Standard Deviation of fixed-odds bet

Question

I'm looking to calculate the standard deviation of a fixed-odds betting proposition. The bet pays 5/6 for a win and you lose your stake for a loss. For a bet of 100, a winner pays 83.33 (and the stake is returned) and a losing bet pays -100. The probability of winning is evens (or very close to evens).

I'm uncertain about the formula to use for Standard Deviation in this case. Is it sqrt(npq) or 2* 1 [1 * sqrt(npq)]? In that case, the standard deviation of a single bet on the game above would be 100*sqrt[1 * 0.5*0.5]. Should it not be 2*100*sqrt[0.5*0.5] or something else, as a binomial distribution assumes a win of 1 and loss of 0 (or a win of 83.33 and a loss of 100 in this case), rather than a win of 1 and loss of -1 in this case.

I saw an example where a bet cost 20 and paid 100 if it was a winner, with the probability being 0.2. The standard deviation for that single bet was given as 100*sqrt[(0.2*0.8)]. I wanted to check whether this was right.

Thanks for any help on this.

Answer 1

The expected value of a single bet is

$$ \mu = wp - l(1-p) $$

and the standard deviation is

$$ \sigma = \sqrt{p(w-\mu)^2 + (1-p)(-l-\mu)^2} $$

where $w$ is the amount you stand to gain, $l$ is the amount is stand to lose, and $p$ is the probability of winning. If you play the game $n$ times all you have to do is multiply the above by $n$ to get the relevant numbers.

If $p = 0.5$, $w = 83.33$, and $l = 100$ then

$$ \mu = 41.67 - 50 = -8.33 $$

and

$$ \sigma = \sqrt{0.5(83.33 + 8.33)^2 + 0.5(-100+8.33)^2} = \sqrt{8402.47} = 91.665 $$

Answer 2

If the probability of winning is $p$, then the expected value $\mu$ is $\sum p_i x_i$, which is $\frac{5}{6} p - (1-p) = \frac{11}{6} p - 1$. The variance is $\sum p_i (x_i - \mu)^2$, which is: $p(\frac{5}{6} - \mu)^2 + (1-p)(-1 - \mu)^2$.

The variance works out to $-\frac{121}{36} (p - 1) p$, and the standard deviation works out to $\frac{11}{6} * \sqrt{ -(p-1)p }$. If $p = \frac{1}{2}$ the standard deviation is $\frac{11}{12}$.

Edit: If the amount of the bet is $b$ the standard deviation is multiplied by the amount of the bet: $\sigma = \frac{11}{6} b \sqrt{p - p^2}$.

Edit #2: Formatted the math. I can't comment on the other answer but I believe he forgot to square the deviations. For $b=100$ I get $\sigma = 91.67$.