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I am trying to answer the question:

Given a game of chance with a "house edge" E (defined below), what is the maximum "amount ahead" A (defined below) when I should exit the game, if I want to achieve probability P that I will not run out of money if I bring bankroll B (defined below)?

E is defined as the average amount lost per bet, as a proportion of the amount of the bet. (e.g. 2%)

B is defined the amount you bring to the game (the maximum you are willing to lose, e.g. $500)

A is the amount you are ahead of your starting amount, e.g. A=$50 means that you have won $50, e.g., you started with $500 and now have $550.

P is the probability of not losing all my money (B) before getting ahead by A that I am willing to tolerate (e.g. 90%)

I am trying to define function f, such that A = f(P,B,E)

For example, in this statement:

With a house edge of 2%, and a bankroll of $500, you should leave when you are $50 ahead, if you want a 90% probability of not running out of money

the variables would be:

E=0.02, B=500, A=50, P=0.9 for A=f(P,B,E)

NOTE: I'm not sure if this question belongs on this site or on mathematics.

JoelFan
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  • @BruceET, maximum expected gain was also not the question. – JoelFan Oct 17 '21 at 20:59
  • Again not an answer. I never asked about overall success, I asked for areas to study related to my statistics question. I'm pretty sure you never even read my question. You saw it was gambling related and it became a moral mission. You also seem to have zero knowledge of gambling or else you would know that being ahead by some amount, at some point, is extremely common ("if that ever happens") – JoelFan Oct 18 '21 at 02:44
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    "Gambler's Ruin" is a topic you should google. Not a moral issue. Done here. – BruceET Oct 18 '21 at 03:44
  • "Gamblers Ruin" refers to increasing bets while ahead, which, yet again, is not related to the question. The question does not mention any change in betting amount, assuming it is equal on every bet – JoelFan Oct 18 '21 at 18:20
  • Broad topic. You must be looking at a special case. Bets can be same size. BTW. What guarantees you'll ever be ahead? Good luck. – BruceET Oct 18 '21 at 19:30
  • As I said, if you had any experience with gambling, you'd know that being ahead (by some amount, at some point) is extremely common. (Even without gambling experience, just a general familiarity with random number distribution should tell you that) Also, on a common sense level, do you think habitual gambling would have any appeal if the gambler was hardly ever ahead? – JoelFan Oct 18 '21 at 22:37
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    I am an experienced gambler. I once won $18.75 on a quarter slot machine in Carson City. – BruceET Oct 18 '21 at 22:45
  • @BruceET, from everything I've read about "Gambler's Ruin", it seems like my question is explicitly suited to minimize that effect as much as possible, since it seeks an optimal exit point, as well as including a defined bankroll in the assumptions – JoelFan Oct 20 '21 at 20:48
  • Well then, one of us misunderstands either your question or how to answer it. Further discussion seems futile. – BruceET Oct 21 '21 at 01:49
  • FYI, I think the answer I was looking for was "standard deviation". This page helped: https://wizardofodds.com/gambling/house-edge/ – JoelFan Dec 16 '21 at 07:59

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