Originally posted by TheMaster37
Hmm, you are stating two situations, but I cannot see the difference. My thoughts were that after finding out what the guard told C the chances for A increase to 1/2 (and he knows that).
Where's the flaw in A's reasoning by thinking: "Myself and C are in the same situation, with the same information. Therefore our chances of survival are equal."
A ...[text shortened]... inal problem I'm easily thinking the wrong way. That's one of the reasons I posted this 🙂
My conception of the problem is as follows:
Each prisoner has an equal chance of survival initially, equal to 1/3. This is equivalent to having a dart board cut into three equal sections, with a single dart throw determining who survives. After prisoner A asks who's going to die, the guard throws the dart. The dart could have ended up anywhere on the board, but when the guard says "B is going to die", he reveals that the survival dart did not end up in B's section. So, prisoner A reasons that the dart had a 1/3 probability chance of ending up in his own section, and a 2/3 chance of ending up somewhere else. Since he knows that it didn't end up in B's section, the entire 2/3 chance gets shifted to C. Right now, prisoner is wishing he could play "let's make deal" with the guard, dropping the soap notwithstanding.
Now, prisoner C asks the same thing, and also finds out that B is going to die. He also correctly deduces that his own chances of survival remain 1/3, while A's go up to 2/3. While each analysis is correct on it's own, when taken together they appear to form a contradiction. However, there is no contradiction because neither A nor C have complete and consistent information about the situation. Therefore, the Law of Total Probability does not apply across both analyses as pointed out by Cribs in the 3 Prisoners thread Thread 11542
When A finds out that C knows B is going to die, A now has a complete picture and must re-evaluate his chances. I believe the main reason is that if A knows that C knows B is going to die, this eliminates some of the possibilities. Initially, the guard could have selected AB, BC or AC as the victims, and reported either B or C in the case of BC. This leaves A safe 1/3 of the time because he knew AC wasn't possible (CB out of AB, BC or CB). However, since he knows that C knows B is going to die, BC with C being reported is no longer a possibility, so A's chances are now 1/2 (CB out of AB, CB). C has no such restriction, 'cause he still don't know jack.
So, at this stage in the game I would say A has a 1/2 chance of surviving. However, overall A's chances are still 1/3 because he had a 2/3 chance of being chosen in the first place (AB, AC out of AC, BC, AC), and a 1/2 chance of making it past the second part where he finds out B is going to die, so (2/3)*(1/2) = 1/3.
I think that's right. Makes sense? Close enough, I hope.