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Death Probability! Really Hard One!


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29 replies to this topic

#11 bonanova

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Posted 28 December 2007 - 02:22 AM

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.
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#12 spoxjox

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Posted 28 December 2007 - 08:34 AM


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.
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#13 Martini

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Posted 28 December 2007 - 09:35 AM

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.
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#14 bonanova

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Posted 28 December 2007 - 09:36 AM

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.
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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell

#15 roolstar

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Posted 28 December 2007 - 10:53 AM

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...
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#16 unreality

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Posted 29 December 2007 - 05:09 AM

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?


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#17 unreality

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Posted 29 December 2007 - 05:21 AM

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
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#18 Martini

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Posted 29 December 2007 - 06:45 AM

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.
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#19 unreality

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Posted 29 December 2007 - 05:28 PM

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 :lol:



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!
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#20 spoxjox

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Posted 29 December 2007 - 06:22 PM


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.
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