Question

(1) Calculate for the game of the craps the probability of

(a) Winning. This means the probability of winning in the 1^st throw, in the 2^nd throw, 3^rd throw, etc. for all the possible games (n = 1, 2, 3, 4, … indefinitely). Again, show how you derive the formula; the formula itself; the fractions in the form p/q; and the decimal expression to 4^th significant digit.

(b) Losing. This means the probability of winning in the 1^st throw, in the 2^nd throw, 3^rd throw, etc. for all the possible games (n = 1, 2, 3, 4, … indefinitely). Again, show how you derive the formula; the formula itself; the fractions in the form p/q; and the decimal expression to 4^th significant digit.

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 + 3²/9 + 4²/10 + 5²/11) = 0.492929293!