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Old 08-07-2002, 05:15 PM
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Default Arrash Zafari question revisited



Earlier, Arrash Zafari asked:


If someone is playing a game with an overall payout of 100.2% and a 16.7% chance of each bet winning, what bankroll would be needed in terms of units to have a 95% chance of not going bankrupt?


I computed a bank of 3758 bets using a formula, assuming a winning bet always paid off the same amount (which I'm sure isn't how the payoff works but it's an approximation). Let's test this answer using the method of Sklanky's latest problem:


Probability of winning is 1/6. Start with a bank of 1 bet, and when you win you increase your bank to 6.012 bets since .012 = .002*6. p is the probability of going bust with a 1 bet bank:


p = 5/6 + (1/6)p^6.012


Tough equation to solve, but let's check my answer. If I was right then p^3758 = .05 so

p = .999203. This does solve the above equation, so 3758 is correct. Pretty good eh?
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