Fair unfavorable games

Fair gambling games should be… fair, right?!

Félix Locas, here at UQAM, introduced me to “fair but unfavorable” gambling games of the following type:

  • you pay \mu to draw a random positive gain X such that \mathbb{E}X = \mu;
  • you can play for as long as you want.

Precisely, we have a sequence X_1, X_2, X_3, \dots of independent and identically distributed random variables representing your gain (minus \mu) at the kth game. Thus X_k \in [-\mu, \infty) and \mathbb{E}X_k = 0.

The surprising fact is that there exists such “fair” gambling games for which the total gain S_n = \sum_{i=1}^n X_i tends to -\infty in probability. More precisely (and even worse), there are games for which

P\left( S_n < -\frac{(1-\varepsilon)n}{(\log n)^{a} }\right) \rightarrow 1

for all small a > 0. (See W. Feller, Note on the law of large number and “fair” games, 1945.)

Playing wisely

The game, as I just described it, can still be favorable to you. In fact, playing it for long enough, you can be certain to make as much money as you want. You just have to know when to stop: for all T > 0, there almost surely exists an indice k such that S_k > T.

But here’s the catch: You can’t (generally) play it for long enough. You could be ruined before you make up for previous losses.

Let’s therefore adjust the model to make it more realistic. The player has only a finite amount M of money. She can play for as long as she is not broke, and can stop at any time.

Winning strategy

Let T be a winning goal for the player. Consider the ruin time r_0 = \inf \{k \,|\, S_k < -M\} and the stopping time \tau = \inf \{k \,|\, S_k > T\}. If the player keeps playing until she is broke or until S_k \geq T, then her expected gains G are

G  = \mathbb{E}[S_\tau] P(\tau < r_0) - M P(r_0 < \tau).

(Note that both \tau and r_0 are almost surely defined unless X_k is constant.)

Proposition.
Suppose that \tau and X_k are of finite variance. Then for all winning goal T > 0, there exists a starting amount M > 0 such that the expected gain G is positive.

Sketch of proof.
Note that G \geq T -(T+M) P(r_0 < \tau) and that

P(r_0 < \tau) = \sum_{j=0}^\infty P(r_0 = j) P(\tau > j) \le \sum_{j=0}^\infty \frac{j \text{Var}(X_1)}{M^2} P(\tau > j)

is of order M^{-2}. Letting M \rightarrow \infty, we find that \lim \inf G \geq T > 0. \Box

Unfavorable games?

Problem.
Are there games such that G < 0 for most amounts M?

Put another way: are there games for which no strategy yield a positive expected gain?

Problems with unbiaised consistent estimators

In close relationship with the total gains S_n = \sum_{i=1}^n X_i is the Monte Carlo estimator \bar{X}_n = \frac{1}{n}S_n. Here, \bar{X}_n is an unbiaised and consistent estimate of \mathbb{E} X_k.

Although the strange behavior of Feller’s game (that S_n \rightarrow -\infty quite fast in probability) is not as striking in this case – consistency obliges – problems may still arise in pratical situations.

Proposition
If X_k is of infinite variance, then for all M > 0

P\left(|\bar{X}_{n} - \bar{X}_{n-1}| > \frac{M}{\sqrt{n}} \quad \text{infinitely often}\right) = 1.

Conversly, if X_k is of finite variance, then for all M > 0 and small \varepsilon > 0

P\left(|\bar{X}_{n} - \bar{X}_{n-1}| > \frac{M n^{\varepsilon}}{\sqrt{n}}\quad \text{infinitely often}\right) = 0.

Proof. See the next post.

“Biased” unbiasedness

In practice, the convergence of the estimator \bar{X}_k is often assessed using an empirical criterion. The estimate \bar{X}_N is taken to be sufficiently good when the criterion is first satistied. For instance, we may stop at the first indice \tau such that the estimated variance of \bar{X}_k is less than some fixed \varepsilon > 0 (this is sometimes eyeballed). The expected estimate of \mathbb{E}[X_k] is then

\mathbb{E} \bar{X}_\tau.

Problem/Conjecture.
For all M > 0, there exists a game such that

\mathbb{E}\bar{X}_\tau < -M.

How bad can things get when the variance of X_k is finite? What is the probability that a significant error is made?

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