## Friday, July 5, 2013

### The Mona Lisa rooms

There is a countably infinite number of red-shirted people and a countably infinite number of blue-shirted people. On Sunday, each of them rolls a die marked "Monday", "Tuesday", "Wednesday", "Thursday", "Friday" and "Saturday", but none of them get to see the result of the roll. It is made sure that each of the 12 possible shirt-color and day combinations is realized in infinitely many cases--if not, the rolls are repeated.

For the next six days, on each day they are paired up, in such a way that a red-shirt and a blue-shirt are together taken into a room with a TV. Every red-shirt is paired with a blue-shirt and every blue-shirt is paired with a red-shirt. Moreover, the pairings change every day, so nobody is paired with the same person twice. Now, the TV is connected to a camera, which either points at the Mona Lisa or at a perfect fake of the Mona Lisa.

But there is a crucial difference between the red-shirts and the blue-shirts. The red-shirts are shown the real Mona Lisa on the day shown on their die, and the fake one on the other five days. The blue-shirts are shown the fake Mona Lisa on the day shown on their die, and the fake one on the other five days.

Everybody knows these rules. No other information is available.

Oh, and there is no conceptual difficulty in setting up these pairings if you have an infinite number of people and rooms. Let's say it's Saturday. Then you have four infinite sets: red-shirts who need to be shown the real Mona Lisa, blue-shirts who need to be shown the real Mona Lisa, red-shirts who need to be shown the fake Mona Lisa, blue-shirts who need to be shown the fake Mona Lisa. You also have some exception rules: for each person, there are now four people they're not allowed to be paired with. No information is available to the participants on how the pairings are arranged.

You're one of the red-shirts. You're in a room with a blue-shirt. What are your probabilities that you're looking at the real Mona Lisa?

The plausible answer seems to be 1/6. But if you say that, then by parity you should say that for the blue-shirt with you, it's 5/6.

Now, you and the blue-shirt get talking. You exchange names, you talk about what you saw on the preceding days. Eventually, you have all the same information. And let's even suppose that all the people in the experiment have the same priors and that they weigh evidence according to the same rational rules.

At this point, either your probabilities will become the same or they will stay different.

If they stay different, then we have a nice argument that it's possible to have two people with exactly the same information, the same priors and the same rules of judgment who disagree. That's an interesting conclusion.

If they become the same, there are two reasonable possibilities. Maybe they become 1/2. Suppose now that everybody is offered a bet where you get \$3 if it's the real Mona Lisa and \$2 if it's the fake, where the bets are all going to be settled up at the end of the experiment. If your probabilities have changed to 1/2, you will take the bet, and then by the end of the experiment, given that you're a red-shirt, you will have received \$3 and paid out \$8. Oops.

The other possibility is that they become an interval of probabilities, presumably [1/6, 5/6]. But that doesn't reflect the fact that there are perfectly clear answers to how you should bet. As the above example shows, clearly you should be betting according to a 1/6 probability of the Mona Lisa.

So, it seems that you need to stay at 1/6. We do get an interesting conclusion about disagreement, but maybe it's not absurd.

Alright, but now let's complicate the matter.

You're an intelligent fly on the wall in one of the rooms on Monday. You know all the above rules, but otherwise nothing more than any of the participants. A red-shirt and a blue-shirt come in. What is your probability that it's the real Mona Lisa on the TV? Surely either 1/2 or [1/6, 5/6].

But then you learn that there is a rule for flies, too. A fair coin is tossed for you. If it's heads, you are assigned to follow the red-shirt for the rest of the week, and if it's tails you are assigned to follow the blue-shirt for the rest of the week. But it's heads and this is announced to you.

At this point, your lot is exactly like the lot of this red-shirt. You will see the real Mona Lisa on exactly one out of six days. But it's still Monday. You've learned something: the coin is heads. Should you change your probability for whether it's the real Mona Lisa?

Presumably not, since in learning that the coin is heads you've learned something that's independent of the die rolls that determine whether it's the real Mona Lisa.

But how are you relevantly different from the red-shirt that you're following? The main difference is that the red-shirt presumably knew she was a red-shirt before she came into the room, while you didn't know that you were assigned to a red-shirt until you came into the room. But surely the order in which you've learned these things doesn't matter.

So it seems that your probabilities should be the same as the red-shirt's. And so they should switch to 1/6 when you learn the coin is heads. Which is absurd.

William said...

There is a subtle selection bias introduced by forcing exactly one of each day of the week in sequence, while fixing on one individual as being required to fit with one of those days, not none of them.

One way to see the selection bias is to consider a blue shirt that gets shown the fake Mona Lisa on the first day of the 6-day "week."

On all subsequent days the probability of that particular blue shirt being shown the real Mona Lisa is 100%, not 5/6 given the 6 days. If the Mon-Sat day of the week chosen to do the next pairing was chosen at random it would be different.

I think the probability for an individual red shirt starts on Monday at 1/2 and ends at 1/6. The probability for an individual blue shirt starts at 1/2 and ends on Saturday at 5/6.

The larger probabilities for reds and decrease in probabilities for blues comes from the selection bias.

Alexander R Pruss said...

Your probabilities don't add up. For 1 = P(ML on Monday) + P(ML on Tuesday) + ... + P(ML on Saturday). But 1/2+...+1/6>1.

So if your first day probability is 1/2, the remaining day probabilities have to be lowered so that they add up to 1. But it is implausible that on Saturday the probability of seeing the (real) ML is less than 1/6.

Rob K said...

This refutes a famous result of Robert J. Aumanns's: Robert J. Aumann The Annals of Statistics, Vol. 4, No. 6. (Nov., 1976), pp. 1236-1239.
I'm curious about which assumption of Aumann's has to go in this context.

Alexander R Pruss said...
This comment has been removed by the author.
Alexander R Pruss said...

In my example some of the information is not measurable. Specifically, that I end up in the same room as y.

By the way, you can't change probabilities in the room on pain of inconsistency, regardless of which shirt color you have.

Alexander R Pruss said...

I can perhaps tighten things up.

Let's consider a red-shirt, Francine, on Saturday. There is no difference between any of the days at this point--in hindsight, they're all on par. (Pace William's comment. To make the parity clearer, imagine that all the days happen at once--perhaps our people have six eyes, each pointing into a different room at the same time.) But she is certain that exactly one of these days has her looking at the real ML. So if she has a numerical probability on Saturday, consistency requires that it be 1/6. So, nobody can change probability.

But now let's see what she thinks about the blue-shirt she meets on Saturday. Call him Sam. Consistency requires that she thinks that the probability that he's looking at the real ML on Saturday is 1/6, since they're looking at the same TV. But she also knows that he sees the real ML on exactly 5 days. Since Sam's other five days are exactly on par to Francine, she must think that on the other days Sam has probability 29/30 of seeing the real ML. That's weird--why should Sam's probabilities for Francine be any different than her probabilities for other blue-shirts. But at least this is consistent.