The "Balls in the Hat" Game
Ranjan Bhaduri writes:
allaboutalpha.com: Welcome to AllAboutAlpha.com: By: Ranjan Bhaduri, special to AllAboutAlpha.com: The word “liquidity” gets bandied about quite a lot, but it is surprising how many portfolio managers take a naïve approach to liquidity. It is well known that one should be compensated for investing in less liquid instruments (liquidity premium), but how much? What is the value of liquidity? It is dangerous in merely trust one’s intuition on the value of liquidity. Consider the following one-person game:
The “Balls in the Hat Game”
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 play! The ability to stop any time is analogous to perfect liquidity (i.e. being able to pull out of an investment at any time without the action having an impact on the value of the investment). This value of liquidity helps overcome the imbalance between the black and white balls, and thus makes this game profitable. This is interesting from a behavioral finance point of view, since it seems to suggest that humans are wired such that they will tend to underestimate the value of liquidity.
The mathematics behind calculating the value of liquidity can be complex as there can be subtle nuances. Niall Whelan of Scotia Capital and I wrote a pair of papers coming out which tackles the above game in the asymptotic case (i.e. hats of infinite size) and connect the value of liquidity to option pricing. Niall is one of the best quants north of the South Pole and much of these papers were hammered out in an all-night bus ride that we were forced to take from NY to Toronto (our flight from La Guardia got cancelled but we both needed to be back in Toronto in the morning for important meetings)...
I suspect that this is wrong. What is going on here, I think, is not so much liquidity as mean reversion: if you sample with replacement the effect goes away. The reason that it makes sense to play even though one might at first glance think the odds are unfavorable is that if you lose in the early stages the chances of winning in the later stages go up--that the ability to keep playing provides a degree of insurance in the cases in which things break badly. (Of course, "liquidity" does play a role: if you could never stop playing that would be offset by the fact that if you won in the early stages the odds would then move away from you.)
In order to see what is going on here, let us write down the value function: V(m,n) is the expected value of playing the game (and dropping out at the optimal point) when there are m white balls and n black balls in the hat. To begin with:
V(1,0) = 1; if there is one white ball left in the hat, the value of the game is 1--you play, and collect the white ball.
V(0,1) = 0; if there is one black ball left in the hat, the value of the game is 0--you don't play at all.
What is the value of V(1,1)? If you do play, then half the time you will draw a black ball--and be down 1--but then you will be playing the game V(1,0), which is worth one, so if you draw a black ball next, you are even. And if you do play, then half the time you will draw a white ball--and be up one--and then you will be playing the game V(0,1), and so you stop and that is worth zero. The value of V(1,1) is therefore:
V(1,1) = (1/2)(-1+V(1,0)) + (1/2)(1 + V(0,1)) = (1/2)0 + 1/2(1) = 1/2
This is interesting: you might initially think that this is a fair game--there is, after all, one white and one black ball, so you have a 50-50 chance of being up after the first draw. But it is rigged in your favor.
V(m,n) = max[0, (m/(m+n))(1+V(m-1,n)) + (n/(m+n))*(-1+V(m,n-1))]