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? PLEASE!!!! HELP ME!!! I don't understand this problem THANKS!!!!
A student can expect to win 1 in every 6 tries. (roll one die, the probability of the second die matching the first, what ever it is is 1/6) So. the expected win is 5*(1/6) = .83333 tickets. If it cost a ticket to play, the expected loss per play is .83333-1 = -.16666. So with 20 plays, a student can expect a loss of 20*.1666 = 3.3333 tickets
To solve this problem, let's break it down step by step:
Step 1: Determine the probability of winning in one game
In one game, the player pays one ticket and rolls two six-sided dice. The goal is to match the numbers on the dice. Since each die has six sides, there are 6 possible outcomes for each die, resulting in a total of 6 * 6 = 36 possible outcomes when rolling two dice. Out of these 36 outcomes, there are 6 outcomes where the numbers match (e.g., rolling a 1 on both dice, rolling a 2 on both dice, etc.). Therefore, the probability of winning in one game is 6/36, which simplifies to 1/6.
Step 2: Calculate the expected win in one game
Since the player wins 5 tickets when the numbers match, the expected win in one game is (5 tickets) * (1/6 probability) = 5/6 tickets.
Step 3: Calculate the expected loss in one game
Since the player pays 1 ticket to play, the expected loss in one game is (1 ticket) - (5/6 tickets) = -1/6 tickets.
Step 4: Calculate the expected loss in 20 games
Since the expected loss per game is -1/6 tickets, the expected loss in 20 games is (20 games) * (-1/6 tickets) = -20/6 tickets, which simplifies to -3.3333 tickets.
Therefore, a player can expect to lose approximately 3.3333 tickets when playing the Doubles Game 20 times.
To solve this problem, we need to calculate the expected number of tickets that the student can win or lose.
First, let's calculate the probability of rolling two matching numbers on two six-sided dice. Since there are six sides on each die, the probability of rolling a specific number on the first die is 1/6. If we want the second die to match the first, there is only one out of six options that can give us a match. Therefore, the probability of rolling matching numbers is 1/6.
Next, we can calculate the expected number of tickets won from one play. Given that the player pays one ticket to play, they can win five tickets if they roll matching numbers. So the expected win from one play is: 5 * (1/6) = 0.83333 tickets.
Since the player plays the game 20 times, we can calculate the expected loss by multiplying the expected loss per play by the number of plays. The expected loss per play is the difference between the win from one play (0.83333 tickets) and the cost to play (1 ticket), which is -0.16666 tickets.
Therefore, the expected loss from 20 plays is: 20 * (-0.16666) = -3.3333 tickets.
In conclusion, the student can expect to lose approximately 3.3333 tickets after playing the Doubles Game 20 times.