Solved – The prisoner paradox

probabilityself-study

I am given an exercise, and I can't quite figure it out.

The Prisoner Paradox

Three prisoners in solitary confinement,
A, B and C, have been sentenced to death on the same day but, because
there is a national holiday, the governor decides that one will be
granted a pardon. The prisoners are informed of this but told that
they will not know which one of them is to be spared until the day
scheduled for the executions.

Prisoner A says to the jailer “I already know that at least one the
other two prisoners will be executed, so if you tell me the name of
one who will be executed, you won’t have given me any information
about my own execution”.

The jailer accepts this and tells him that C will definitely die.

A then reasons “Before I knew C was to be executed I had a 1 in 3
chance of receiving a pardon. Now I know that either B or myself will
be pardoned the odds have improved to 1 in 2.”.

But the jailer points out “You could have reached a similar conclusion
if I had said B will die, and I was bound to answer either B or C, so
why did you need to ask?”.

What are A’s chances of receiving a pardon and why? Construct an
explanation that would convince others that you are right.

You could tackle this by Bayes theorem, by drawing a belief network,
or by common sense. Whichever approach you choose should deepen your
understanding of the deceptively simple concept of conditional
probability.

Here's my analysis:

This looks like the the Monty Hall problem, but not quite. If A says I change my place with B after he is told C will die, he has 2/3 chances to be saved. If he doesn't, then I would say his chances are 1/3 to live, like when you don't change your choice in the Monty Hall problem. But at the same time, he is in a group of 2 guys, and one should die, so it is tempting to say that his chances are 1/2.

So the paradox is still here, how would you approach this. Also, I have no idea how i could make a belief network about this, so i'm interested to see that.

Best Answer

Initially there are three possibilities with equal probabilities:

A will be freed (prob $1/3$)
B will be freed (prob $1/3$)
C will be freed (prob $1/3$)

With the promise of the message, there are four possibilities with different probabilities:

A will be freed and A is told B will be executed (prob $1/6$)
A will be freed and A is told C will be executed (prob $1/6$)
B will be freed and A is told C will be executed (prob $1/3$)
C will be freed and A is told B will be executed (prob $1/3$)

Conditional on "A is told C will be executed" this becomes

A will be freed and A is told C will be executed (prob $1/3$)
B will be freed and A is told C will be executed (prob $2/3$)

So after the message A would like to swap with B (the Monty Hall problem) but cannot and so keeps the original $2/3$ probability of being executed.

Related Solutions

Solved – The prestige magician paradox

This mistake was put in evidence in written conversations among Fermat, Pascal, and eminent French mathematicians in 1654 when the former two were considering the "problem of points." A simple example is this:

Two people gamble on the outcome of two flips of a fair coin. Player A wins if either flip is heads; otherwise, Player B wins. What are player B's chances of winning?

The false argument begins by examining the set of possible outcomes, which we can enumerate:

H: The first flip is heads. Player A wins.
TH: Only the second flip is heads. Player A wins.
TT: No flip is heads. Player B wins.

Because Player A has two chances of winning and B has only one chance, the odds in favor of B are (according to this argument) 1:2; that is, B's chances are 1/3. Among those defending this argument were Gilles Personne de Roberval, a founding member of the French Academy of Sciences.

The mistake is plain to us today, because we have been educated by people who learned from this discussion. Fermat argued (correctly, but not very convincingly) that case (1) really has to be considered two cases, as if the game had been played out through both flips no matter what. Invoking a hypothetical sequence of flips that wasn't actually played out makes many people uneasy. Nowadays we might find it more convincing just to work out the probabilities of the individual cases: the chance of (1) is 1/2 and the chances of (2) and (3) are each 1/4, whence the chance that A wins equals 1/2 + 1/4 = 3/4 and the chance that B wins is 1/4. These calculations rely on axioms of probability, which were finally settled early in the 20th century, but were essentially established by the fall of 1654 by Pascal and Fermat and popularized throughout Europe three years later by Christian Huyghens in his brief treatise on probability (the first ever published), De ratiociniis in ludo aleae (calculating in games of chance).

The present question can be modeled as 100 coin flips, with heads representing death and tails representing survival. The argument for "1 in 100" (which really should be 1/101, by the way) has exactly the same flaw.

Simpson’s Paradox – Understanding the Basics of Simpson’s Paradox

I think A and E aren't a good combination, because A says you should pick Mercy and E says you should pick Hope.

A and D have the virtue of advocating the same choice. But, lets examine the line of reasoning in D in further detail, since that seems to be the confusion. The probability of success for the surgeries follows the same ordering at both hospitals, with the A type being most likely to be successful and the E type being the least likely. If we collapse over (i.e., ignore) the hospitals, we can see that the marginal probability of success for the surgeries is:

Type     A     B     C     D     E     All  
Prob   .81   .78   .56   .21   .08     .52

Because E is much less likely to be successful, it is reasonable to imagine that it is more difficult (although in the real world, other possibilities exist as well). We can extend that line of thinking to the other four types also. Now lets look at what proportion of each hospital's total surgeries are of each type:

Type     A     B     C     D     E  
Mercy  .08   .39   .06   .44   .03  
Hope   .09   .54   .23   .09   .05

What we notice here is that Hope tends to do more of the easier surgeries A-C (and especially B & C), and fewer of the harder surgeries like D. E is pretty uncommon in both hospitals, but, for what it's worth, Hope actually does a higher percentage. Nonetheless, the Simpson's Paradox effect is going to mostly be driven by B-D here (not actually column E as answer choice D suggested).

Simpson's Paradox occurs because the surgeries vary in difficulty (in general) and also because the N's differ. It is the differing base rates of the different types of surgeries that makes this counter-intuitive. What is happening would be easy to see if both hospitals did exactly the same number of each type of surgery. We can do that by simply calculating the success probabilities and multiplying by 100; this adjusts for the different frequencies:

Type     A     B     C     D     E     All  
Mercy   81    79    60    21    09     250  
Hope    80    76    51    14    04     225

Now, because both hospitals did 100 of each surgery (500 total), the answer is obvious: Mercy is the better hospital.

The Prisoner Paradox

Best Answer

Related Solutions

Solved – The prestige magician paradox

Simpson’s Paradox – Understanding the Basics of Simpson’s Paradox

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