlie, on the other hand, now has a 2/3 chance of making it.

Interesting ... because if it had been Charlie who had asked the question,

and the guard had told Charlie that poor old Bob was gonna bite the big one,

then Al would have the 2/3 chance.

Right?

[roolstar]

Let me just add that I started this new puzzle as a way to answer this old one:

viewtopic.php?f=7&t=325

Can anybody see the similarities (or differences) between the two?

Didn't we move from:

Before the answer:

Al - Bob ==> Dead

Al - Charlie ==> Dead

Bob - Charlie ==> Dead

After the answer:

Al - Charlie

Bob - Charlie

So from a probability of 1/3 of survival for Al to 1/2 now?

This is just a question for discussion and not necessarily the final answer to this riddle.

PS: And bonanova, nice call on the Monty Hall reference...

[roolstar]

Before the answer:

Al - Bob ==> Dead

Al - Charlie ==> Dead

Bob - Charlie ==> Dead

After the answer:

Al - Charlie

Bob - Charlie

Just to be clearer on the above quote

Before the answer there were 3 possibilities:

Al - Bob will die

Al - Charlie will die

Bob - Charlie will die

This means Al has a 2/3 probability of dying (in 2 cases out of 3 above)

Now after the answer:

Al - Charlie will die

Bob - Charlie will die

(Bob & Al cannot both die or else the guard couldn't have said Charlie)

Hence, Al now has 1/2 probability of dying (in 1 case out of 2 above)

What's the error in this reasoning? (If there is one)

[Guest]

Hence, Al now has 1/2 probability of dying (in 1 case out of 2 above)

What's the error in this reasoning? (If there is one)

spoxjox already answered this and the replies of others seem to indicate that they agree. It's usually at this point that the OP either posts that he agrees or disagrees with the conclusions given.

[roolstar]

I do agree with the reasoning presented in this post... This is in fact my own reasoning...

I'm just the kind of people who start with a puzzle and move to an even bigger application or project...

Example: From this puzzle we can make a reference like Monty Hall for example in the show where out of 3 doors, the guest picks one and then the host opens one of the remaining 2 (the one that has nothing behind it) and then the guest was presented with the choice to either keep his choice unchanged or open the remaining door instead. (Only one door had a prize behind it)

And we can compare that with the "Deal or No Deal" program today to see the differences between both!

Personal opinion: In the case of Monty Hall, the guest should always change his initial choice. This will in fact double his chances of winning (2/3 instead of 1/3).

While in Deal or No Deal, switching the box at the end will not affect the probabilities one bit. This also suggests that every time they open a box with a small or big amount of money in it, the guests should neither feel happy, nor nervous. He/She is not changing anything whatsoever in His/Her probabilities of success!!

Now where does "One Girl, One Boy" post fit into all this? And where in the calculations does the fact that the host knows which are the empty boxes (Or who will die the next day) affect the probabilities?

As far as spoxjox answer: I consider that the fact that Charlie will be among the executed IS new information (New Information: Charlie will die) and does not answer my dilemma of moving from a system of 3 possibilities to a system of two!

The remaining possibilities are:

Alfred and Charlie die

Bob and Charlie die

To be clear, I share the opinion of the rest of the people who replied on the post. But how can I clearly refute the other ones.

Thank you for your patience.

[bonanova]

My comments in red.

in Deal or No Deal ... every time they open a box with a small or big amount of money in it, the guests should neither feel happy, nor nervous.

He/She is not changing anything whatsoever in His/Her probabilities of success!!

Since the offered amount is calculated from the undisclosed amounts, your payout is increased if large amounts remain undisclosed.

Now where does "One Girl, One Boy" post fit into all this?

It's not a direct fit.

The condition [one of the children is a girl] affects the outcome for a different reason from your puzzle, which is Monte Hall in disguise.

The outcome differs if the sex of a particular child is disclosed; and opening a door knowing [or not] that it's a loser affects the odds.

But that's the extent of the similarity. The odds change for different reasons.

And where in the calculations does the fact that the host knows which are the empty boxes

(Or who will die the next day) affect the probabilities?

In your puzzle [and in Monte Hall] if the guard [host] does not know, then there's no advantage to swapping:

1/3 of the time the opened door will have the car and 1/3 of the time you will have chosen the door with the car.

But if it is certain that a losing option is revealed, then your odds of winning double if you swap.

To be clear, I share the opinion of the rest of the people who replied on the post. But how can I clearly refute the other ones.

Give a clear explanation of the correct answer.

Thank you for your patience.

[Guest]

As far as spoxjox answer: I consider that the fact that Charlie will be among the executed IS new information (New Information: Charlie will die) and does not answer my dilemma of moving from a system of 3 possibilities to a system of two!

You may consider it new information, but it is not. To be more specific: It is not relevant new information. I mean, Alfred could have been told that Charlie's wife is pregnant, and that's new information, too -- but it doesn't help Alfred know whether he is going to be executed.

The remaining possibilities are:

Alfred and Charlie die

Bob and Charlie die

Here is the crux of the problem. You are assigning names to the possibilities, but in this case, the names don't matter.

Possibility One: Alfred and the first of the other two get executed.

Possibility Two: Alfred and the second of the other two get executed.

Possibility Three: Both of the other two get executed.

As we see, Alfred has 2 chances in 3 of being executed. Now, Alfred asks the guard to name which of the other two is being executed (or to pick one of the other two, if both are to be executed). But Alfred already knows that at least one of them is to be executed! Finding out the name of the victim doesn't give him any relevant new information. To put it another way: Alfred doesn't know whether Charlie is "the first" or "the second" of the "other two" in the possibilities listed above. The exact same three possibilities still exist, even after Alfred gets a name out of the guard.

[bonanova]

To put it another way: Alfred doesn't know whether Charlie is "the first" or "the second" of the "other two" in the possibilities listed above.

The exact same three possibilities still exist,even after Alfred gets a name out of the guard.

Alfred knows: Charlie is either the first or the second.

Alfred knows: One of those possibilities is in fact gone.

Alfred knows: The liklihood of the other got doubled.

[Guest]

To put it another way: Alfred doesn't know whether Charlie is "the first" or "the second" of the "other two" in the possibilities listed above.

The exact same three possibilities still exist,even after Alfred gets a name out of the guard.

Alfred knows: Charlie is either the first or the second.

Which he already knew -- obviously, Charlie is either the first or the second, and Ben is the other.

Alfred knows: One of those possibilities is in fact gone.

False. Alfred knows that TWO of those possibilities are in fact "gone", because only ONE possibility is true -- that is, only two people are getting executed. The problem is, HE DOESN'T KNOW WHICH ONES ARE FALSE. Finding out that Charlie is to be executed does not give Alfred any new information at all to tell him which of the original possibilities are false.

Alfred knows: The liklihood of the other got doubled.

Incorrect. No probability "got doubled". The probabilities remain the same.

[Guest]

Incorrect. No probability "got doubled". The probabilities remain the same.

If bonanova meant that the probability of the other living doubled, then he is correct.

Here are the possible scenarios (the first symbol represents Alfred) L=live, D=die

D D L

D L D

L D D

Each man has a 2/3 chance of dying. Alfred asks the guard to tell him which one (or randomly pick one of the two if they are both to die) of the other two will die. Remove the D from the first set, the other lives. Remove the D from the second set, the other lives. Remove either one of the Ds from the third set, the other dies. The probability of the "other" man living ("other" meaning the one of the two others that the guard did not mention) went from 1/3 to 2/3 with the new information.

[bonanova]

Possibility One: Alfred and the first of the other two get executed.

Possibility Two: Alfred and the second of the other two get executed.

Possibility Three: Both of the other two get executed.

As we see, Alfred has 2 chances in 3 of being executed.

Now, Alfred asks the guard to name which of the other two is being executed (or to pick one of the other two, if both are to be executed).

But Alfred already knows that at least one of them is to be executed!

Finding out the name of the victim doesn't give him any relevant new information. [* - see below]

To put it another way: Alfred doesn't know whether Charlie is "the first" or "the second" of the "other two" in the possibilities listed above.

The exact same three possibilities still exist, even after Alfred gets a name out of the guard.

Agree.

Alfred knows - and we know - either

[1] Charlie is "the first" - in which case Possibility Two has been eliminated - i.e. determined to be false. or

[2] Charlie is "the second" - in which case Possibility One has been eliminated - i.e. determined to be false..

Can you clarify why you say a possibility has not been eliminated?

"The exact same three possibilities still exist ... "

An option that has a zero fraction of the total probability [i.e. is false] is usually thought of as having been eliminated.

The remaining possibilities - the options with non-zero probabilities - are:

[1] Charlie and Alfred die - probability is 2/3.

[2] Charlie and Bob die - probability is 1/3.

If you want to hang onto your third possibility, then:

[3] Albert and Bob die - probability is 0/3.

[* - you can argue it's not relevant to Alfred, but it's certainly relevant to Bob.]

Since Bob could have been named, but was not, Bob's chances of survival just doubled.

Alfred's chances remained the same.

Proof:

Alfred's chances [of survival] stayed at 1/3.

Charlie's chances went to zero.

Bob's doubled, to 2/3.

Since one and only one survives, these chances must sum to unity.

This is why in Monty Hall you make the swap.

[roolstar]

The thing is that this puzzle isn't very intuitive. I mean it can be best explained by rules of probability.

I am french educated myself and I'm not familiar with the English terms involved; so I'm gonna write down my idea in the real international language, Mathematics.

Let A be the event that Al will be executed

Let C be the event that the guard said that Charlie is going to be executed

The simple question from this puzzle is as follows: “What is the probability of Al being executed, knowing that the guard said that Charlie will be executed?”. All this sentence can be summarized by “Calculate P(A/C)”. And this not at all easy to calculate especially that the relationship between A & C is the whole quest of this thread. But “Bayes” gives us a way to solve this with his famous equation:

P(A/C) = P(C/A) x P(A) / P© = (1) x (2) / (3)

This equation can be “read” this way:

The probability of Al being executed, knowing that the guard said that Charlie will be executed = The probability that the guard said that Charlie will be executed knowing that Al will be executed (1) x The probability of Al being executed regardless of the guard information (2) Over the probability of the guard saying that Charlie will be executed regardless of Al situation (3).

Easy isn’t it?

(1): The probability that the guard said that Charlie will be executed knowing that Al will be executed = P(C/A) = 50% = 1/2. Because if Al will be executed, Bob and Charlie have the same probability of being the second guy to go. In other words, the 2 possible events if Al will be executed are:

Al & Bob will be executed

Al & Charlie will be executed

(2): The probability of Al being executed regardless of the guard information = 2/3

(3): The probability of the guard saying that Charlie will be executed regardless of Al situation = 50%. Simple reasoning: since Bob and Charlie have the same probability of dying, the probability of the guard “choosing” either one is the same.

P(A/C) = (1/2) x (2/3) / (1/2) = 2/3

And the probability of Al being executed knowing that the Guard said Charlie is the same as the probability of Al being executed before this information.

And that’s why this is not a NEW information.

I came to this conclusion after posting my reply yesterday, when I set out to define what the statement “NO NEW INFORMATION” means mathematically: it simply means that the two events are independent and P(A/C) = P(A) ==> C is irrelevant for A. And this does not come by intuition in the case of the prisoners.

Now following the above calculations, I also went after Bob’s fate. What happens to Bob after the guards declaration?

Can you find out? And if you were Al would you trade your fate for Bob’s if you were presented the opportunity? (Back to Monty Hall)

PS: I know that bonanova already stated that Bob's probability of survival will double and that it will be of benefit to trade Al's fate with him. But try to calculate it if you like...

[unreality]

All of this (not just roolstar's post, the whole discussion) is assuming that the two to die were picked at random- they are picked beforehand, the guard knew which two would die and which would live. I don't think they rolled some dice to determine who would live. Whoever was making the decision would do it on a reason... based on their crime or their sentence and how much the judge (or whoever) likes Alfred, Bob and Charlie, you know what I mean.

but assuming their chances of survival are equal and random then there are three scenarios, and one of these 3 MUST HAPPEN and have equal chances:

* Al dies, Bob dies, Charlie lives: 1/3

* Al dies, Bob lives, Charlie dies: 1/3

* Al lives, Bob dies, Charlie dies: 1/3

But then we get confirmation from the guard that Charlie dies no matter what

these are the only two options:

* Al dies, Bob lives, Charlie dies: 1/2

* Al lives, Bob dies, Charlie dies: 1/2

There's no reason to weigh Bob over Al or vice versa, and one of those two MUST happen, so isn't it 1:1 chances?

LIke Al said, it's him or bob. I'm probably wrong and there are many people here with more probability knowledge so I would appreciate it a lot if someone could point out the error in my reasoning

Alfred, Bob and Charlie (A, B, C) are three prisoners.

One day, one of the guards came and told them: "Tomorrow, 2 among you will be executed, and 1 will be set free. I know who will die myself but I don't want to tell you now, I prefer to tease you during your last night here."

That night while the other prisoners were sleeping, Alfred was unable to: he was calculating the probaility of being executed the next day. According to his calculations, he had a 2/3 (66.67%) probability of being executed tomorrow.

But then "EUREKA"!

He called the guard and said: "I don't want you to tell me if I'm gonna die tomorrow, but can you tell me which among the other two will die?"

The guard answered him: "Mmmm, ok. Charlie will be one of the two prisoners who will get executed tomorrow. But I will not tell you if the second one will be you or Bob!"

Alfred with a smile: "Thank you so much man!'

Guard: "But I don't understand how this will make you feel any better."

Alfred replied with a huge smile: "Well, now I have a 1/2 (or 50%) chance of being executed instead of the 2/3 (66.67%) I had before your answer: either me or Bob will die tomorrow so 1 out of 2!"

Is Alfred right in his reasoning?

[unreality]

One million people, including yourself, purchase lottery tickets of which there will be three winners. If you could get someone with inside info to rule out 999,994 losers, other than yourself if you should be one, is your probability of being a winner 1/2?

I would say yes... but then I realized I read over the "OTHER THAN YOURSELF IF YOU SHOULD BE ONE" part. Oops. It doesn't say if you are picked or not. Silly me. So the answer is no. Cuz you could be part of the losers and not told. Your chances are the same.

But does that relate to the riddle? He doesnt ask about himself, he asks about the other two people- and the answer DOES give him new information

Take this example (i know it doesnt seem to directly relate... im not sure) :

There are 6 people, who bet on a different number on a dice. You are #2. If they switch out the dice at the last minute with a 5-sided die, the sixth person is ruled out, and now the chances are 1-5, all equal. Your chances went from 1-in-6 to 1-in-5, right? However if you are #6 and you are never told of the die switch, you *think* you have a 1/6 chance of winning when in fact you have no chance.

Are these just random shots in the dark at understanding? How does the 1:1 become 2:1? what makes alfred more special than bob? I'm confused

[Guest]

All of this (not just roolstar's post, the whole discussion) is assuming that the two to die were picked at random- they are picked beforehand, the guard knew which two would die and which would live. I don't think they rolled some dice to determine who would live. Whoever was making the decision would do it on a reason... based on their crime or their sentence and how much the judge (or whoever) likes Alfred, Bob and Charlie, you know what I mean.

It doesn't matter if they were picked randomly or not. It could have been done based on their crime, who the judge likes best, or who is oldest. It really doesn't matter because the riddle tells us "According to his calculations, he had a 2/3 (66.67%) probability of being executed tomorrow". Whether or not they were picked randomly, if Alfred was not aware what method was used to pick which two die (and that he concluded he had a 2/3 chance of dying is evidence that he wasn't), the choice was essentially random (some would argue that true randomness doesn't exist, anyway).

these are the only two options:

* Al dies, Bob lives, Charlie dies: 1/2

* Al lives, Bob dies, Charlie dies: 1/2

There's no reason to weigh Bob over Al or vice versa, and one of those two MUST happen, so isn't it 1:1 chances?

There is reason to weigh Bob over Al. If you think about my lottery example again, it may be more intuitive:

One million people, including yourself, purchase lottery tickets of which there will be three winners. If you could get someone with inside info to rule out 999,994 losers, other than yourself if you should be one, is your probability of being a winner 1/2?

There is now you and five others left. Do you have a 1/2 chance of being a winner? How about the other five?

I would say yes... but then I realized I read over the "OTHER THAN YOURSELF IF YOU SHOULD BE ONE" part. Oops. It doesn't say if you are picked or not. Silly me. So the answer is no. Cuz you could be part of the losers and not told. Your chances are the same.

But does that relate to the riddle? He doesnt ask about himself, he asks about the other two people

Exactly. That's HOW it relates to the riddle. In the lottery example you also don't ask about yourself. You ask about the others which doesn't help you narrow down your chances of winning.

- and the answer DOES give him new information.

Well, it does give him new information, but as spoxjox pointed out, "But Alfred already knows that at least one of them is to be executed! Finding out the name of the victim doesn't give him any relevant new information." In the lottery example we get new information too. But nothing that helps us narrow down your chances of winning.

Are these just random shots in the dark at understanding? How does the 1:1 become 2:1? what makes alfred more special than bob? I'm confused

I explained this in an earlier post:

Here are the possible scenarios (the first symbol represents Alfred) L=live, D=die

D D L

D L D

L D D

Each man has a 2/3 chance of dying. Alfred asks the guard to tell him which one (or randomly pick one of the two if they are both to die) of the other two will die. Remove the D from the first set, the other lives. Remove the D from the second set, the other lives. Remove either one of the Ds from the third set, the other dies. The probability of the "other" man living ("other" meaning the one of the two others that the guard did not mention) went from 1/3 to 2/3 with the new information.

Do you see why Alfred still has a 2/3 chance of being executed? When the guard told Alfred the name of one of the other two that would die, nothing changed for him. He already knew that at least one of the other two would die.

Same in the lottery example. You have a 3/1,000,000 chance of being a winner. If you ask a lottery official to eliminate 999,994 losers from the "others", you still only have a 3/1,000,000 chance of winning. You already knew that there would be 999,997 losers, eliminating 999,994 others in which you're already excluded from doesn't help you. Likewise, the information the guard gave Alfred was about "others". The other five people have great reason to be excited, though.

[unreality]

ah, I see now. Though my reasoning was:

he knows that at least one of the other two is already going to die, yes, but knowing which one changes it.

But I guess not.

(some would argue that true randomness doesn't exist, anyway).

I believe in true randomness

But does that relate to the riddle? He doesnt ask about himself, he asks about the other two people

Exactly. That's HOW it relates to the riddle. In the lottery example you also don't ask about yourself. You ask about the others which doesn't help you narrow down your chances of winning.

yep I see now. Thanks man!

[Guest]

Finding out the name of the victim doesn't give him any relevant new information. [* - see below]

To put it another way: Alfred doesn't know whether Charlie is "the first" or "the second" of the "other two" in the possibilities listed above.

The exact same three possibilities still exist, even after Alfred gets a name out of the guard.

Agree.

Alfred knows - and we know - either

[1] Charlie is "the first" - in which case Possibility Two has been eliminated - i.e. determined to be false. or

[2] Charlie is "the second" - in which case Possibility One has been eliminated - i.e. determined to be false..

Can you clarify why you say a possibility has not been eliminated?

Because the name is decoupled from the possibility. Knowing that "Charlie" is the name of one person being executed does not change Alfred's probability tree.

Another way of looking at this, which is perhaps equally valid, is that Al's probability doesn't change, but Bob's does. I don't like this way of viewing things, though; if Bob secretly went and asked the guard the same question, the guard would likely say "Charlie" -- which would then mean that Al's probability just doubled, too! Now each man has a 2/3 chance of survival, giving almost even odds that neither will be executed. This violates the conditions of the puzzle.

The fact is that two will be executed, so there is only one of the three "possibilities" that will take place. The other two "possibilities" are false, and are in fact not possible at all. They only seem possible because we are ignorant. Alfred knows that one or both of his compatriots will be offed. Finding out the name of one of them who will be killed doesn't change his state of ignorance. That's all I meant.

[bonanova]

[1] Alfred will survive - 1/3 chance; or [2] Bob will survive - 2/3 chance.

I don't see why that's [a] incorrect, or incomplete.

[roolstar]

The fact is that two will be executed, so there is only one of the three "possibilities" that will take place. The other two "possibilities" are false, and are in fact not possible at all. They only seem possible because we are ignorant. Alfred knows that one or both of his compatriots will be offed. Finding out the name of one of them who will be killed doesn't change his state of ignorance. That's all I meant.

The whole point is that the sentence in red cannot be taken out of context.

Let me clarify:

Finding out the name of one of them who will be killed doesn't change his (Al's) state of ignorance if he was not among the possible choices for the guard!

Finding out the name of one of them who will be killed does change his (Al's) state of ignorance if he was among the possible choices for the guard!

And that is the source of the confusion!!!

In fact take this example:

Altered version of the puzzle :If the guard came to the prisoners and said: "Tomorrow, 2 among you will die. And Charlie, since I don't like you, I'm gonna tell you now that you are one of them. As for the two of you (Al and Bob), I'm gonna leave you with some hope for the night!"

The probabilities for Al and Bob are now 1/2!!!

Now why is that so?

The only thing that had changed between this version and the original puzzle, is the position of Al between the possible choices for the guard: in the second version above, the guard could have just as easily said: "Al, you're gonna die tomorrow!". But in the original puzzle that was not a possibility, and this is the source of the mis-balance between Al and Bob: THEY ARE NOT ON EQUAL FOOTING IN THE FIRST PUZZLE!

One million people, including yourself, purchase lottery tickets of which there will be three winners. If you could get someone with inside info to rule out 999,994 losers, other than yourself if you should be one, is your probability of being a winner 1/2?

Martini gave the lottery tickets example to introduce big numbers and make it even more absurde to think that your probability changed so dramatically. And that is a cool intuitive technique BTW.

However, the answer to his lottery tickets puzzle depends on the interpretation of the part in RED above: "If you should be one". 2 possible interpretation for me:

Case 1: If, APART FROM YOURSELF, this guy with inside info ruled out 999,994 losers from the remaining 999,999 lottery ticket holders, this will keep your initial probability of 3/1,000,000 unchanged. (Same case as the initial puzlle)

Case 2: If this someone with inside info, simply removed 999.994 losers among the 1,000,000 WITH YOU INCLUDED, and you "stayed" among the last 6 remaining, that would actually give you 50% chance of winning.

And why is that so? Well here's the best explanation I can find:

In Case 2, I "survived" 999,994 times from being among the losers! While in Case 1, I was not in any “danger” of being present among these 999, 994 names. And that is exactly why in Case 2, my probability of winning is greatly increased from 3/1,000,000 to 1/2 !!!

That same logic applies with the prisoners' situation.

In the original puzzle, Al did not “survive” any danger of being picked. Therefore, he is not “granted” any better chance (stays 1/3). Bob, however did “survive” the danger of being the picked one, and therefore is “granted” extra chances of survival (now 2/3)!

[Guest]

I'd like to see what people think of this. I'm not sure I make sense.

I agree that the information is not new, but I think it is relevant. I think Alfred had a 50% chance of being killed from the beginning. If he knows that one of the other guys is going to die no matter what than that person can be eliminated (names don't matter) Therefore it comes down to him and one other guy being killed.

[roolstar]

I'd like to see what people think of this. I'm not sure I make sense.

I agree that the information is not new, but I think it is relevant. I think Alfred had a 50% chance of being killed from the beginning. If he knows that one of the other guys is going to die no matter what than that person can be eliminated (names don't matter) Therefore it comes down to him and one other guy being killed.

In probability you simply CANNOT say that if

- 2 people will die

- I may be one of them

==> This means that I have a 50% chance of dying.

Otherwise the more people may die, the lower the possibility of dying. Let me illustrate, if:

- 100 people may die

- I may be one of them

==> I have 1% chance of dying (According to your reasonning)

And if:

- 1 pereson may die

- I may be this person

==> I will die for sure 100% (according to your reasonning)

The correct reasonning in probability is:

The possibilities are:

1- A & B will die or

2- A & C will die or

3- B & C will die

So Probability of A dying = Number of possibilities where A dies (1 & 2) / Total number of possibilities (3)

P(A) = 2 / 3 = 66.6666%

I hope this answers your query.

Edited February 5, 2008 by roolstar

[Guest]

It seems to me that the answer to this question depends upon the question you are asking. If we are asking about the probability that Alfred will survive based on the initial selection process, then he has a 1/3 chance of survival, as does Bob and Charlie. This holds true even after we know that Charlie will die. If the question is what was the probability of Alfred surviving, the answer is 1/3 (same for Bob and Charlie). If instead we are asking about the probability that Alfred will survive based on the fact that Charlie will die for sure and that one of either Bob or Alfred will die, then he has a 1/2 chance of survival. The question and the setup have changed. Probabilities change when new parameters are given. To say that Bob now has a 2/3 chance and Alfred has a 1/3 chance of survival is mathematical absurdity and it not grounded in reality.

[Guest]

Someone in all this mess got this right, but was shoved aside by the statisticians among you.

Here's the rub: The guard says "I know who will die myself but I don't want to tell you now"

Thus, statistics do not apply to who is going to die, but merely to Al's ability to guess. His knowledge of one of the others being a "chosen" one has absolutely NO bearing on who the two to die are, because this is a predetermined condition that DOES NOT CHANGE.