Re: Fair Game (Expected Value) Question

Robert Israel wrote:

In article <b9mdnTy1LoVDU_nZnZ2dnUVZ_v2dnZ2d@xxxxxxxxxxxx>,
Leonard M. Wapner <lwapner@xxxxxxxxxxxx> wrote:


The problem below appears on page 8 of David Gale's book, Tracking the
Automatic Ant. I give it below, essentially as it appears in the book. I
would appreciate any help in solving the problem posed. (I may have posted
this some years ago, but I can't recall the solution, if one was given.)

__________________________________________________________

In a certain casino one can play the following game. The house posts a
positive integer n. In this game it is you the customer who are invited to
toss a fair coin until it first falls tails. If you tossed n-1 times, then
you pay the house
8^(n-1) dollars, but if you tossed n+1 times, you win 8^n dollars from the
house. In all other cases the payoff is zero. Since the probability of
tossing exactly n times is 2^(-n), your expected winnings are 4^(n-1), for
n>1, and 2 for n=1. So your expected gain, which is the house's expected
loss, is positive. But now it turns out that the house arrived at the
number n by tossing its own fair coin and counting the number of tosses up
to and including the first tails. Thus, you and the house are behaving in a
completely symmetric manner. Each of you tosses a coin, and if the number
of tosses happens to be the consecutive integers n and n+1, then the
n-tosser pays the (n+1)- tosser 8^n dollars. But we have just seen that the
game is to your advantage as measured by expectation no matter what number
the house announces. How can there be asymmetry in a completely symmetric
game?



On the other hand, if you count m tosses, then given that result
(without knowing the house's n) you have a disadvantage.



My questions -

1) Is this a fair game?

2) If so, is the player's expectation zero?



In the symmetric version of the game, the payoff has no expected value
because its absolute value has expected value infinity. So fairness
for this game is undefined.

If X is the result of the house's coin-tossing, Y the result of the player's coin-tossing, and u(X,Y) the resulting payoff to the player,

sum_{x=1}^infty sum_{y=1}^infty u(x,y) Prob(X=x,Y=y)
is different from
sum_{y=1}^infty sum_{x=1}^infty u(x,y) Prob(X=x,Y=y)

That can happen for a double sum that does not converge absolutely.



Or, to express it slightly differently, let W be the net winnngs to the player. Since E[W^+] and E[W^-] (expectations of the positive and negative parts) are both infinite, EW is undefined. I would still say the game is fair by symmetry, but it is impossible to put a price on what you would be willing to pay to play this game without some additional features (e.g., imposing utility on the winnings).

This example points out that EX better be defined before you apply the law of conditional expectaion, viz.,
EX = EE[X | Y]. This also offers one explanation that in the formal definition of conditional expectation, one requires E|X| finite.

--
Stephen J. Herschkorn sjherschko@xxxxxxxxxxxx
Math Tutor on the Internet and in Central New Jersey and Manhattan

.

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