We are playing a game at a casino. We pay \$50,000 to press a button. $p\%$ of the time, we win \$150,000 (for a net gain of \$100,000). The other $(1-p)\%$ of the time, the house keeps our money. We want to know what the probability of winning should be such that if we play ten rounds of the game, we can be 95% confident that we will walk away from the table with a profit.

I think that we should try using Chevyshev's inequality, but I'm not quite sure how to do this.

It costs $\$50,000\cdot10 = \$500,000$ to play ten rounds. Of those ten rounds, we must win at least $6$ to make a profit.

Let $X_i = 1\{\text{The ith game pays out}\}$. Let $A_n = \frac{X_1 + X_2 + ... + X_{10}}{10}$.

We want to estimate $\Pr[|A_n - p| \geq 0.05] \leq \frac{Var[A_n]}{0.05^2} = 400\cdot Var[A_n]$.

Each indicator, $X_i$, is i.i.d., so the variance can be distributed linearly. This yields $Var[A_n] = \frac{1}{10}\cdot p(1-p) = \frac{p(1-p)}{10}$.

Substituting the variance of $A_n$, we now have a bound of $\Pr[|A_n - p| \geq 0.05] \leq 40p(1-p)$.

But this gives puts $p$ at something like ~99.8%. This doesn't make sense, since I think that $p$ should be somewhere around 70%. It also seems incorrect that I haven't specified that we need 6 wins anywhere in the inequality -- but I don't see where I can do that, since that is what we want the expectation of $A_n$ to be. Any guidance would be much appreciated.

  • $\begingroup$ Small point: Do you think this problem depends upon the base value of a bet, $50,000? $\endgroup$ – David G. Stork Apr 12 '17 at 22:58
  • $\begingroup$ @DavidG.Stork I that the base value of a bet matters only in the context of determining the number of wins required to break even. Am I mistaken? $\endgroup$ – greedIsGoodAha Apr 12 '17 at 23:20
  • $\begingroup$ Suppose you called $50,000 one greedBuck. What would change in your presentation/calculation? $\endgroup$ – David G. Stork Apr 12 '17 at 23:26
  • $\begingroup$ Your error is that you believe $A_n = {X_1 + X_2 + \cdots + X_{10} \over 10}$. Recall that if you win you get three times your wager. $\endgroup$ – David G. Stork Apr 13 '17 at 0:58
  • $\begingroup$ I'm not quite sure if that makes sense to me. I thought about the greedBuck case: 1GB to buy in, with a return of 0GB on loss and 3GB on success (net gain of 2GB). It costs 10GB to play all of the rounds in our experiment. We still need to win 6 times, and since we can calculate the number of successes needed to describe the whole undertaking a success in advance, I can't quite see why it's a mistake to use $X_i$ as an indicator, exclusively looking at the probability that would be required to give a 95% CI for at least 6 successes in 10. Thank you for your reply and follow-up! $\endgroup$ – greedIsGoodAha Apr 13 '17 at 1:56

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