13 Jan 05
4 edits
Originally posted by AThousandYoungFormulate a specific question about your confusion and I will answer it.
Interesting problem. I am confused. Both answers make sense - that his chances increased and that they did not increase.
Both answers can be correct, depending on the guard's knowledge. The spirit of the problem is typically taken to be that the guard must have complete knowledge in order to reveal the name of one prisoner who will be executed. In this case, A's chances of survival remain at 1/3.
It is consistent with the wording of the problem, although not with the intended spirit, that alternatively, the guard could have had no prior knowledge and happened to pick one name to be executed uniformly, and it happened to not be A. In this case, A's survival chances do increase to 1/2, but that is not the process that the problem intends to convey.
Distinguishing between these two cases of the guard's information is essential to the analysis. Those who believe that 1/2 is the correct answer to the first case are simply wrong and conclude 1/2 for the wrong reasons. Those who don't understand the difference between these two cases probably don't fully understand either answer.
In short, 1/3 is the correct solution to the intended problem, although an alternate and consistent, but unintended, reading of the problem leads to a solution of 1/2.
Whichever your favorite interpretation the problem is, it is trivial to empirically verify that the solutions above are correct using pieces of paper with the names on one side. I'll describe each variety of experiment further if it's not clear to you.
If you understand the Monty Hall problem, an identical analysis applies here.
13 Jan 05
Originally posted by DoctorScribblesIf the guard has complete knowledge, then the prisoners chances
Formulate a specific question about your confusion and I will answer it.
Both answers can be correct, depending on the guard's knowledge. The spirit of the problem is typically taken to be that the guard must have complete knowledge in order to reveal the name of one prisoner who will be executed. In this case, A's chances of survival remain a ...[text shortened]... t clear to you.
If you understand the Monty Hall problem, an identical analysis applies here.
remain 1/3, right. Doesn't it also hold that, if he changes outfits
with the other prisoner (not the one the guard mentions), his chances
go to 2/3s?
Nemesio
13 Jan 05
Originally posted by NemesioThat's correct, for the same reasoning that the Monty Hall contestant should always switch doors.
If the guard has complete knowledge, then the prisoners chances
remain 1/3, right. Doesn't it also hold that, if he changes outfits
with the other prisoner (not the one the guard mentions), his chances
go to 2/3s?
Nemesio
14 Jan 05
Prisoner A's chances of survival are, and always were, one half. If two out of three prisoners were going to be executed, then one out of B or C is definately going to be killed. It is just a fifty-fifty chance whether or not it is A or the remaining prisoner out of B or C that would be killed.
14 Jan 05
1 edit
Originally posted by kitripperAs DoctorScribbles noted so eloquently, it really does depend on how you conceive of the problem.
Prisoner A's chances of survival are, and always were, one half. If two out of three prisoners were going to be executed, then one out of B or C is definately going to be killed. It is just a fifty-fifty chance whether or not it is A or the remaining prisoner out of B or C that would be killed.
If the problem is akin to a lottery, where the guard picks a picks the first prisoner's name randomly in response to the question, then A's chance of survival go up to 1/2. This is because of the 3 initial possibilities:
1. AB
2. BC
3. AC
only 2 remain after B's execution is announced. Essentially the second pick has been reduced to a 50/50 coin toss between A and C. This is the solution that conditional probability gives:
P(A given B) = P(A and B)/P(B) = (1/3) / (2/3) = 1/2
Conceptually, this can be thought of as removing a section from the set of possibilities.
If, on the other hand, both prisoners to be executed have already been selected, then when the guard says "B", what he really means is that of the three initial possibilities:
1. AB
2. BC
3. AC
he has selected either "AB" or "not AB", and that while "AC" is obviously no longer possible, it was considered during the selection process. The probability of "AB" being picked is 1/3, therefore the probability of "not AB" being picked is 1-1/3 = 2/3. Prisoner A knows that AC being picked didn't happen, so prisoner A's chances remain 1/3, while prisoner C's chances (unbeknownst to him) have increased to 2/3.
Conceptually, this can be thought of as covering up a section of the set of possibilities (meaning that at the time of selection that section was a valid possibility, but in light of subsequent information that section is no longer a reasonable choice).
Does that help?
14 Jan 05
It should be noted, however, that prisoner A's overall chances remain 1/3 no matter how you read the problem. This is because in the lottery version, A's chance of getting past the first round is 2/3, and his chance of getting past the second round is 1/2. Therefore, his overall chance is (2/3)*(1/2) = 1/3.
14 Jan 05
Originally posted by kitripperThis is nonsense. Every question of probability in the world could be answered 1/2 under this reasoning. Surely you don't think everything has a 50% chance of occuring, do you?
Prisoner A's chances of survival are, and always were, one half. If two out of three prisoners were going to be executed, then one out of B or C is definately going to be killed. It is just a fifty-fifty chance whether or not it is A or the remaining prisoner out of B or C that would be killed.
14 Jan 05
Originally posted by DoctorScribblesWait a minute. When I buy lottery tickets, I figure I've got a 50-50 shot, because I can either win or lose. Are you telling me I'm wrong 😛?
This is nonsense. Every question of probability in the world could be answered 1/2 under this reasoning. Surely you don't think everything has a 50% chance of occuring, do you?
14 Jan 05
4 edits
Originally posted by royalchickenOh, no, your conclusion for that problem is correct. I'm not saying there are no events with a 50% chance of occurance. I'm saying that not all events have that likelihood, so it must be the case that the reasoning that I called nonsense is in fact nonsense, for it leads to a contradiction.
Wait a minute. When I buy lottery tickets, I figure I've got a 50-50 shot, because I can either win or lose. Are you telling me I'm wrong 😛?
In the lottery case, you're right. It's basically a coin toss. You can think of it that way if you want. (Assuming of course that there's only one ball with only two possible values: Win or Lose.)
14 Jan 05
1 edit
Originally posted by DoctorScribblesAu contraire. I claim that all events have probability 50%, because they can either happen or not happen. This is what's so fascinating about probability. Consider an infinite event space partitioned into a countable infinity of nonempty events. Each of these events can happen or not happen, so each has probability 1/2. Since the probability of their union is the probability of the certain event, and the sum of their probabilities,
Oh, no, your conclusion for that problem is correct. I'm not saying there are no events with a 50% chance of occurance. I'm saying that not all events have that likelihood, so it must be the case that the reasonin ...[text shortened]... course that there's only one ball with only two possible values.)
1/2 + 1/2 + 1/2 +... = 1
Thus 1 + 1 + 1 + 1 + 1 + ... = 2.
There are two important consequences of this. The first, an aside, is that:
1 + (1+1) + (1+1+1) + ... = lim(n->infinity) n(n+1)/2 = 2.
The second is that:
1 + 1 + 1 + 1 +... = 2 + 1 + 1 +... = 2.
Thus 1 + 1 + 1 + 1+...= 0, so 0 = 2. Dividing both sides by 4 gives
0 = 1/2, so the probability of all events is 0, so nothing ever happens.
😛
14 Jan 05
1 edit
Originally posted by royalchickenSure, I follow your reasoning perfectly. Now define event E to be:
so the probability of all events is 0, so nothing ever happens.
E: royal chicken will post "so the probability of all events is 0, so nothing ever happens." in the tired old 3 Prisoners problem.
You have conclusively shown that P(E)=0.
The emprical evidence is that P(E)>0.
What do we do now?
Have we shown the relations "is equal to" and "is greater than" to be equivalent?
14 Jan 05
2 edits
Originally posted by DoctorScribblesConclude that 0 > 0, naturally, but empirical scientists do worse than that every day 😉.
Sure, I follow your reasoning perfectly. Now define event E to be:
E: royal chicken will post "so the probability of all events is 0, so nothing ever happens." in the tired old 3 Prisoners problem.
You have conclusively sho ...[text shortened]... E)=0.
The emprical evidence is that P(E)>0.
What do we do now?
Besides, you only showed empirically that P(E|E) > 0, not that P(E) > 0. Since P(E) = 0 and P(E|E) = P(EE)/P(E) = P(EE)/0, P(E|E) is undefined!
14 Jan 05
1 edit
Originally posted by royalchickenHoly Christ! P(E|E) is now undefined? What have we done?
Conclude that 0 > 0, naturally, but empirical scientists do worse than that every day 😉.
Besides, you only showed empirically that P(E|E) > 0, not that P(E) > 0. Since P(E) = 0 and P(E|E) = P(EE)/P(E) = P(EE)/0, P(E|E) is undefined!
We started with the solution that a 1/2 survival chance is the solution to the problem. We have now shown that the prisoner can not only not estimate his survival chances, but even after he's dead, he can't even ask the question of whether he died or not!
Well, on second thought, I guess that does make sense. After all, we wouldn't want P("I died"| "I died" ) to be 1, since the question can't even be pondered by a dead man, so I suppose it is right and proper that that probability is undefined.
14 Jan 05
1 edit
Originally posted by DoctorScribblesEven better, since we proved at FW that conditional probability is a probability function, we can conclude that since all probabilities are 1/2, P(E|E) = 1/2, so we have 00 = 1/2 = 2. Thus 1 = 4, so, 1/4 = 4 etc 😛.
Holy Christ! P(E|E) is now undefined? What have we done?
We started with the solution that a 1/2 survival chance is the solution to the problem. We have now shown that the prisoner can not only not estimate his survival chances, ...[text shortened]... suppose it is right and proper that that probability is undefined.