# What is the optimal strategy for this game?

You are playing a game where you put in a certain amount of money $m$. A random number in $[0, 1]$ is chosen. If the number is greater than $p$, you now have k% more money, otherwise, you lose all of your money. You can play the game as many times as you want, but you have finite and finite time. What strategy would optimize your return?

Explicit example:

Let's say the probability of winning is 0.6.

You have \$$10. You put \$$1$in and win 15%. So you have \$$1.15. You play again with that \$$1.15$, but this time you lose. \$$1.15 -> \$$0.00$. But you still have \$$9 left. You put \$$2$ in and win 15%, (\$$2.30). You keep it in and win again (\$$2.645$). And you play one more time and win again (\$$3.04165). You decide to withdraw from the pot, so your total is \$$12.04165$.

Obviously, if you put the whole \$$10 and play until you lose, your expected value is \$$0$. But the expected value of just one game is positive... • It is not clear that you have any strategy at all. Do you have any control over$m$?$p$?$k$? the meaning of random? Mar 2, 2013 at 5:53 • Assuming m, k, p given; and you want to maximise expectation? You can decide to play or not at each round? Mar 2, 2013 at 6:18 • @Nick Remember, that the $ means that you want something in LaTeX. Therefore, if you "$10 texttext$5", it will take that as you want everything within the \$$sign as LaTeX. Therefore, use this next time \$$number$ Mar 2, 2013 at 6:19 • Are you trying to describe the following game? _____$p$and$k$are fixed numbers. You start with$N$dollars. You will play$T$rounds of a game. Each round, you select any real number$0 \leq M \leq N$. With probability$p$, you win$kM$dollars. With probability$1-p$you lose$M$dollars. _____ Note you don't need any special caveats for "stop playing" and "lose all your money": both cases are handled by simply choosing$M=0$for all of the remaining rounds. – user14972 Mar 2, 2013 at 6:44 • @Hurkyl, I think at each round you select (as your bet size) any real number from zero to your current holdings (which generally won't be the$N\$ you started with). Mar 2, 2013 at 12:18

Technically speaking, there's only one option, which is to keep stepping forward. Mathematical games, as far as I understand it, don't include not playing as an option. There's no way to optimize or formulate any strategy, because the other player in the game has perfect play as their only move, and always wins with perfect play.

So I guess the optimal strategy is to keep playing until you lose all your money, since it's the only strategy.

I would look at the definition of a game at this Wikipedia article

See the Kelly Criterion. It appears that your game is essentially what Hurkyl described (now user14972):

_____ $$p$$ and $$k$$ are fixed numbers. You start with $$N_0$$ dollars. You will play $$T$$ rounds of a game. Each round, you select any real number $$0≤M≤N_n$$. With probability $$p$$, you win $$kM$$ dollars, so that $$N_{n+1} = N_n + kM$$ with probability $$p$$. With probability $$1−p$$ you lose $$M$$ dollars, so that $$N_{n+1} = N_n - M$$ with probability $$1-p$$. _____

In the case that $$T$$ approaches infinity, the criteria that maximizes $$N_n$$ is the Kelly Criterion. Here is an example:

$$f^* = \frac p a - \frac q b$$

$$f^* = \frac p {kM} - \frac {1-p} M$$

The Kelly Criterion gives the fraction of your total assets that is optimal to apply to the investment. If $$f^*$$ is negative, it isn't optimal to invest at all.

So for cases where M will optimally be positive:

$$\frac p {kM} - \frac {1-p} M > 0$$