Losing your wallet at an economics conference:

You go to an economics conference. The speaker takes wallets from two members of the audience and offers to combine the total and give it to whomever had the lesser sum. You stand to gain the larger sum (if you had the smaller) or lose the smaller sum (if you had the larger).

You should expect a certain distribution of wallets. You should be willing to play the game if :

sum-of-richer-wallets > your wallet * number of poorer people. The median person in the audience should be happy to play this game -- if there are (R-1)/2 people richer and (R-1)/2 people poorer, then the first sum exceeds the second sum, since each richer-wallet is richer than your wallet. But of course, the audience as a whole doesn't get richer if we play... just the median person.

Can we design a group of players for whom all of them should play except the top one? Yes, of course: we give the poorest person 0. The next person has 0 too. If we ever rise above 0, then that person's rank, from least to greatest, is r. This first nonzero wallet value is 1, say. The next wallet must have value r, the next r(r+1), and so on. A super-exponential distribution of wallets insures that everyone but the richest person is willing to play. But this is hardly a paradox... if the richest person is carrying an industrial nation's wealth in traveller's checks, this character might *strongly* resist playing this game with a bunch of people with relatively empty wallets.

Suppose there are (R-1) other people in the audience.

We choose another member of the audience randomly from R-1 people.

You should begin by guessing a distribution of wallets. Just guess some numbers and assign them to the audience. Calculate for each person his or her rank r in the list of wallet-values w, ordering people from the richest to the poorest. Let R be the maximum value of r, i.e., the number of people. Let c be the cumulative sum of the wallets so that to each rank r we assign c(r) = the wealth in the biggest r-1 wallets. Let C be the sum of all wealth. The person with rank r has R-r ways to lose w and (r-1) ways to gain an expected c/(r-1). The expected benefit to the player with rank r is: c/(R-1) - w * (R-r)/(R-1). If w is normally distributed with respect to r, then the majority of players expect a small benefit, but a few players expect a small benefit.

Or if R = 10 and the wallet-sizes are 5,4,4,4,3,3,3,2,2,2, we can arbitrarily choose which 4 should lose to which 4, and get a linear order:

[r] 1 2 3 4 5 6 7 8 9 10 -- rank.

[w] 5 4 4 4 3 3 3 2 2 2 -- wealth or wallet.

[c] 0 5 9 13 17 20 23 26 28 30 -- what you can win; the sum of the wallets from people who will have to pay you.

If I am the fourth-highest wallet-holder then with chances 3/9 I will win 13/3. I.e., my expected winnings are 13/9. With chance 6/9 I will lose 4. I.e., my expected loss is w/(R-1)...

Tim Harford in his book _Undercover Economist_ mentions the following twist: We now action the prize w1+w2 to the two owners of the two wallets. What is their winning strategy in an auction (with English-auction rules)? Each player makes an estimate of the other player's wallet: e1 and e2 are the estimates of w1 and w2. Now player 1 expects the value to be w1+e2 and player 2 expects

the value to be e1+w2. If player 1 bids w1+e2 and player 2 drops out, then player 1 pays w1+e2 to earn w1+w2. He has paid e2 to win w2, which he perceives as a random variable with expected value e2. The information available to him in the action -- that player 2 continued bidding until w1+e2, can only improve his estimate of e2. Thus, each player should be happy to obtain the

other player's wallet at its expected value. Perhaps they should bid a little higher in order to take into account each other's information.

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