Problem 9, 15 Points. The game of craps is played by rolling two balanced dice.

Question

Problem 9, 15 Points. The game of craps is played by rolling two balanced dice. A first roll of a sum of 7 or 11 wins; a first roll of a sum of 2, 3, or 12 losses. To win with any other first sum, that sum must be repeated before a sum of 7 is thrown. Suppose that a player rolls a sum of 8 on the first roll. Determine the probability that the player wins. Hint. Define W, a sum of 8 occurs on the nth roll of the dice and that neither a sum of 8 nor a sum of 7 occurs before that. E,- a sum of 8 on the jth roll. S - a sum of 7 on the jth roll. You can also apply your own setup other than this hint.

Explanation / Answer

We know that the player rolled a 8 on the first die

So to win he needs to roll an 8 or not roll a 7 on the next die

P(8 on 2nd throw) = 5/36, P(7 on 2nd throw) = 6/36 or 1/6, P(rolling again) = 25/36

From this logic

P( 8 on nth throw) = 5/36 + 5/36*25/36 + ....+ (25/36)^n * 5/36 = 5/36*(1/1-25/36) = 5/11 (Applying GP formula a/1-r)

Probability that 8 is the first throw and you win = 5/36 * 5/11 = 25/396

Probability that you win given that 8 was the first throw = 5/11

P

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