Riddle That Will Screw With Your Brain

Okay, before you read this and decide to embark upon trying to solve this brain-teasing mother, make sure that you don't have anything else to do in the next few hours. Because trust me, not being able to solve this WILL be killing you.

All right, first let me give my mini-disclaimer by stating that I didn't create this (haha, that'd be a riot), nor did I even discover it firsthand. In fact I just read this on another discussion forum a while ago, and I suffered (along with many others) trying to solve it.

3 Prisoners are to be shot at dawn. Instead of calling them Prisoners A, B, and C, we're gonna give them names. They are now Randy Orton, Hayden Panettiere, and Jack Bauer, respectively.

After finding out that 1 of them will be pardoned and spared, Bauer asks the warden:

''As only one of us is going to be spared, it is clear that at least one of the other two of us is going to die. If you tell the name of one of those who is going to be shot for sure, I'll give you access to CTU's encrypted files. You're not giving away any secret, because, it is mathematically certain that at least one of the other two will be shot. Just give me the name of one of the two who is certain to be shot and I'll give you the information.''

The guard thinks for a moment. He is convinced that this will have no effect on anything; nothing is changed by telling Jack. So he says, "The Legend Killer Randy Orton will die.''

Jack is now happy because where before he had a one-in-three chance of surviving, he now has a 50/50 chance of being spared.

QUESTION: IS JACK BAUER'S REASONING CORRECT? (If not, what is the probability of him surviving and the probabilities of each of the other two surviving?)

NOTE: If it was already certain that (at least) one of the other two must die, how is it possible that just knowing which of the other two will die will improve Jack's chances of survival?[/b]

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And before anyone makes any retarded, smart-ass remarks such as "Hayden can't die, she's Claire and she'll just regenerate herself" or "Jack would never compromise CTU and the safety of the peace-loving world," save it.

Also note that this is not a matter of linguistics, meaning the answer to this riddle does not lie within the wording of the problem.

If you actually decided to try and solve this Monty Hall-like riddle, then God help you. LMAO at you, BTW.

And yeah, if anyone actually gets the answer, I'll be pleasantly shocked.

[supersnake52]

god...

i know the obvious awnser is 50/50.

but imm guessing thats too simple.

Hints?

im pretty sure it has something to do with the names.

Nope, absolutely nothing to do with the wording of the problem. You have to think about how would this problem change if Bauer had flipped a coin and gotten an answer from that instead? Or if it had been one of the other prisoners who had asked the guard for an answer? If you say nothing would have changed, then obviously the answer of a 50/50 chance that is given CANNOT be right.

The question is, why?

[supersnake52]

=(

im confused now

All right, I posted this riddle on a variety of places, including other discussion forums, myspace, websites, facebook, etc., and only one person was able to figure out the whole thing correctly.

There's a longer explanation for it and a shorter one. Let me just say right now that I didn't write either of them, I'm just posting the answers. I can't take credit for writing the explanations.

The answers to the question are as follows:

Jack Bauer still only has a 1/3 chance of surviving
Hayden Panettiere now has a 2/3 chance of surviving
Randy Orton has a 0/3 chance of surviving

Still adds up to a total of 1.




Here's the short explanation:

Assuming the pardon was random, he had a 1/3 chance. I don't see how this has changed, other than that the riddle actually tells us the name of one of the condemned. So, let's break down the possibilites, keeping the weight equal amongst the three. If Jack was pardonoed, the guard could pick either other prisinor, so we split that into two scenarios. If Jack is not pardoned, the guard must list the other name, but we will list that twice, to keep the 1/3 probability in effect.

1 - Jack Pardoned - guard says Orton will be killed
2 - Jack Pardoned - guard says Panitierre will be Killed
3 - Orton Pardoned - guard says Panitierre will be killed
4 - Orton Pardoned - guard says Panitierre will be killed
5 - Panitierre Pardoned - guard says Orton will be killed
6 - Panitierre Pardoned - guard says Orton will be killed

Now, the gaurd tells Jack that Orton will be killed. thus eliminating scario 2, 3, and 4. Leaving scenarios 1, 5, and 6:

1 - Jack Pardoned - guard says Orton will be killed
5 - Panitierre Pardoned - guard says Orton will be killed
6 - Panitierre Pardoned - guard says Orton will be killed

Jack still has only a one in three chance of being pardoned
Panitierre has a 2/3 chance of being pardoned
and Orton is gonna die. (if we can trust the guard)[/b]

And here's the much longer one:

The correct reply is as follows: Jack's chances of being spared is and remains one in three, but, thanks to the information provided by the guard, the probablility of the remaining prisoner, Hayden Panettiere, being spared has now risen to 66 percent, or a two-thirds probability.

Jack Bauer has strayed in to one of our mental tunnels; he has fallen in to a probability trap. The only one who gains from the guard revealing this information is Hayden, who may know nothing at all about the transaction and has been sleeping in her cell. If it seems to you quite mad that the girl who plays the indestructible cheerleader's chances of survival have objectively increased, thanks to something that happened in which she took no part and of which she knows nothing, then yours is a cognitive illusion. An increased probability is not a sort of fluid that can pass in a jolt from one prisoner to another and produce material change.

We dislike the idea that Hayden's, who also plays Kairi in Kingdom Hearts, chances of survival have been increased by an event in which the beneficiary took no part, and about which she knows nothing. Symmetrically, we dislike the idea that the chances of everyone's favorite CTU Field Agent, who has elicited and received the piece of information, has not changed at all. It is our view of what a probability is that betrays us. We think of probability not as an abstract mathematical entity but as a ''thing,'' a ''process;'' to us it has real weight and body.

The reader can (I hope, at least) work out the solution to the problem as follows:

Only one of the three prisoners is to be spared. Until the guard intervenes, each prisoner has the same probability of being spared. That is, each has a one in three chance of surviving (and a two-thirds chance of being executed the following morning). The Legend Killer Randy Orton and Hayden Panettiere together have a cumulative probablilty of two thirds or 66 percent. Now a cumulative probability does not determine the individual probabilities; RKO and Hayden together have a two thirds probability that one of them will be spared. The moment the guard says that the Legend Killer is certain to be executed, the whole of that cumulative probability passes to Hayden; by herself she now enjoys a 66 percent probability of survival. That she neither knows nor feels it is totally irrelevant.

Let's, in fact, pay attention to the piece of implicit reasoning that the guard has to follow. The piece of reasoning isn't specified explicitly in the statement of the problem, but it is obvious, and mandatory, even though it is left implicit, In fact, it is one of our mental tunnels here to disregard totally the strategy that the guard has to adopt in giving his answer. Here is the strategy, and its heavy, inevitable, probabilistic consequences.

If Bauer happens to be one of the two who will die for sure, then the guard will be forced to name the only other prisoner who will also die for sure - in this case, Orton. Then we know for sure that Panettiere is saved, regardless of the fact that she himself does not know this. How probably is it that this is the case? The obvious reply is two thirds, or 66 percent. If, on the contrary, Jack happens to be the lucky one, then the guard will indifferently name either of the other two prisoners, say, by tossing a coin. How often will this be the case? The equally obvious reply is one third, or 33 percent. As a consequence of the strategy imposed on the guard by the logic of the situation, it matters not a whit to Jack Bauer whether the guard names Randy Orton or Hayden Panettiere. His chances were and still are one third. But it matters a lot to 17-year old Hayden that the guard has named the youngest World Heavyweight Champion in history, Randy Orton, and not her. In fact, as we have just seen, the answer has two thirds chance of implying that young Hayden will be saved.

The objective probability of an event involving any given individual may well be affected by another taking place anywhere in the world, and quite without the individual knowing it. Foolish Jack thinks his chances of survival have jumped from one third to one half because he suffers from the same illusion we saw in the three box problem: He believes that in any given situation about whose outcome he is uncertain, probability is subdivided equally and becomes 50/50. But here again, the only link that rigorous probability calculation allows is that the sum of two probabilities is 1, or 100 percent; probability theory does not claim the sum is necessarily equally divided between the two remaining boxes, or the two undesignated prisoners.

There is an elegant reasoning that also shows the absurdity of Agent Bauer's argument, and why his probability of survival cannot be improved. Suppose the guard says it is sweet, innocent Claire/Kairi/Lizzie/Hayden rather than the cocky, arrogant Orton who is sure to die.

This would make no difference at all to Jack Bauer because all he wants, following a strange way of thinking, is to rule out one of the other two as certainly condemned. Any one of the other two will do for him. In fact, the subjectively experienced boost in probability, from one third to one half, is tied for him to the certainty that now there are only two candidates left who could possibly benefit from the pardon. Therefore, as it does not matter at all to him which prisoner is actually named by the guard, and because he knows for sure that at least one of them will die, he may as well toss a coin himself and pick one of the other two at random. In other words, he may as well imagine that there is a guard whom he manages to bribe. All this rigmarole might as well take place in the private chamber of his imagination.

Thus Jack can self-boost his chances from one third to one half. If, I say if, we assume that the line of reasoning of Bauer is probabilistically right, then he has a one half chance of being saved. But then an identical piece of reasoning is open to any one of the three prisoners. Thus each of them would have a probability of one half of being saved. This is, of course, a monstrosity from the point of view of the most basic probability theory: Their cumulative probabilities would add up to more than 100 percent (in fact, to 150 percent). This is a little proof per absurdum that shows that the reasoning of CTU Agent Jack Bauer cannot be right.[/b]

Credit for writing those explanations go to other people. I just changed the names a little, hehe.