## The puzzle of the hats

Luc Bovens and Wlodek Rabinowicz (2010 and 2011) present the following puzzle:

Three people are each given a hat to put on in the dark. The hats' colours, either black or white, has been decided by three independent tosses of a fair coin. Then the light goes on and everyone can see the hats of the two others, but not their own. All of this is common knowledge in the group.

Let's call the three players X, Y and Z. There are eight possible distributions of hat colours, each with probability 1/8:

 X Y Z B B B W B W W W B B W B W B W W B W B B W W B W

Now consider the hypothesis

(A) Not all hats have the same colour,

which corresponds to the middle six columns in the table. At the beginning of the game, each player rationally assigns probability 3/4 to (A). When the light goes on and (A) is true, then one player will see two hats with the same colour and the other two will see hats with different colours. If (A) is false, all players will see hats with the same colour. In either case, there is at least one player who sees two hats with the same colour. This player can thereby rule out four of the six (A) possibilities, leaving her with a rational credence of 1/2 in (A).

It then looks as if the group can be Dutch Booked:

1. Before the light goes on, we offer each player a bet for $3 that pays$4 in case of (A) and otherwise nothing. All players regard the bet as fair. We sell it to one of them and pocket the $3. 2. When the light is on, we offer each player to sell to us a bet for$2 that pays $4 in case of (A). Since there is at least one player who assigns credence 1/2 to (A), there will be at least one player who regards this bet as fair. We buy it from her, paying the$2.
3. We now have $1. If (A) is true, we have to pay out$4 on the first bet and receive $4 from the second. If (A) is false, no further transactions take place. We've made a sure profit of$1.

This is the puzzle, as presented by Bovens and Rabinowicz. They go on to argue by a game theoretic analysis that the second bet should not be accepted and conclude that this is an unusual situation in which betting odds and credences come apart.

I disagree. I think Bovens and Rabinowicz have misrepresented the bets in their scenario. If we look at the actual options offered to the players, we can see why the second bet should be rejected.

The crucial issue that Bovens and Rabinowicz neglegt is that the bookie doesn't really offer the relevant bets to all the players. Consider the first stage. Does player X have an option that would cost her $3 and pay$4 in case (A) is true? No, she doesn't. The bookie asks her whether she is willing to buy the corresponding bet, but if she says yes', she can't be sure that she gets it. In fact, she knows that the other two players also say yes' to the alleged offer, and only one of them will be sold the bet -- chosen at random, let's say.

If everyone who said yes' to an offer actually got the offer, there would be no Dutch Book. The bookie would sell three instances of the first bet. If (A) is false, she would also sell three instances of the second, for a total payoff of 3 x $-3 + 3 x$2 = $-3. If (A) is false, she would sell the second bet only once, with net result 3 x$1 + $-2 =$1. In the long run, the average payoff is $0, since (A) is three times as likely as not-(A). To evaluate whether a player should say yes' or `no' in the original story, we have to get clear on what exactly they are offered. 1. At the first stage, each player knows that all of them are willing to buy the bet. Let (X chosen) be the proposition that the bookie will make a deal with player X; similarly for (Y chosen) and (Z chosen). The real offer made to player X is a conditional bet, conditional on (X chosen). Accepting this offer pays$1 in case of (X chosen) & (A), $-3 in case of (X chosen) & not-(A) and$0 in case of not-(X chosen). Since (X chosen) is independent of (A), X's probability for (X chosen) & (A) is 1/3 x 3/4 = 1/4; her probability for (X chosen) & not-(A) is 1/3 x 1/4 = 1/12. So the offer has expected payoff 1/4 x $1 + 1/12 x$-3 = $0. 2. The action happens at the second stage. We know that at least one player sees two hats of the same colour. Without loss of generality, let it be player X. Again let 'X chosen' be the proposition that the bookie will actually make a deal with X. The offer made to X therefore looks like this: she gets$-2 if (X chosen) & (A), she gets $2 if (X chosen) & not-(A), and$0 if not-(X chosen). But this time (X chosen) is not independent of (A). If (A) is true, then the others see different colour hats, so they won't be willing to enter into a deal with the bookie. If (A) is false, they will. So P_X((X chosen) & (A)) = P_X((X chosen) / (A)) x P_X((A)) = 1 x 1/2 = 1/2; and P_X((X chosen) & not-(A)) = P_X((X chosen) / not-(A)) x P_X(not-(A)) = 1/3 x 1/2 = 1/6. The expected utility of the offer is therefore 1/2 x $-2 + 1/6 x$2 = \$-2/3.

# on 13 March 2013, 15:14

This is the "winner's curse", basically.

Buyers need to condition their valuation of an item on the proposition that they are allowed to buy it... otherwise in an auction, that happens if their estimate of its value is highest, which often means they've overestimated its value.

Same thing here, but the form is more disguised... if the people were allowed to bid for the second bet in an auction, and A, then X would overbid because he overestimated the bet's value, giving the bookie some free cash.