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I have the following game: you flip a coin (heads with probability $p$), and if you get heads you earn $1000$ dollars, and you can decide if you want to flip again. If not, you keep the money. However, if you get tails, the game is over and you go home with nothing.

I am having trouble calculating an optimal strategy for this game: I would like to maximize the winning amount by stopping at turn $f(p)$, but... what is $f(p)$?

If you did not lose all the money when tails appears, the expected winning value would just be a standard geometric distribution $\sum_{i=0}^\infty i p^{i-1} (1-p) = \frac{1}{1-p}$, but I am having trouble evaluating the risk at each turn.

Clearly never stopping is a good strategy only when $p=1$, but intuitively when $p$ is very close to $1$ then stopping after the first win is not the best strategy. I am confused on how to model this phenomenon.

FifteenPointOne
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    If you have $N$ in the bank, the expected value of the next toss is $p\times 1000-(1-p)*N$. Stop when this is negative. – lulu Nov 01 '19 at 11:10
  • @lulu - seems to maximise winnings if you can play over-and-over again, but for a one-off game the chances of winning anything at all can be as little as 37% – Cato Nov 01 '19 at 12:15
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    @Cato absent a specified utility function, I don't see what else one can hope to say. Sure, if you attach a value to the time it takes you to realize the expectation, you can get a different result...but we don't have that sort of information. – lulu Nov 01 '19 at 12:22

2 Answers2

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You have to calculate the expected value of continuing. Suppose you have $x$ dollars in earnings, if you continue you can expect to earn $1000$ dollars at chance $p$ and lose $x$ dollars at chance $(1-p)$, so your expected earnings are $1000p-(1-p)x$. If this is value is positive carry on, if it is not quit. In this case this means you should stop if $x\geq 1000\frac{p}{1-p}$.

Floris Claassens
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If p is your probability of heads and x then number of turns you take, then the best number of turns to take will be $p^x=0.5$ , as less than half will mean tails is more common.

To solve for $x$ you take the natural log of both sides and move down the exponent, and then divided by $ln(p)$ to get this:

$x=ln(0.5)/ln(p)$

Now you just plug in p and get expected number of turns. This formula lines up with expected results such as;

p --> 1/2 then you should quit after 1 turn, which makes sense

p --> 0 approaches 0, which follows common sense

p --> 1 approaches infinity (meaning you play forever)

Jakub Skop
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