Question
On May 23, 2009, a craps player held the dice for 154 rolls at the Borgata casino in Atlantic City (source). What is the probability of going 154 rolls or longer in craps? Please give an expression of the answer and/or a numeric answer.
Answer
The answer is P1 + P2 + P3 + P4, where
[ P1 ] [12 3 4 5]^153 [1] [ P2 ] [ 6 27 0 0] [0] [ P3 ] = (1/36)^153 × [ 8 0 26 0] × [0] [ P4 ] [10 0 0 25] [0]The exact answer is:
P1 = 3.12763 × 10-11
P2 = 4.63460 × 10-11
P3 = 4.95558 × 10-11
P4 = 5.17044 × 10-11
The sum of these probabilities is 1.78882 × 10-10 = 1 in 5,590,264,072
Acknowledgement: My thanks to BruceZ for this help with this problem.
Michael Shackleford, ASA — June 2, 2009