### Brain Teasers

# Confess or Not

Fun: (2.35)
Difficulty: (2.64)
Puzzle ID: #8823

Submitted By: lesternoronha1 Corrected By: Winner4600

Submitted By: lesternoronha1 Corrected By: Winner4600

Probability
Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.

You and your partner in crime are both arrested and questioned separately. You are offered a chance to confess, in which you agree to testify against your partner, in exchange for all charges being dropped against you, unless he testifies against you also. Your lawyer, whom you trust, says that the evidence against both of you, if neither confesses, is scant and you could expect to take a plea and each serve 3 years. If one implicates the other, the other can expect to serve 20 years. If both implicate each other, you could each expect to serve 10 years. You assume the probability of your partner confessing is p. Your highest priority is to keep yourself out of the pokey, and your secondary motive is to keep your partner out. Specifically, you are indifferent to you serving x years and your partner serving 2x years. At what value of p are you indifferent to confessing and not confessing?

### Answer

If you confess, there is probability p of two confessions and (1-p) probability that only you implicate your partner. If you don't confess, there is probability p that neither of you implicate each other and a (1-p) probability that only you are implicated. Let s denote one unit of suffering. One year in jail yourself would cause you 2 units of suffering, and one year of your partner serving would cause you 1 unit of suffering.This shows the total suffering given the four possibilities:

Both of you confess: 10*2+10=30

Your partner confesses and you don't: 2*20+0=40

You confess and your partner doesn't: 0+20=20

Neither of you confesses: 2*3+3=9

So given the probability p of your partner confessing if you confess, the expected, or average, amount of suffering would be 30p + 20(1-p)=10p+20. If you don't confess the expected amount of suffering would be 40p + 9(1-p)=31p+9. To find the indifference point equate the two expressions:

10p+20 = 31p+9

11 = 21p

p=11/21 =~ 52.4%.

Thus if you think your partner's probability of confessing is greater than 11/21, you should confess; otherwise, you should keep your mouth shut.

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## Comments

are you mr. nash?

Ah, but of course it could also be said that, whatever happens, you are better off confessing.

If your partner confesses, and you don't you get 30 years. If you do you get ten years. So you are better off confessing.

If your partner doesn't confess and you don't, you get three years. If you do, you get no years, so again, you are better of confessing.

If your partner confesses, and you don't you get 30 years. If you do you get ten years. So you are better off confessing.

If your partner doesn't confess and you don't, you get three years. If you do, you get no years, so again, you are better of confessing.

This seems like an unlikely answer given that the conditions for both of you are identical and yet we are told that the point of indifference is when your partner serves double your time. An answer of about 50% should see you both doing equal time on the average. I'm still working on it!

i would keep my mouth shut cus im not a rat

MY BRAIN HURTS!!!!!!!!!!

It's a nice teaser. I agree with the given answer - I don't see why it troubles some people.

AAAhhhrrgghh ! don't even want to try to solve this one.

Nash had less to do with the Prisoner's Dilemma than von Neumann. In any case, the answer is wrong. There is no point in calculating a Nash equilibrium point for a single game that is not zero-sum.

It is the classic Prisoner's Dilemma -- Given that my partner has chosen one choice or the other, and his choice is not influenced by my choice, there is no reason for me not to confess. Of course, I'd LIKE my partner to think that I might not confess, thereby making a chance that he doesn't. But whether he has confessed or not, I always get a lesser sentence by confessing, so it is foolish for me ever not to.

When the game is repeated, as in von Neumann's experiments, it makes sense not to confess (or "betray" in his terms). But in a single game (or the last round of a multi-game) there never is a reason, assuming I am acting purely selfishly.

It is the classic Prisoner's Dilemma -- Given that my partner has chosen one choice or the other, and his choice is not influenced by my choice, there is no reason for me not to confess. Of course, I'd LIKE my partner to think that I might not confess, thereby making a chance that he doesn't. But whether he has confessed or not, I always get a lesser sentence by confessing, so it is foolish for me ever not to.

When the game is repeated, as in von Neumann's experiments, it makes sense not to confess (or "betray" in his terms). But in a single game (or the last round of a multi-game) there never is a reason, assuming I am acting purely selfishly.

Question wording is too complicated. Should be made simpler.

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