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

Question

2. 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 oint) 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?

Explanation / Answer

(a) probability of tossing a natural = probability of getting a 7 or 11 on the first roll of the dice = probability of getting any of the following combinations (1,6),(2,5),(3,4),(4,3),(5,2),(6,1), (5,6),(6,5) out of 36 possibilities = 8/36 = 2/9.

(b) Outcomes for event A: 2,3,12

Outcomes for event B: 7,11

Outcomes for event C: 9,10,11

(i) we can see that events A & B and A & C are mutually exclusive since they don't have any common elements.

(ii) Independence can refer to the simulteneous occurence of 2 events. In our case, only B and C have a common element, so events B and C can occur simulteneously with outcome 11.

P(B) = P(7 or 11) = 2/9, as determined in (a)

P(C) = P(9,10 or 11) = P((3,6),(4,5),(5,4),(6,3),(4,6),(5,5),(6,4), (5,6),(6,5)) = 9/36 = 1/4

P (B and C) = P(11) = P((5,6),(6,5)) = 2/16 = 1/18

P(B) x P(C) = (2/9) x (1/4) = 2/36 =1/18

So, we have P(B and C) = P(B) x P(C)

so we can conclude that B and C are independent.

(c) A player can throw a point in the following ways: (1,3), (2,2),(3,1),(1,4),(2,3),(3,2),(4,1),(1,5),(2,4),(3,3),(4,2),(5,1),(2,6),(3,5),(4,4), (5,3),(6,2),(3,6),(4,5),(5,4),(6,3),(4,6),(5,5),(6,4)

So probability of throwing a point = 24/36=2/3

(d) probability of winning, given that point was thrown on the first toss = probability of throwing 4 on the first toss, given that point was thrown in the first toss x probability of a 4 on the second toss + probability of throwing 5 on the first toss, given that point was thrown in the first toss x probability of throwing 5 on the second toss +.. So on for 6,8,9,10.

probability of throwing 4 on the first toss, given that point was thrown in the first toss = 3/24 (3 possible cases (1,3), (2,2),(3,1) out of the 24 possibilities for point throw)

probability of throwing 4 on the second toss = 3/36

probability of throwing 5 on the first toss, given that point was thrown in the first toss = 4/24

probability of throwing 5 on the second toss = 4/36

probability of throwing 6 on the first toss, given that point was thrown in the first toss = 5/24

probability of throwing 6 on the second toss = 5/36

probability of throwing 8 on the first toss, given that point was thrown in the first toss = 5/24

probability of throwing 8 on the second toss = 5/36

probability of throwing 9 on the first toss, given that point was thrown in the first toss = 4/24

probability of throwing 9 on the second toss = 4/36

probability of throwing 10 on the first toss, given that point was thrown in the first toss = 3/24

probability of throwing 10 on the second toss = 3/36

So the probability a player wins on the second toss, given that he throws a point on the first toss = (3/24) x (3/36) + (4/24) x (4/36) + (5/24) x (5/36) + (5/24) x (5/36) + (4/24) x (4/36) + (3/24) x (3/36) = (9+16+25+25+16+9) / (24x36) = 100/864 = 0.116

(e) Probability of winning in the second toss is as determined in (d) = 0.116

Probability of winning in the first toss is as determined in (a) = 2/9 = 0.222

So the probabibility of winning in 2 or fewer tosses = 0.116 + 0.222 = 0.338

Related Questions

Navigate

• Browse (All)
• Browse 2
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.