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# Death Probability! Really Hard One!

### #21

Posted 30 December 2007 - 01:11 PM

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

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

- Bertrand Russell

### #22

Posted 31 December 2007 - 11:50 AM

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

seempossible 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)!**

### #23

Posted 04 February 2008 - 07:03 PM

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.

### #24

Posted 05 February 2008 - 07:18 PM

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 by roolstar, 05 February 2008 - 07:19 PM.**

### #25

Posted 18 April 2008 - 06:01 PM

### #26

Posted 18 April 2008 - 08:10 PM

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.

### #27

Posted 18 April 2008 - 08:56 PM

That is what's being asked and the chance of survival for Alfred stays at 2/3 and the reasons for it have been given many times in this thread. Just claiming it's a "mathematical absurdity and it not grounded in reality" and not giving a mathematical reason why is not how things work around here.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.

Look at Martini's example using the lottery:

"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?"

According to your rationale you would now have a 1/2 probability of winning the lottery. But why should you? All the lottery official did in his example was rule out losers

*not including yourself*. Because you were excluded from the list of possible losers, your chances remain the same- three in a million. You already knew he could rule out 999,994 losers. However, five others are left from all of the ones that were excluded randomly- they have an awesome chance of being winners.

It's the same in this riddle. Alfred already knew that one of the others was going to be executed, so knowing which one it was doesn't help his chances. But Bob wasn't excluded from the selection process. He's in the same boat as the five others in the lottery example. His odds of living just get better (see post #13).

I'm not sure what you're saying. Are you agreeing that the probability is the same for Alfred but has increased for Bob, or are you saying something else?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.

### #28

Posted 18 April 2008 - 09:41 PM

If I know that I have two red marbles in my pocket, but I tell you to guess what color they both are and tell you that they are primary colors, does it change the fact that the two marbles are red? No. Probability only matters when it comes to your guess. You will guess r/r r/b r/y b/b y/y b/y. IF I then tell you one is red, does it change the colors? NO! The probability of the ACTUAL color changing is zero. They have been and always will be red. Only the probability of your guessing correctly changes.

Probability only matters when you have an unknown state and more than one solution. Here, there is ONLY ONE SOLUTION. The guard KNOWS who is living and dying. It's a fixed state with no chance of changing. The only thing that changes is Al's probability of guessing correctly, not the probability he will live or die.

### #29

Posted 21 April 2008 - 05:34 PM

"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?"

According to your rationale you would now have a 1/2 probability of winning the lottery. But why should you? All the lottery official did in his example was rule out losersnot including yourself. Because you were excluded from the list of possible losers, your chances remain the same- three in a million. You already knew he could rule out 999,994 losers. However, five others are left from all of the ones that were excluded randomly- they have an awesome chance of being winners.

I think you are missing my point. I believe it still depends upon the question. If the question is, what are your chances of winning the lottery, the answer would be three in a million. Even after the official ruled out the 999,994 losers the answer would still be three in a million. But if the question is, based on the information you now know what are your chances of winning, the answer WOULD be 1/2. The fact that the lottery offcial merely ruled out 999,994 losers did not change anything in reality, but, probability depends on the question you are asking and the parameters that are given. Thus if the question is, what were Alfred's chances of survival based on the initial conditions, the answer is 1/3. If the question is, between him and Bob (now that we know that Charlie will die and that either Bob or Alfred will die) what are his chances of survival, the answer is 1/2. To say that between the two of them one has a higher probabiility than the other, to me, does not make sense. I could be wrong, but to me that just sounds like it's not grounded in reality and that we are just playing with the numbers with no regard to common sense.

**Edited by baker.uva, 21 April 2008 - 05:37 PM.**

### #30

Posted 21 April 2008 - 06:13 PM

I think you are missing my point. I believe it still depends upon the question. If the question is, what are your chances of winning the lottery, the answer would be three in a million. Even after the official ruled out the 999,994 losers the answer would still be three in a million. But if the question is, based on the information you now know what are your chances of winning, the answer WOULD be 1/2. The fact that the lottery offcial merely ruled out 999,994 losers did not change anything in reality, but, probability depends on the question you are asking and the parameters that are given. Thus if the question is, what were Alfred's chances of survival based on the initial conditions, the answer is 1/3. If the question is, between him and Bob (now that we know that Charlie will die and that either Bob or Alfred will die) what are his chances of survival, the answer is 1/2. To say that between the two of them one has a higher probabiility than the other, to me, does not make sense. I could be wrong, but to me that just sounds like it's not grounded in reality and that we are just playing with the numbers with no regard to common sense.

I read up on the Monty Hall problem on answers.com and I am wrong in my reasoning. Alfred has a 1/3 chance of survival and Bob has a 2/3 chance.

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