The game of craps is played with two dice. A player throws both dice, winning un

Question

The game of craps is played with two dice. A player throws both dice, winning unconditionally if he/she produces a natural (the sum of the numbers showing on the two dice is 7 or 11), and losing unconditionally if he/she throws craps (a 2, 3, or 12). However, if the sum of the two dice is 4, 5, 6, 8, 9, or 10 (each of these is known as a point) the player continues to throw the dice until the same outcome (point) is repeated (in which case the player wins), or a 7 occurs (in which case the player loses). For example, if a player’s first toss results in a 5, the player continues to roll the dice until a 5 or 7 occurs. If a 5 occurs first, the player wins, if a 7 occurs first, the player loses.

(a) What is the probability that a player will toss a natural on the first roll of the dice?

(b) Consider the following events for the first toss of the dice:

A: {Player throws craps}

B: {Player throws a natural}

C: {Player throws 9, 10, or 11}

(i) Which pairs of events, if any, are mutually exclusive?

(ii) Which pairs of events, if any, are independent?

(c) What is the probability a player throws a point on the first toss?

(d) If a player throws a point on the first toss, what is the probability the player wins the game on the next toss?

(e) What is the probability that a player wins the game in two or fewer tosses?

Explanation / Answer

a)

Outcomes counting as Naturals = { (1,6) (2,5) (3,4) (4,3) (5,2) (6,1) (5,6) (6,5) }

Thus,

P( Naturals) = 8 / 36 = 2/9

b)

i) A and B; A and C are mutually exclusive

ii) The events are not independent of each other as occurence of one definitely guarantees non-occurence of other.

c)

24 outcomes count as points

So, P (Point) = 24/36

= 2/3

d)

Here the first outcome is important because, the probability of the first outcome will determine the next point outcome.

Thus, required probability

= (3/36 * 3/36) + (4/36*4/36) + (5/36*5/36) + (3/36 * 3/36) + (4/36*4/36) + (5/36*5/36)

= 25 / 324

e)

P ( of winning in two or fewer tosses)

= P (X=1) + P(X=2)

= 2/9 + 25/324

= 97 / 324

Hope this helps.

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