Thanks for sharing this with us.

It seems as though the balls in the hat example could be related to contrarian trading too.

Whenever you remove a ball, you can think of this as a time step so you get a stochastic process with some mean reversion built in. E.g., when you take out a black ball at t, the probability of a white ball goes up at t+1, etc.

This is similar to the situation where there is a fixed no. of potential buyers and sellers in a security, of fixed size. Whenever someone buys, they drop out so that the no. of sellers is proportionately larger. In theory this may lead to a mean reverting process.

Thanks again,

Hari

One minor quibble: In the essay you write “Hedge funds that have the opposite type of liquidity mismatch (trade liquid instruments but have onerous redemption terms) often claim that they are trying to avoid “hot money” of fund of funds. However hedge fund managers always retain the right to refuse a potential investor, so that does not really wash.”

In practice, there is significant information ineffeciency, and while many investors often profess to not be “hot money” the manager never really knows for sure until the investor redeems. A better solution to this problem (rather than lockup) is some sort of redemption fee for redemptions on shorter notice.

There are also some special cases worth considering such as newly launched funds, where lockup is required to secure the stability of the business (although often in these cases the early investors will appropriately trade longer lockup for lower fees).

Thanks for your note. I’m not familiar with Shepp’s work, but I’ll be sure to check out the reference that you gave. Just from the title it looks interesting. Thanks!

Best Regards,
Ranjan

I wonder if you know of the work of Lawrence Shepp (Rutgers University) on this problem. See page 999 of:

Explicit Solutions to Some Problems of Optimal Stopping

The Annals of Mathematical Statistics, Vol. 40, No. 3. (Jun., 1969), pp. 993-1010

Roger Purves

Thanks for your comment. It’s easy to make an error in the stats here (probability calculations can often be delicate). I did not give the calculation in the piece that I did, so your question is a good one. Below I give some more details:

The best way to calculate the expected winnings from a given hat is to realize that it can be done so in a recursive nature. Consider, a hat with m black balls and n white balls. Suppose you pick a white ball, then you are left with m black balls and (n-1) white balls, and the decision of whether or not to play is determined by the expected winnings from this “new, smaller” hat. Indeed,

E (winnings from m black balls, n white balls) =
max { m/(m+n) [-1 + E(n,m-1)] + n/(m+n) [1 + E(n-1,m)] , 0 }

The above formula shows that we should calculate the expected winnings from the “small” hats and build our way up. This recursion lends itself well to a computer language, and can be done on excel in minutes. I will email you the excel sheet separately later.

The full details of the solution of this game, including exploring other strategies, will appear in the January 2008 issue of Wilmott Magazine.

Cheers,
Ranjan

Thus the expected value of the game (1,1) is 1/2 even though you might think initially that the game is fair.

This generalizes, and this is what gets you the positive value for the game (4,6)

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The game consists of a hat that contains 6 black balls and 4 white balls. The player picks balls from the hat and gains $1 for each white ball, and loses $1 for each black ball. The selection is done without replacement. At the end of each pick, the player may choose to stop or continue. The player has the right to refuse to play (i.e. not pick any balls at all). Given these rules, and a hat containing 6 black balls and 4 white balls, would you play? (Why?)

Mathematically one can prove that there is a POSITIVE expected value (of 1/15) in playing this game, so one SHOULD