We play a game. The pot starts at $\$0$. On every turn, you flip a fair coin. If you flip heads, I add $\$100$ to the pot. If you flip tails, I take all of the money out of the pot, and you are assessed a "strike". You can stop the game before any flip and collect the contents out of the pot, but if you get $3$ strikes, the game is over and you win nothing. What is the expected value of your winnings if you follow an optimal strategy?
EDIT: Okay, here's what I've done. I think we want to maximize $$100n\left(1 – \left(\sum_{i = 1}^n \left({1\over2}\right)^i\right)^3\right) = 100n\left(1 – \left(1 – \left({1\over2}\right)^n\right)^3\right) = 100n\left({1\over2}\right)^n\left(\left({1\over2}\right)^{2n} – 3\left({1\over2}\right)^n + 3\right),$$where $n$ is the number of flips we commit to doing no matter what i.e. either we win $100n$ and then pull out, or we lose. Let me explain how I got the expression. If I flip $n$ times and get $n$ heads, I get $100n$ dollars. I only lose if I don't flip $n$ consecutive heads in $3$ attempts. With each attempt, the probability of not flipping $n$ consecutive heads is $\sum_{i = 1}^n \left({1\over2}\right)^i$, so the probability of not flipping $n$ consecutive heads across all $3$ attempts is $(\sum_{i = 1}^n \left({1\over2}\right)^i)^3$, and so the probability of winning $100n$ dollars is:$$1 – \left(\sum_{i = 1}^n \left({1\over2}\right)^i\right)^3 = 1 – \left(1 – \left({1\over2}\right)^n\right)^3 = \left({1\over2}\right)^n\left(\left({1\over2}\right)^{2n} – 3\left({1\over2}\right)^n + 3\right)$$Hence the expected amount we win is:$$100n\left(1 – \left(\sum_{i = 1}^n \left({1\over2}\right)^i\right)^3\right) = 100n\left(1 – \left(1 – \left({1\over2}\right)^n\right)^3\right) = 100n\left({1\over2}\right)^n\left(\left({1\over2}\right)^{2n} – 3\left({1\over2}\right)^n + 3\right)$$Now I have two questions:
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Without using WolframAlpha, what's an easy way to maximize the expression for $n = 1, 2, 3, \ldots$?
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I made the assumption that we follow the same strategy across all three attempts, but didn't justify it. How can one justify that? Or is that wrong?
Best Answer
You should certainly not follow the same strategy on all the attempts. Your strategy for each attempt should be of the form "I will flip $n$ times and quit if they are all successes". The $n$ will change depending on how many strikes you have left. Mike Earnest's comment describes how you should think about finding $n$. When you are down to the last strike, if you have $k$ in the pot, you can either take $k$ or have $\frac 12$ chance of $k+100$ and $\frac 12$ chance of $0$. Then find the expected value of playing the last strike. Now when you have two strikes, instead of $0$ you get the expected value of playing your last strike. That may increase the stopping value because failing is not as bad. Find the threshold and the expected value of playing the two strikes game. Finally, one more time around for the three strikes game.