Question

Calculate for the game of the craps the probability of

(a) Winning. This means the probability of winning in the 1st throw, in the 2nd throw, 3rd throw, etc. for all the possible games (n = 1, 2, 3, 4, … indefinitely). show how you derive the formula.

(b) Losing. This means the probability of winning in the 1st throw, in the 2nd throw, 3rd throw, etc. for all the possible games (n = 1, 2, 3, 4, … indefinitely).

Answer #1

The dice game craps is played as follows:

The player throws 2 dice, and if the sum is 7 or 11, he/she wins. If the sum is 2, 3, or 12, he/she loses. If the sum is anything else, then he/she continues throwing until that number appears again or throws a 7, where the game ends in a loss.

P(winning) = P(w|2)P(2) + P(w|3)P(3) + ........+ P(w|12)P(12) = 0 * (P(2) + P(3))

prob of the first throw,

nP(n) = (n-1) / 36

P(w|n) = p / (p+1/6)

= [(n-1)/36] / [(n-1)/36 + 1/6]

= (n-1) / [(n-1)+6]

= (2/36) (4 + 3^{2}/9 + 4^{2}/10 +
5^{2}/11) = 0.492929293!

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