# Response to Data Science Imposters’ Prisoner’s Dilemma

In one of their recent podcast episodes, the Data Science Imposters discussed the prisoner’s dilemma. It goes like this:

Andy and Bon (that’s me) commit a crime. Not anything terrible, just something gently bad. (No sense in killing folks for a blog post, right?)

We get caught. (I know – it gets worse!) And they separate us into two rooms for interrogation.

The detective says to me, “If you confess, you may get some leniency.”

Bon: Just how much of this “leniency” can I get?

Detective: Well, if you both both confess, you both get reduced sentences of 5 years.

Bon: What if I don’t confess?

Detective: You get 10 years, if Andy confesses.

Bon: What if Andy stays quiet?

Detective: That’s the great part. If he keeps quiet and you confess, you get NO time!

Bon: But what if we both refuse to confess?

Detective: We have your DNA, so you automatically get 3 years each.

### What should I do?

The Data Science Imposters gave the classic solution:

Suppose I have an attorney and he’s somehow in the know. If he comes and says, “Andy confessed,” what should I do?

1. If I stay quiet, I will do 10 years.
2. If I confess, I’ll do 5 years.

So if Andy confesses, I will too, so I only get 5 years.

But what if Mr. Know-it-all-attorney comes to me with, “He’s sticking to his guns – Andy is NOT confessing!”

1. If I stay quiet, I will do 3 years.
2. If I confess, I’ll do 0 years.

Again, if Andy zips his lips, it’s still far better for me to confess.

Either way, I confess.

### I can decide with probability and expected values.

I’ve been teaching expected value in my statistics class these past few weeks, so I began to think of the problem from that perspective.

Suppose I don’t know what Andy is doing. And since Andy is flaky, it’s a total coin toss on what he’ll do. In other words, the probability of him doing either is 0.5.

Let’s say I choose to be a loyal friend (maybe he saved my mom from a crazed escaped alligator from the zoo). So I’m going to stay quiet.

We can assign a random variable X to the outcomes of what Andy chooses (as his are totally random):

1. If Andy chooses to confess, the random variable X = -10 (ten years hard time for me).
2. If Andy chooses to stay quiet, X = -5 (only 5 years in the big house).

Since each of these are equally likely, the expected value of the random variable X is the average: -7.5.

On the other hand, let’s say Andy gets on my nerves and I saw him spit in my mom’s alligator gumbo once. So I choose to rat on him.

We assign the random variable Y to the outcomes like this:

1. If Andy chooses to confess, the random variable Y = -3 (a 3 year sentence).
2. If Andy chooses to stay quiet, Y = 0 (I get off scott free!).

Again, these are equally likely, so the expected value is the average: -1.5.

### Do I still confess?

One way to determine if you should play a game is by examining the expected value. If you play the game a bazillion times, the expected value is what you’ll win on average for each game.

In the game of what-will-Andy-do, the expected value of my confession (1.5 years in prison) is better for me than the expected value of staying quiet (7.5 years).

So I’ll still confess.

### But what about loyalty?!

Where there are humans, there is variability. The reality is that my partner in crime could be known for being a snitch or being very loyal. Let’s do some crimes with two other people: Charlie and Dean.

Charlie is a great friend. But not only that, out of the 10 times he’s been in police custody, only 1 time did he turn on his partner.

Empirically, the probability that he will confess is 0.10. And the probability that he will stay quiet is 0.90.

If I stay quiet, and Charlie chooses to

1. Confess, then X = -10 with P(X) = 0.10.
2. Stay quiet, then X = -5 with P(X) = 0.90.

So the expected value is

If I confess, and Charlie chooses to

1. Confess, then Y = -5 with P(Y) = 0.10.
2. Stay quiet, then Y = 0 with P(Y) = 0.90.

Sadly, even with a super loyal friend, I should still turn him in!

### How about that weasel, Dean?

Dean has been in police custody 10 times and 7 of those he’s thrown his partner under the bus.

Empirically, the probability that he will confess is 0.70. And the probability that he will stay quiet is 0.30.

If I stay quiet, and Dean chooses to

1. Confess, then X = -10 with P(X) = 0.70.
2. Stay quiet, then X = -5 with P(X) = 0.30.

So the expected value is

If I confess, and Dean chooses to

1. Confess, then Y = -5 with P(Y) = 0.70.
2. Stay quiet, then Y = 0 with P(Y) = 0.30.

As suspected, I still should confess, but maybe with Dean I do so very quickly!

Share your thoughts in the comments, on Twitter or on Facebook.

The Data Science Imposters podcast is one of my very favorites. Jordy and Antonio offer just the right amount of intelligent self-depreciation balanced with great information. I highly recommend them!

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### One Response to Response to Data Science Imposters’ Prisoner’s Dilemma

1. Wil says:

I LOVE this thought experiment! When I come across it, I find myself wondering what certain people who pop into my head would decide if I was in that predicament. There is a team/trust building exercise similar to this that relies on everyone considering the greater good. Just did a search online but came up empty. I’ll have to talk with Bon to see if we can find it and maybe she could do a follow-up on this post.

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