Question

One player will wager $1 on one of two results from rolling two dice: outcome A (2, 3, 9, 10, 11, 12) or outcome B (4, 5, 6, 7, 8.) If the player predicts the correct outcome he or she will win $2 (the $1 wagered plus $1 from the house;) otherwise he or she will lose the $1 bet.

If the player chooses Outcome A an infinite number of times, how much will he or she win or lose on average, every hand?

Answer 1

When a player wagers on Outcome A an infinite number of times, based on the probability of the outcome and the expected value, they will lose an average of $0.0833 per hand.

Step-by-step explanation:

If a player chooses Outcome A, which includes rolling a 2, 3, 9, 10, 11, or 12 with two dice, they must calculate the probability of winning and the expected value to determine their average win or loss per hand when playing this game an infinite number of times.

First, let's find out the probability of Outcome A occurring:

• The sum of 2 has 1 possibility (1+1).
• The sum of 3 has 2 possibilities (1+2 and 2+1).
• The sum of 9 has 4 possibilities (3+6, 4+5, 5+4, 6+3).
• The sum of 10 has 3 possibilities (4+6, 5+5, 6+4).
• The sum of 11 has 2 possibilities (5+6 and 6+5).
• The sum of 12 has 1 possibility (6+6).

Total possibilities = 1 + 2 + 4 + 3 + 2 + 1 = 13
The probability of Outcome A = 13/36 (since there are 36 total outcomes when rolling two dice).

Next, we calculate the expected value (EV) for Outcome A. The player wins $2 for a correct prediction and loses $1 for an incorrect one.

EV = (Probability of winning) x (Winnings) - (Probability of losing) x (Losses)
EV = (13/36) x $2 - (23/36) x $1 = 26/36 - 23/36 = 3/36 or $0.0833

Therefore, for Outcome A, the player would experience an average loss of approximately $0.0833 every hand played.