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Imagine I am betting on the outcome of flipping $n$ biased coins $C_i$. I know exactly what the probability of each coin landing on heads is $P(H)$ and exactly the probability of each coin landing on tails is $P(T)$.

For every flip I can place one bet on either (but never both) outcome at odds with 5% value i.e. the probability implied by the odds is 5% lower than $P(H)$ or $P(T)$. Because I am getting value on each bet, I know that in each case my expected value is positive.

If I choose to bet on $n$ coin flips, what is the probability that I will make a net loss?

Cheers,

Pete

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    So you really mean that the returns are computed as if the odds of winning were lower?2011-01-01

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If P(H) varies by coin, it will be hard to do better than adding up all the cases. If P(H) is the same for all coins and P(T)=1-P(H) you can use the binomial theorem. First calculate how many coins need to come heads to make a profit. For example, if P(H) = 0.5 you should win 1/.45 -1= 11/9 for heads (is this what you mean for 5% value? I could argue for various other figures, but let's use this) and lose 1 for tails. To break even, then, you need to get tails 11/20 of the time and to lose you need more tails than that. So the probability of loss in $n$ throws is $\frac{1}{2^n}\sum_{k=0}^{\lceil \frac{9k}{20} \rceil -1} \binom{n}{k}$ If the coin is not fair, you will need to bring the P(H) into the sum.

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    If there are many coins, you can just add the expected values, variances, and use the Gaussian approximation2011-01-02