Question

There is a dice game at the casino that costs 3 to play. A player rolls two dice. If the player gets a sum of 2 or 12 he or she gets back 28 (wins 25). If the person gets a 7, he or she gets back 5 (wins 2). If the player rolls anything other than a 2, 12, or 7 he or she gets nothing back (loses 3). How much would the player need to be charged to play the game in order to make this a fair game? 1) 0 2) 3 3) 5 4) 25

Answer 1

To make the casino dice game fair, a charge of approximately $3.16 should be required per play. This ensures the expected value of the game is zero, meaning that in the long run, neither the player nor the house will gain or lose any money.

Step-by-step explanation:

The subject of your question is probability in Mathematics, specifically concerning a dice game in a casino. In order to determine how much a player would need to be charged to make the game fair, we first need to define what a 'fair game' is in probability theory. A 'fair game' is one in which the expected value of the game is zero, meaning that in the long run, neither the player nor the house gains or loses any money.

First, we need to calculate the probability of each outcome and its respective payoff. There is only one way to sum up 2 (1 and 1) and 2 ways to get 12 (6 and 6) in two dice. So, the probability of getting a 2 or 12 is 3/36. For 7, there are 6 combinations in two dice, hence the probability is 6/36. For other numbers, there are 27 combinations, so the probability is 27/36.

Now, the expected giveaway from the game for each play can be calculated as follows: ExpectedGiveAway = (28*(3/36)) + (5*(6/36)) + (0*(27/36)) = 2.33+0.83+0 = $3.16.

Since for a fair game, we need expected value to be zero, the charge needs to be the expected giveaway, which is $3.16.

Learn more about Fair Game in Probability