More Prisoner Problems
Out of three prisoners scheduled to be put to death, Tom, Dick, and Harry, one of them will be pardoned. Tom asks the warden to tell him the name of one of the others who will be executed. As the question is not directly about Tom's fate, the warden obliges — either naming the other prisoner to be executed, in case Tom was too, or secretly flipping a coin to decide which of the remaining names to give Tom if Tom is the one being pardoned. Assuming the warden's truthfulness, there are now only two possibilities for who will be pardoned: Tom, and whichever of Dick or Harry the warden did not name. Did Tom gain any information as to his own fate, that is, does he change his estimate of the chances he will be pardoned?
If the warden says "Dick will be executed" and Tom could switch fates with Harry, should he?
If the warden says "Dick will be executed" and Tom could switch fates with Harry, should he?
Labels: logic
posted by ToM at
8:49 AM
23 Comments:
Assuming that each has the same probability of being pardoned (1/3), if the warden says "Dick will be executed" than Tom should change his fate with Harry, as this will double his probability of being pardoned (2/3).
This is easy to understand if you think of 100 prisioners instead of 3. at the beginning, Tom has a 1/100 probability of being pardoned and the rest 99 have a probability of 99/100; if the warden tells you the name of 98 prisioners that will be executed, than Tom keeps the 1/100 and the other that was not named keeps the 99/100. Of course Tom should switch his fate.
Well.. I'm going to go out on a limb here and say he has a 100% chance of being executed.
The reasoning here, is because one of these three were already executed back on Feb 25, 2009 (see here http://trickofmind.com/2009/02/tom-dick-and-harry.html). They just forgot to remove the corpse, which is now just a skeleton on the floor. So now the other two get executed.
if dick is executed, tom should switch fates with the other guy. here is why
each man has a 33.33% chance of living. each has a 66.66 chance of being executed. when you reveal one person who will be executed, the odds arent 50-50. in the begining, the odds were that tom would die. so it is more likely that the other guys fate is to live. maybe this will explain the reasoning better
if you are on a game show, and there are 1000 boxes in front of u. 999 of them are empty and 1 has $1000 in it. the host tells u to chose 1 box. (lets say box #647). then the host removes 998 boxes. the ones remaining are the one u picked and another box. the host tells you that one of these two boxes has $1000. you are allowed to keep the box you picked, or trade boxes. the odds are that you picked an empty box!
I'm impressed. Some great minds have been know to get this wrong.
If you are allowed to swap fates, always do so. You double your chances of survival.
See: http://en.wikipedia.org/wiki/Monty_Hall_problem for a full exposition.
I understand the reasoning nd know the math.... but tell me this if dick is being executed nd out of both tom and harry one will live.... So they both have a 50-50 chance of survival nd lets say that tom was orignally being pardoned nd he changes places wid harry then he will be executed so y change places at all....
A more direct explantion can be seen here: http://en.wikipedia.org/wiki/Three_Prisoners_problem
It took me ages to accept the solution when I first came across it (only a few weeks ago) - but under the guise of the Monty Hall problem. Computer simulations were required before I really accepted the result.
I wasn't happy to post it (Monty Hall) as a problem until I felt 100% comfortable that I could explain it very clearly and convincingly in my own words.
I can only completely convince myself that Tom continues to have a 1/3 chance of being executed after he learns the name of one of the other prisoners who is going to be executed.
What happens if tom and harry went to the warden at the same time and learned that dick will die?-how can they BOTH better their chances by switching with each other?
ha!
Hi Knightmare. If that happened, then they can't improve their odds by switching. One (and only one) of them is going to die - I assume that's what you assume. If the warden was unable to name who was going to die, then they are both going to die.
Hi again Knightmare. Sorry, I should have acknowledged that you were joking:)
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There are three equally likely cases. First, if Tom is going to live then the warden can name Dick or Harry. It actually doesn't matter if the warden is biased to say Dick or to choose randomly. Nor does it matter if Tom knows that the warden is biased to say Dick. This is because, the unnamed man is scheduled to die anyway. In the other two cases, the warden has no choice about the name to give - so the unnamed man is going to live.
So considering separately the three cases in the order Tom, Dick, Harry to live.
Probability of unnamed man going to live = (1/3)*0 + (1/3)*1 +(1/3)*1 = 2/3
So Tom should swap with the unnamed man to maximize his chance of living.
I practically guarantee that that is the nicest solution you'll find anywhere.
Sorry about all the deletes. My previous post made them all redundant.
aaarggh. I hadn't stated the obvious - Tom has learned nothing about his own fate. He knew that at least one of the others was going to die when he initially assessed his risk at 1/3.
Would you believe that I have an even better way to see it. Everyone's initial risk is 1/3. Once the warden has named a prisoner who will die,then it is certain that either of Tom or the unnamed prisoner (but not both) will live. So as Tom's chance of survival is 1/3, the unnamed prisoner's chance of survival is 1-(1/3) = 2/3.
Small error is last post, I meant Tom had assessed his likelihood of survival at 1/3.
Doh! same error twice in a row. I meant everyone's initial chance of survival is 1/3.
Having ried this on my dad, he still doesn't accept the reasoning. I have just sent him a better explanation, here it is:
Consider separately the three case of Tom, Dick and harry scheduled to live. All three cases are equally likely.
1. Tom to live, Dick and Harry to die. Whichever name the warden gives, the other is still going to die. Tom shouldn’t swap with the unnamed man.
2. Dick to Live, Tom and Harry to die. The warden can only name Harry. Tom should swap with Dick (the unnamed man).
3. Harry to live, Tom and Dick to die. The warden has to name Dick. Again Tom should swap with Harry (the unnamed man).
So, on two out of three (equally likely) occasions, Tom should swap with the unnamed man.
kicking my last post through
Hello cyberspace. I’ve just realised where (some) confusion comes in. As far as the swapping goes, what we have to decide is what is Tom’s fixed strategy going to be. Tom is either (always) going to stick with his original fate, or he’s (always) going to swap with the unnamed man. He could of course toss a coin each time (in fact he’ll then get to 50-50 for living) – but I’m not going to analyse that strategy. I’m only going to analyse the case of whether or not Tom should swap.
Tom never swaps: Then he may as well not even trouble the warden with the question as he’s not going to act on the information. So Tom will live 1/3 of the time. I won’t analyse this situation further, as that would simply be recursively examining the whole problem.
Tom always swaps: Then 1/3 of the time, he will have done the wrong thing (that’s when Tom was scheduled to live and the unnamed man was scheduled to die). On the other 2/3 occasions, (when Tom is scheduled to die) one of the others must be scheduled to live. The warden names the only one who is scheduled to die, so the unnamed man is scheduled to live.
It follows that the naming process doesn’t change Tom’s (initial) expectancy of living. It does however, allow Tom to assign a probability of the unnamed man being scheduled to live to 2/3.
If e.g. Harry overheard that “Dick is going to die”, but didn’t know that Tom was not allowed to be named, he’d incorrectly (in point of fact) calculate that his and Tom’s odds were 1/2 each. Harry would have no way of knowing that his odds were as high as 2/3.
That is definitely my last post on this subject (I think).
he should.
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