Q:

Miguel is playing a game in which a box contains four chips with numbers written on them. Two of the chips have the number 1, one chip has the number 3, and the other chip has the number 5. Miguel must choose two chips, and if both chips have the same number, he wins $2. If the two chips he chooses have different numbers, he loses $1 (–$1). Let X = the amount of money Miguel will receive or owe. Fill out the missing values in the table. (Hint: The total possible outcomes are six because there are four chips and you are choosing two of them.) Xi 2 –1P(xi) What is Miguel’s expected value from playing the game? Based on the expected value in the previous step, how much money should Miguel expect to win or lose each time he plays? What value should be assigned to choosing two chips with the number 1 to make the game fair? Explain your answer using a complete sentence and/or an equation.

Accepted Solution

A:
1) The only way Miguel wins the 2 dollars is if he pulls the two chips with the number 1. There are four chips in total, so his probability of winning is:
2/4 * 1/3 = 1/6

The odds of losing a dollar is the compliment of this number, so 5/6.

2) Expected Value = (1/6)2 + (5/6)(-1) = -1/2

Miguel's expected loss is $-1/2 dollars each time he plays, although since this isn't actually a number he can lose, he should expect to lose a dollar each time he plays, which has a 5/6 chance of happening.

3) Let's set the expected value equal to zero, which implies a fair game, and Miguel' payout equal to X:
0 = (1/6)(x) + (5/6)(-1)          Add 5/6 to both sides
5/6 = 1/6 (x)                         Multiply both sides by 6
30/6 = x
x = 5
The payout should be $5 in order to make the game fair.