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Lets say you can bet on the outcome of a random event in which one outcome has a 50.5% chance of happening and the other outcome has a 49.5% chance of happening, but you would be able to get even odds on this bet from anyone because almost anyone would think this is clearly a 50:50 proposition.
If you could make this bet 300 times per hour you should on average make 3 bets per hour. What would an acceptable risk of ruin be for a bet like this? and Given a $1000 Bankroll for this best what would be the highest amount you could bet per outcome while maintaining an acceptable risk of ruin?
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Your risk of ruin ror for constant bets is given exactly by:
ror = (49.5/50.5)^B
where B is your bankroll in bets. Solving for B gives:
B = log(ror) / log(49.5/50.5)
You decide what risk of ruin you are willing to accept. For example, if you want a 1% risk of ruin, then B = log(0.01) / log(49.5/50.5) = 230 bets. So if you have $1000, you would make bets of $4.34.
For a derivation of these formulas, see the first part of the derivation in
this post. I also recently answered a similar question
here.