Download AMS 80A: Gambling and Gaming (Spring 2014)

AMS 80A: Gambling and Gaming (Spring 2014)
Homework 1 solutions
1. Exercises 2, 4, and 6 from Chapter 1 of the Rodriguez and Mendes draft book.
Solution: Using in all cases the multiplication rule for counting, the answer is: 5 × 12 × 3 for
exercise 2; 268 for exercise 4; and 13 × 12 × 11 × ... × 2 × 1 for exercise 6.
2. Suppose that a coin is tossed five times. Let A be the event that a head is obtained on the first
toss, and B be the event that a head is obtained on the third toss. Are events A and B disjoint?
Solution: No, for example, outcome (H,T,H,H,T) is in the intersection of A and B.
3. Exercises 1 and 2 from “Exercise Set D”, Chapter 14 of the Freedman, Pisani and Purves book
(page 250).
Solution: For exercise 1, the combinations of bets and corresponding winning outcomes are:
(a) with (i); (b) with (i) and (ii); (c) with (iii); (d) with (ii) and (iii); (e) with (i) and (ii); and
(f) with (i).
For exercise 2, bets (a) and (f) are the same; also the same are bets (b) and (e). Finally, bet
(d) is better than (c).
4. Consider an experiment in which a fair coin is tossed once and a balanced die is rolled once.
Describe the sample space for this experiment. Assuming equally likely outcomes, what is the
probability that a head will be obtained on the coin and an even number will be obtained on
the die?
Solution: The sample space contains 12 outcomes, where each outcome comprises a pair with
the first element H or T (from the toss of the coin), and the second element a number from 1 to
6 (from the roll of the die). Event A = (head on the coin and even number on the die) includes 3
outcomes, specifically, (H,2), (H,4), and (H,6). Under the assumption of equally likely outcomes
for this finite sample space, Pr(A) = 3/12 = 1/4.
5. Suppose that four fair coins are tossed (assume that all possible outcomes are equally likely).
What is the probability that all four faces will be the same?
Solution: There is a total of 16 = 24 possible outcomes, where each outcome is a combination
(of size 4) of heads (H) or tails (T). Only 2 of these combinations have all four faces the same
(outcomes (T,T,T,T) and (H,H,H,H)). Because this is a finite sample space with the uniform
probability distribution, the probability that all four faces will be the same is 2/16 = 1/8.
6. Suppose that a fair coin is tossed 10 times (again, assume that all possible outcomes are equally
likely). What is the probability of observing at least one head?
Solution: This is an example where the simpler approach to the solution involves working with
the complement of the event of interest. If A = (at least one H in 10 coin tosses), then the
complement, Ac = (no H in 10 coin tosses). Now, Ac includes only one outcome (T 10 times)
among the 1024 = 210 possible outcomes of the sample space, which consists of combinations
(of size 10) of H or tails T. Therefore, Pr(Ac ) = 1/1024, and Pr(A) = 1 − Pr(Ac ) = 1023/1024.
Note that an alternative way to obtain Pr(Ac ) involves the (natural) assumption of independence across the 10 coin tosses. In particular, Pr(Ac ) = Pr((not H on first toss) and (not H on
second toss) and ... and (not H on tenth toss)) = (1/2)10 = 1/1024 (using the multiplication
rule for independent events).
7. Consider the experiment of rolling two balanced dice, and assume that all outcomes for the
pair of numbers are equally likely. Obtain the probability for each possible value of the sum of
the two numbers. Compute Pr(the sum is 7 or 11) and Pr(the sum is 2, 3, or 12), which are
respectively the probabilities of winning and losing at the first stage of the game of craps.
Solution: Let pi = Pr(sum of the roll is i), for i = 2, 3, ..., 12. Counting the number of outcomes
in the original sample space (comprising the 36 pairs of numbers from the two dice) that result
in each possible value for the sum, we obtain p2 = p12 = 1/36, p3 = p11 = 2/36, p4 = p10 = 3/36,
p5 = p9 = 4/36, p6 = p8 = 5/36, and p7 = 6/36. Therefore, Pr(winning) = p7 + p11 = 8/36
(the events of observing a sum of 7 and a sum of 11 are disjoint). Similarly, Pr(losing) =
p2 + p3 + p12 = 4/36.
8. Under the same experiment with problem 7, what is the probability that the sum of the two
numbers is odd?
Solution: Under the notation above and since the corresponding events are disjoint, this probability is given by p3 + p5 + p7 + p9 + p11 = 18/36.
9. The birthday problem. Disregarding the possibility of a February 29th birthday, suppose that a
randomly selected individual is equally likely to have been born on any one of the 365 days of
the year. If 10 people are randomly selected, what is the probability that at least two of them
have the same birthday?
Solution: The answer is 1 − {(365 × 364 × ... × 357 × 356)/(36510 )} = 1 − 0.883 = 0.117. For
details, refer to the solution discussed in class.
10. Prove the following properties for probability.
• For any event A, Pr(Ac ) = 1 − Pr(A).
• For two events A and B in a sample space such that A is a subset of B, we have Pr(A) ≤ Pr(B).
• Addition rule for the probability of the union of two events:
Pr(A ∪ B) = Pr(A) + Pr(B) − Pr(A ∩ B)
for any events A and B in the same sample space.
Solution: Refer to the detailed derivation given in class.