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I have a very simple probabilistic process which I have to deal with in the software project I'm involved in, yet I can't figure how to do it. I can describe the situation as a money game (in reality it is not, but this is the most simple way to go):

At each moment $t$ my program gets a probability value $P_t$ on input. I have to output the stake $X_t$ ($X_t$ can be, say, from 0 to 10) and then with probability $P_t$ I will win (get $XA$, where $A > 1$) and with probability ($1 - P_t$) I will lose (get nothing, lose my stake $X$).

What strategy can be considered reasonable here? Make stake proportional to $P$? What should I read to understand the topic?

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What decides the gain? Is it proportional to the probability, is it equal to Xt, or is it something else? – SiliconCelery Aug 13 '12 at 19:28
@SiliconCelery: The gain is proportional to X, I've updated the question, defined return as X*A, where A > 1. – izhak Aug 13 '12 at 19:31
What is known about the behavior of $P_t$? How long are you permitted to play for? Can you "bust", i.e., lose your entire bankroll and be forced to stop playing? Is the goal to maximize the expected return, or the median return, or what? – mjqxxxx Aug 13 '12 at 19:35
up vote 3 down vote accepted

It rather depends on your utility function for wealth.

To maximise your expected return, bet everything when the odds benefit you, i.e. when $AP_t \gt 1$. This also gives you a good chance of losing all your money.

For a safer strategy which adjusts your bet size to the extent the odds are in your favour and avoids losing everything, read about the Kelly criterion, for example at Wikipedia. This is largely based on logarithmic utility.

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@Ilmari Karonen: Thank you for the edit to reflect the changed question – Henry Aug 13 '12 at 22:12

Updated: Your expected gain is $X_t(P_tA- 1).$ If this is greater than zero, you are winning and should bet all you can. If less than zero, you are losing and bet the least you can. This presumes that the desirability of money is linear and you are far from the ends (losing so you can't play any more or winning so much your opponent quits).

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yes, the gain is proportional to X, I've updated the question, defined return as X*A, where A > 1 – izhak Aug 13 '12 at 19:30
This holds if $A$ is the amount won in addition to the original stake (in which case your condition $(P_tA+P_t-1)X_t>0$ simplifies to $(A+1)P_t>1$, assuming that $X_t>0$). If $A$ includes the original stake, then the correct condition for betting is simply $AP_t>1$. – Ilmari Karonen Aug 13 '12 at 21:46
@IlmariKaronen: you are right. The condition changed with OP's update. I will fix. – Ross Millikan Aug 13 '12 at 21:49

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