Question

a popular gambling game played in casinos throughout the world. The player rolls two dice and plays against the house. If the first roll is 7 or 11, the player wins immediately; if it is 2, 3, or 12, the player loses immediately. If the first roll results in 4, 5, 6, 8, 9, or 10, the player continues to roll until either the same number appears, which constitutes a win, or a 7 appears, which results in the player loosing. What is the player’s probability of winning the game ?

Answer 1

Using it's concept, it is found that the player’s probability of winning the game is of 0.2083 = 20.83%.

What is a probability?

A probability is given by the number of desired outcomes divided by the number of total outcomes.

Each roll has 12 outcomes, hence the total number of outcomes is given by:

T = 12² = 144.

The outcomes resulting in a win are as follows:

• (7,x) -> 12 outcomes.

• (11,x) -> 12 outcomes.

• (4,4), (5,5), (6,6), (8,8), (9,9), (10,10) -> 6 outcomes.

That is, there are 12 + 12 + 6 = 30 desired outcomes. Hence the probability of winning is given by:

p = 30/144 = 0.2083.

More can be learned about probabilities at https://brainly.com/question/14398287

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