I saw this question here Compare two coin tossing games and wanted to figure it out myself, and I had some trouble with variance.
I can see easily that both games have expected value $\$500$. But my variance numbers make no sense to me.
$1000\text{Var}(X) = 1000(E[X^2] - E[X]^2) = 1000((\frac{1}{2}(\$1^2) + \frac{1}{2}(\$0^2)) - \$0.50^2) = \$250$
$\text{Var}(X) = E[X^2] - E[X]^2 = (\frac{1}{2}(\$1000^2) + \frac{1}{2}(\$0^2)) - \$500^2 = \$250000$
Is this right?
Yes, your calculation is correct. The second game has much higher variance than the first (although to be pedantic, the units of variance are not dollars, but dollars squared).
Intuitively, let's think about the distribution of possible outcomes of each game if both are played at the same expected value. In the first case, the random total payout $X$ can be expressed as a function of a binomial random variable, which in turn is approximately normally distributed with mean $\mu = 500$ and variance $\sigma^2 = 250$.
On the other hand, in the second game, the random total payout is a function of a Bernoulli random variable, with $$\Pr[X = 1000] = \Pr[X = 0] = 1/2.$$
If we superimpose these probability mass functions atop each other, you can see what this looks like:
The blue "curve" is a plot of the first game's probability distribution; the orange lines represent the second game's probability distribution. You can see that, for the first game, it is virtually assured (although not absolutely guaranteed) that your total payout will be somewhere between $450$ and $550$. Indeed, the probability that it will not be in this narrow range is only $0.00139174$, and the chance that it will be less than $400$ or more than $600$ is tiny: only about $1.80168 \times 10^{-10}$.
Yet, in the second game, the only outcomes are $0$ and $1000$, and there's a $50\%$ chance of either of these. You can't even get any other result! And because variance is a measure of central tendency--that is to say, the extent to which a random outcome will be observed far from its mean--we see why the second game has much higher variability than the first.