In the Doubles Game, students can win carnival tickets to spend on games and food. A player pays one ticket to roll two number cubes(six-sided dice). If the numbers match, the player wins five tickets. If a player plays this game 20 times, about how many tickets can he or she expect to win or lose?
P(match) = 1/6
so,
E = 20/6 * 5 - 20 ≈ -3
surprise, surprise -- the house wins. You only win 5/6 ticket per game. But it costs a whole ticket to play.
To find out how many tickets a player can expect to win or lose in the Doubles Game, you need to calculate the expected value.
The first step is to determine the probability of rolling a double on two six-sided dice. Since each die has six sides, there are 6 * 6 = 36 possible outcomes. Out of these outcomes, there are 6 possible doubles (1-1, 2-2, 3-3, 4-4, 5-5, and 6-6). So, the probability of rolling a double is 6/36 = 1/6.
Next, let's calculate the expected value. Each game costs 1 ticket to play, and if a player wins, they get 5 tickets. So the expected value for winning a game is (1/6) * 5 = 5/6 tickets.
If a player plays the game 20 times, we can multiply the expected value of winning a single game by 20 to find the overall expected value:
(5/6) * 20 = 100/6 = 16.67
Therefore, a player can expect to win or lose approximately 16.67 tickets in the Doubles Game over 20 rounds.