Question

A game of "Doubles-Doubles" is played with two dice. Whenever a player rolls two dice and both die show the same number, the roll counts as a double. If a player rolls doubles, the player earns 5 points and gets another roll. If the player rolls doubles again, the player earns 5 more points. Whenever the player rolls the dice and does not roll a double, they lose points. How many points should the player lose for not rolling doubles in order to make this a fair game? Three-fifths 1 5 Three-fifths.

Answer 1

Using the expected value of a discrete distribution, it is found that the amount of points the player should lose for not rolling doubles in order to make this a fair game is of 1.

What is the expected value of a discrete distribution?

The expected value of a discrete distribution is given by the sum of each outcome multiplied by it's respective probability.

In this problem, considering 6 out of 6^2 = 36 outcomes are doubles, we have that the distribution is:

P(X = 5) = 1/6.

P(X = x) = 5/6.

A fair game means that the expected value is of 0, hence:

$5\frac{1}{6} - \frac{5x}{6} = 0$

$\frac{5x}{6} = \frac{5}{6}$

$x = 1$

The amount of points the player should lose for not rolling doubles in order to make this a fair game is of 1.

More can be learned about the expected value of a discrete distribution at https://brainly.com/question/24855677

Answer 2

Answer:

the player should lose 1 point for not rolling doubles

B. 1

Step-by-step explanation:

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