Question

Finding an Expected Value In Exercises 37 and 38, find the expected value E(x) to the player for one play of the game. If x is the gain to a player in a game of chance, then E(x) is usually negative. This value gives the average amount per game the player can expect to lose.

In American roulette, the wheel has the 38 numbers, 00, 0, 1, 2, . . ., 34, 35, and 36, marked on equally spaced slots. If a player bets \$1 on a number and wins, then the player keeps the dollar and receives an additional \$35. Otherwise, the dollar is lost.

Accepted Answer

Step 1: Understand the problem. The expected value E(x) represents the average gain or loss per game for the player. To calculate E(x), we use the formula E(x) = Σ [x * P(x)], where x is the outcome (gain or loss) and P(x) is the probability of that outcome. Step 2: Identify the possible outcomes. In this game, there are two outcomes: (1) The player wins, gaining $$35 in addition to keeping their 1 $$bet, for a total gain of $$36. (2) The player loses, losing their 1 $$bet, for a total loss of $1. Step 3: Determine the probabilities of each outcome. There are 38 equally likely slots on the roulette wheel. The probability of winning (P(win)) is 1/38, as there is only one winning slot. The probability of losing (P(lose)) is 37/38, as there are 37 losing slots. Step 4: Calculate the expected value using the formula E(x) = Σ [x * P(x)]. Substitute the values: E(x) = (36 * P(win)) + (-1 * P(lose)). Replace P(win) with 1/38 and P(lose) with 37/38. Step 5: Simplify the expression. Combine the terms to find the expected value. This will give you the average gain or loss per game for the player. Note that the result is typically negative, indicating an expected loss for the player.

Verified video answer for a similar problem:

Key Concepts

Expected Value

Probability

Game of Chance