Tom, Dick and Harry
Tom, Dick and Harry are in prison. One of them has been randomly selected to die in the morning, and the other two will be set free. Their guard knows which one will die, but none of the prisoners does. The guard is under strict instructions not to divulge the identity of the doomed man.
Tom is desperate for any information beyond the fact that his probability of death is one in three. He begs the guard to throw him an informational bone. Finally, to shut him up, the guard agrees to reveal only the following: the name of one of Tom's fellow prisoners who will be set free rather than killed. The guard then says that Dick will be set free.
After receiving this information from the guard, what is the most accurate calculation Tom can make of the probability that he is the doomed man?
Thanks for your time
-- Josh Mudie
Tom is desperate for any information beyond the fact that his probability of death is one in three. He begs the guard to throw him an informational bone. Finally, to shut him up, the guard agrees to reveal only the following: the name of one of Tom's fellow prisoners who will be set free rather than killed. The guard then says that Dick will be set free.
After receiving this information from the guard, what is the most accurate calculation Tom can make of the probability that he is the doomed man?
Thanks for your time
-- Josh Mudie
Labels: logic, mathemagic, SharedPuzzle
posted by Rajesh Lal at
2:01 PM
30 Comments:
ah the marilyn vos savant paradox.. i remember it was 1/3 but its really confusing why this is the case..
well, the obvious answer is 50%. But, something tells me that I'm missing something.
The information does not help him at all.. the probabilty is still 1 in 3 from the original selection.
All 3 already know that at least one of their prison mates will be set free, since they are only executing one. It doesn't help him to know which one.
The way I see it, either Dick or Harry (or both) are always going to be set free. So saying it is Dick doesn't change anything.
So the odds are still 1/3.
50%
There was 3 unknowns, but now there are 2 unknowns. As more is known the chances of random events are decreased.
BUT, knowing it's 50% does not actually change anything. He had a 33% chance of being free when the die was cast does not change anything. His chances of being killed or set free both went up.
oh, I meant to say he had a 67% of being freed initially, not 33%.
Sadly, there is 100% chance that Tom will die as he was told the name of ONE OF HIS FELLOW PRISONERS to be set free meaning both will be set free...
1/3 is the right answer this is a famous problem, creative try tom^ but thats not what the wording implies
I seem to remember the Marylin vos Savant puzzle from a game show involving a donkey or prizes behind closed doors and whether you should swap choices after one is revealed. It is also done with 3balls in separate boxes, 2 of one colour and one of another.
Anyway,in this puzzle, having the name of the prisoner to be set free means that Tom has a 2 in 3 chance of dying.
the odds are obviously 1-2 since dick is eliminated from the choices. It's as simple as that. In my opinion, there is no way to figure out who was to be executed. Saying it is Dick that will be set free does give us a better idea of who will be executed.
If Tom, Dick and Harry are the only three prisoners, in the prison.I believe it is now a 50/50 chance for Tom. The model has to be modified to fit the new data.
Prior to speaking with the guard the chances of Tom (T), Dick (D) and Harry (H) being FREED fall into three sets of paired prisoners:
T/D
T/H
D/H
ONE of the three above sets will be the FREEDOM combination. Tom (T) has a 67% chance of being in one of those three sets (freedom), and a 33% chance of NOT being in one of the three sets (death).
After speaking with the guard he learns there is a 100% probability that the Tom/Harry (T/H) set will NOT occur because it does not contain (D), leaving the following two possible sets for freedom:
T/D
D/H
Tom has 50/50 chance of being in one of those two sets.
Or, you could look at it like this...The die was always cast for Dick (D) to go free. He was a ruse. He can be zeroed-out, and might as well be one of the guards. Tom and Harry are the only ones in jepordy (i.e. 50/50.)
try analyzing all the outcomes relative to Tom.
2 scenarios are possible: tom lives or tom dies
in the case where tom dies, harry and dick live
in the case where tom lives either harry lives and dick dies or dick lives and harry dies
we are told that dick will live forsure
this leaves us with 2 outcomes:
dick lives harry dies or dick lives tom dies
this give tom a 50/50 chance to live
Precisely, Dick does not have a door to death. No matter what happens tomorrow he lives. He is a ruse...a false choice meant to confuse you.
He could just as well be the prison guard.
It's 50%, and neither of the other two matter. it's the simple fact that either he dies or he doesn't
100% probability he will die, it says the guard was going to give the name of one of tom's fellow prisoners that will be set free sounding like there is more than one of his fellow prisoners will be set free.
It has to be 50 percent for Tom.Since dick is not going to be killed , there is 2 more person left and so the probability become for tom from 1/3 to 1/2.Hence i think it has to be half chance for him.
It is obvious that he will not die because the person to die volunteered and Dick will be set free so shall Tom be set free. Because obviously he didn't volunteer. 0% chance of dying.
shoot them all
Ignoring the most likely scenario where Tom seals his fate by continuing to be a pest and the guard finally just shots him and lets the other two go...
...then Toms chances of dieing were 1/3 before the guard said anything, are still 1/3, and will continue to be 1/3 right up until additional information is revealed.
Now, Harry's prospects are a bit bleaker. If he was listening and has a IQ over about 120 he will suddenly realize that his prospects of taking a bullet in the morning are now 2/3. I suspect he is quite pleased with Tom at this point.
-h
This problem is tricky for reasons not well understood by psychologists. Some characteristic of human reasoning makes it difficult to grasp the logic, even once it has been laid out openly.
I find that providing an analogous illustration helps facilitate people's understanding of this problem. Assume that there are 100 children and 100 boxes. Each box has a different one of the children's names on it. I pick one at random to put a gold star inside; inside all the rest I put a silver star. I then seal all the boxes.
Now suppose that Tom is one of the children. Because I selected the box at random, as far as he knows there is a 1% chance that he received the gold star in his box. That is, if we did this whole "experiment" 100 times, in approximately 1 run of the experiment Tom would get the gold star and the other 99 boxes would contain silver stars; and in approximately 99 runs of the experiment Tom would get a silver star, 98 other boxes would contain a silver star, and 1 other box would contain a gold star. Note that in all 100 runs, at least 98 boxes other than Tom's box contain silver stars; that will be true no matter where the gold star is.
Of course, _I_ know which box contains the gold star. I can therefore always generate a list of 98 names (always excluding Tom's name) that correspond to boxes containing silver stars. I can do this if Tom's box contains the gold star and I can do this even if Tom's box does not contain the gold star.
Suppose that I compile such a list for Tom's benefit, having assured him in advance that the list will contain precisely 98 names of persons other than Tom whose boxes contain silver stars; in other words, that no matter which box contains the gold star the list will reduce the candidate gold-star boxes to two, Tom's and one other.
HAS TOM'S PROBABILITY OF RECEIVING THE GOLD STAR ROCKETED FROM 1% TO 50%? CONSIDER THAT IF WE "RUN" THE "EXPERIMENT" 100 TIMES, WE ONLY EXPECT TOM TO RECEIVE THE GOLD STAR ONCE, AND YET I CAN PRODUCE A LIST OF 98 NON-TOM SILVER-STAR NAMES IN _EVERY SINGLE RUN_. WILL GENERATION OF A 98 NON-TOM SILVER-STAR-NAME LIST CAUSE TOM TO RECEIVE THE GOLD STAR FIFTY PERCENT OF THE TIME? OBVIOUSLY NOT; I CAN GENERATE SUCH A LIST WITHOUT AFFECTING THE PROBABILITY. TOM STILL ONLY HAS A 1% CHANCE OF RECEIVING THE GOLD STAR ON ANY PARTICULAR "RUN."
I hope the correspondence to the Prisoner Problem is obvious. Change the number of boxes from 100 to 3, and the gold star to a death warrant, and the logic is identical across the two situations. The information Tom was given had no effect on his own probability.
Note that although Tom's probability remain 1 in 3, Harry's probability of being the one to be killed can now be rated at 2 in 3. This is because of the conditions the guard imposed: he said he was generating a one-person non-Tom list of people not being killed; he didn't promise to make a non-Harry list, so the new information that, of persons who are not Tom, Dick is safe doubles the probability that Harry is going down.
Sam has explained it very well, and for all of you who said 1/3, that is the correct answer.
Well......... Its a 50/50 shot for tom, but why not kill all of them?
tom did have a 1 in 3 chance prior to being told. this is now 1 in 2 - 50%
Answer is 2/3. For anyone suggesting 1/2 or 1/3, check your math pages.
Amen
such creative names...
any reason for them?
tom...harry....dick..???
perhaps trying to get a point across?
well anyways...
this ones for you will..
...42
...FORTY TWOOOOOOO......
uhm
ya
42
100% CHANCE because the guard named ONE of TOM's fellow prisoners who will be set free rather than killed. He didn't say one of the three, he applied them to Tom so it obviously him.
HERE IS THE CORRECT ANSWER WITH EXPLANATION
Before being told that Dick will be set free there was a 1/3 chance Tom will die, thus a 2/3 chance either Dick or Harry will die. The guard knew Dick will not die so he told him he would be released, this does not change the fact that there is still a 2/3 chance that Dick or Harry will die, but now the odds are all on Harry, so Tom still has a 1/3 chance of dying and Harry now has a 2/3 chance of dying.
Whoever said people need to check their math is wrong and needs to check their own math.
he has a zero percent chance of dying, because dick will be set free, wich is a term related to being killed.yur mom is being setfree to heavan, for example
he has to wait till the morning
They all die, in time.
Post a Comment
Links to this post:
Create a Link
<< Home