Craps. In this game a player throws two dice and observes the sum. A throw of 7or 11 is an immediate win. A throw of 2, 3, or 12 is an immediate loss. A throw of 4, 5, 6, 8, 9, or 10, becomes the player's point. In order to in the game now, the player must continue to throw the dice, and obtain the point before throwing a 7. The problem is to calculate the probability of winning at craps. Let X0 represent the first sum throw. The basic idea of the calculation is first to calculate p(Win|X0 = X) for every possible value x of X0, then use the law of average conditional probabilities to obtain p(Win)
a) Show that for X = 4, 5, 6, 8, 9, 10,
P(Win|X0 = x) P(x)/[P(x) + P(7)]
Where P(x) = P(Xi = x) is the probability of rolling a sum of x. (Refer to Example2).
b) Write down P(Win|X0 = x) for the other possible values x of X0.
c) Deduce that the probability of winning at craps is