one of the wagers in roulette is to bet that the ball will stop on a number that is a multiple of 3. (both 0 and 00 are not included) if the ball stops on such a number the player wins double the amount bet. if a player bets $1 compute the player’s expectation

Answer 1

To compute the player's expectation, we need to calculate the expected value of the outcome.

First, let's determine the probability of the ball stopping on a number that is a multiple of 3. On a roulette wheel, there are 38 numbers (1-36, 0, and 00). Out of these numbers, 12 are multiples of 3 (3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36). Therefore, the probability of the ball stopping on a multiple of 3 is 12/38.

If the player bets $1 and wins, they will receive double the amount bet, which is $2. If the player loses, they will lose the amount bet, which is $1.

Now, let's calculate the overall expectation:

(Probability of winning * Amount won) + (Probability of losing * Amount lost)
= (12/38 * $2) + (26/38 * -$1)
= ($24/38) - ($26/38)
= -$2/38
≈ -$0.0526

The player's expectation is approximately -$0.0526. This means, on average, the player can expect to lose about $0.0526 for every $1 bet on this wager.