Why Doesn't X + X = 2X?


This blog post comes from AP Stats guru Daren Starnes. Daren has taught AP Statistics since the course launched in 1996-97. He has served as a reader, table leader, and question leader for the AP Statistics exam for 20 years. Daren has led over 200 workshops for statistics teachers, both online and in-person. He is also lead author of two popular statistics textbooks.


In Algebra, x + x = 2x. Does a similar result hold in Statistics? That is, if X is a random variable, does X + X = 2X? Let’s investigate with a simple game of roulette.


An American roulette wheel has 38 slots numbered 1 through 36, plus 0 and 00. Half of the slots from 1 to 36 are red; the other half are black. Both the 0 and 00 slots are green. Suppose that a player places a $1 bet on red. If the ball lands in a red slot, the player gets the original dollar back, plus an extra dollar for winning the bet. If the ball lands in a different-colored slot, the player loses the $1 bet. Let X = the player’s profit on a single $1 bet on red. Because there is an 18/38 chance that the ball lands in a red slot, the probability distribution of X is as shown in the table.

The random variable X has mean




Interpretation: The player is expected to lose about 5 cents, on average, when placing a $1 bet on red in a game of roulette. (And the casino expects to win about 5 cents per game from such a player.)


The standard deviation of X is






Interpretation: The player’s profit when placing a $1 bet on red in a game of roulette typically varies by about $1 from the expected loss of 5 cents.


Combining Random Variables: X + X

Suppose a gambler plays two games of roulette, each time placing a $1 bet on red. What can we say about the player’s total profit in two games, X + X?

  • If the player wins both games, the total profit is $1 + $1 = $2. The probability that happens is (18/38)(18/38) = 0.2244, since the outcomes of individual games are independent.

  • If the player loses both games, the total profit is −$1 + −$1 = −$2. The probability that happens is (20/38)(20/38) = 0.2770.

  • If the player wins one game and loses the other, the total profit is $1 + −$1 = $0. There are two ways this can happen—win then lose or lose then win—so the probability is (18/38)(20/38) + (20/38)(18/38) = 0.4986.

Here is the probability distribution of X + X.

The player’s expected profit in two games is



The standard deviation of X + X is

Interpretation: The player’s total profit when placing a $1 bet on red in two games of roulette typically varies by about $1.41 from the expected loss of 10.5 cents.


Note that the mean of X + X can also be found as follows: