Suppose you play a game in which you are dealt one card from a standard deck of 52 cards; you win $10 if the card is an ace, and lose $1 otherwise. What is the approximate variance of this game?

Answer 1

We have this probability distribution function, where x represents the amount of money won or lost

x: +10 -1

p(x) 4/52 48/52 because there are 4 aces and 48 other cards.

E(X) = +10 * (4/52) + (-1) * (48/52) = -8/52 This is the expectation (which is the same as the mean µ)

To find the variance, use either of these formulas:

Var (X) = E(X²) - (E(X))² or

Var (X) = E( (X - E(X))² )

Using the first formula, we calculate the expectation of X²

x: +10 -1

p(x) 4/52 48/52

x² 100 1

So E(X²) = 100 * 4/52 + 1 * (48/52) = 448/52

and thus Var (X) = E(X²) - (E(X))²

Var (X) = 448/52 - (-8/52)² = 448/52 - 64/2704 = 23232/2704 = 8.592

So, in summary, Var (X) = 8.592 to 3 decimal places