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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? –  Raskolnikov Jan 1 '11 at 18:55

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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 $P(H)$ does vary, is there any approach which could work? –  Peter Jan 1 '11 at 21:36
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What makes it hard is that the amount you win and lose is different for each coin, so there isn't a simple way to decide which combinations of wins are profitable. If the winnings are quantized, say you always win some multiple of .1 or lose 1, you can just go through the coins in order and after each toss keep track of the probability of each level of profit. The secret to avoid doing all the 2^n possibilities is to find some regularity so you can ignore a class of detail. In my example above, you only care how many tosses you win, not which ones. –  Ross Millikan Jan 1 '11 at 23:46
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If there are many coins, you can just add the expected values, variances, and use the Gaussian approximation –  Ross Millikan Jan 2 '11 at 2:14

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