Dusting off some high school stats and getting confused over the following:
Two betting games:
- Pick right side of coin, even-money bet ($p = 0.5$, $q= 0.5$),
- Pick right value in a 10-sided die, pays 9x original bet ($p = 0.1$, $q=0.9$),
where $p$ is probability of picking the right choice, $q$ is probably picking wrong.
After $n$ bets, the winnings follows a binomial distribution for both games. The variance I believe should $$npq,$$
which is $0.25n$ for game 1 and $0.09n$ for game 2.
Both games has expected winnings of 0, but intuitively I know game 2 has higher variance, so what's wrong with the variance calculation?
I am sure I missed something really obvious.
The number of games won is binomially distributed, not the winnings. For instance, the variance of the number of coin toss games won out of $n$ games is $$ n \times \frac{1}{2} \times \frac{1}{2} = 0.25 n. $$ Further, the variance of the number of dice games won out of $n$ games is $$ n \times \frac{1}{10} \times \frac{9}{10} = 0.09n. $$
But the payout is $2b$ for each coin toss game and $10b$ for each dice game, where $b$ dollars is your initial bet. Therefore, the variance in the payout for the coin toss game is $$ (2b)^2 \times 0.25 n = b^2 n, $$ and the variance in the payout for the dice game is $$ (10b)^2 \times 0.09 n = 9b^2 n. $$