Hypothetically speaking, assume there was a side bet that paid 25-1 and the probability of winning was 5% under the given conditions (lose other 95%, no ties). What would be the optimal bet amount (% of bankroll)?
The limited background I have would says the optimal bet is your advantage times your bankroll. But this doesn't seem correct in this case, as the standard deviation is really large.
Thanks in advance,
el_jefe
el_jefe,
Your question cannot be answered because you have not defined what you mean by "optimal".
Usually (but not always), AP's use the term "optimal" to mean "the bet that maximizes the log normal growth of the bankroll"; if this is what you mean, then you can find the answer to your question in the concept of "Kelly Betting" (see link below). On the other hand, if by "optimal" you mean something else, then your answer will be different.
Dog Hand
> Hypothetically speaking, assume there was a
> side bet that paid 25-1 and the probability
> of winning was 5% under the given conditions
> (lose other 95%, no ties). What would be the
> optimal bet amount (% of bankroll?
> The limited background I have would says the
> optimal bet is your advantage times your
> bankroll. But this doesn't seem correct in
> this case, as the standard deviation is
> really large.
As per Dog Hand's comments, I'll assume that, by "optimal," you mean "Kelly optimal," which is a fair assumption. In that case, you're right: You can't just multiply your bank by your edge, because the 25-1 payoff comes into play. Instead, you multiply your bank by your edge, divided by the ratio of a winning payoff (25) to a losing one (1), which is to say that you divide your edge by 25 in this instance.
As you might know, your edge for the example you give is 30%. On average, you win once every 20 bets (5% of the time). But, when you win, you find yourself with 26 chips in front of you, for a net gain of 6/20, or 30%. Optimal Kelly bet would thus be (30/25)% * bank, or 1.2% of your bank.
Don
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