A 1-dollar bet on craps has an expected winning of −.0141. What does the Law of Large Numbers say about your winnings if you make a large number of 1-dollar bets at the craps table? Does it assure you that your losses will be small? Does it assure you that if n is very large you will lose?
Law of Large Numbers
Let \(X_1, X_2, …, X_n\) be an independent trials process, with finite expected value μ = \(E(X_j)\) and finite variance \(σ^2\) = \(V(X_j)\).
Let \(S_n=X_1+X_2+⋯+X_n\). Then for any \(ϵ > 0\),
\(P(|\frac{S_n}{n} - \mu| \ge ϵ)\to 0\)
as \(n\to \infty\), Equivalently,
\(P(|\frac{S_n}{n} - \mu| \ge ϵ)< 1\)
Since \(X_1, X_2, …, X_n\) are independent and have the same distributions, we obtain
\(V(S_n)=nσ^2 ,\)
and
\(V(\frac{S_n}{n}) = \frac{σ^2}{n}\)
Also we know that,
\(E(\frac{S_n}{n}) = \mu\)
By Chebyshev’s Inequality, for any ϵ > 0,
\(P(|\frac{S_n}{n} - \mu| \ge ϵ)\le \frac{σ^2}{nϵ^2}\)
Thus, for fixed ϵ,
\(P(|\frac{S_n}{n} - \mu| \ge ϵ)\to 0\)
as \(n\to \infty\), or equivalently,
\(P(|\frac{S_n}{n} - \mu| < ϵ)\to 1\)
as \(n\to \infty\).
Therefore, The Law of Large Numbers gives the convergence result for mean of sample to theoretical mean of the population.
What does the Law of Large Numbers say about your winnings if you make a large number of $1 bets at the craps table?
So, if you have expected winning of -0.0141, on an average in large number of bets, you are expected to loose -0.0141.
Does it assure you that your losses will be small?
Since it converges to the mean, this does not in any way assures that the losses will be small since the random variable has variance which can take the value to large loss or large win as well!
Does it assure you that if n (the number of trials) is very large you will lose?
It does since law of large numbers converges with probability one (1) which implies that you will loose with probability 1 in large number of bets.