A certain slot machine is rigged to pay out, on average, once every 10 games. It costs 1 dollar to play, and the machine pays out 11 dollars if you win. On average, then, you will be up one dollar every ten games. On average, mind you.

However, there is a catch. The slot machine is an up-and-coming machine and takes a credit card. If you win, it locks out your card. So you can only play once.

Now, a certain person has particularly bad credit and bad health. They have missed some payments on their debt and have multiple unpayed hospital bills. They have only one credit card. Suppose they've played the slot machine four times already. If they win they cannot play again - the bank will not issue them another card. They are four dollars down. Is it profitable to keep playing until they win?

Edit: Suppose they lose their fourth and fifth game. What then?

• Welcome to Math.SE! We usually expect questions to include some important context like what the asker has tried already and which part of the problem they're having difficulty with. You'll probably get higher-quality answers if you edit your question to include these things. – dbx Oct 11 '18 at 19:20
• I muffed the math of it. (oops) I edited the question, hopefully it is more clear now. – Jonathan Graef Oct 11 '18 at 21:52

In the question "should they keep playing", two terms are ambiguous: "should", and "keep playing".

Let's say "keep playing" means "play exactly one more game". Then they stand a one in ten chance of ending up with $$9$$ dollars and a nine in ten chance of ending up with $$-4$$ dollars. Is that better or worse than walking away now with $$-3$$ dollars? It depends on the meaning of "worse". I would probably take the gamble. The expected value of the gambler's profit if they play is $$9\frac1{10}+(-4)\frac9{10}=-2.7$$, which is negative, which some might consider to mean that you "shouldn't" play.

Let's say "keep playing" means "keep playing until you win". Then the gambler's profit will be $$-3-(N-1)+12=10-N$$, where $$N$$ is the number of times the gambler plays before winning. The distribution of $$N$$ is geometric with parameter $$\frac1{10}$$, so the expected value is $$0$$ (better than the $$-3$$ you get by walking away now). The probability of ending up with $$-3$$ dollars or more is $$P(N\leq13)=1-(\frac9{10})^{13}>0.74$$, so maybe it's worth it.

• Okay. So even if you're down, you stand a good chance of gaining more than you lose if you keep playing. Makes sense. I have a new question now, I'll ask it. – Jonathan Graef Oct 11 '18 at 22:04