Probability Theory—Solutions to Chapter 1 Practice Problems

Required course reading, both for the solution methods and hints and the method of exposition.

1. A poker hand is defined as five cards randomly drawn from a standard 52-card deck. Note that each 52-card deck has 13 ranks (2, 3, …, 10, jack, queen, king, ace) and each rank is represented in 4 suits (hearts, diamonds, clubs, and spades).

a. How many possible poker hands are there?

In card games such as poker, a hand is considered unordered (that is, whichever way you order the cards, it’s still the same “hand”). Also, the cards are dealt without replacement. (This statement applies to all parts—a through d—of this solution.) Hence, the number of poker hands is simply . [I included the numerical calculation for completeness sake, but this is not necessary for your homework solutions. The numerical calculations will become more interesting when we calculate probabilities.]

b. How many ways are there to get a “full house” (i.e., three cards of one rank and two cards of another rank)?

The outcome of a “full house” can be broken into the following separate tasks: choose a 3-of-a-kind rank, then choose 3 cards from that rank (unordered); choose a 2-of-a-kind rank, then choose 2 cards from that rank (unordered). Hence, by the basic principle of counting, the number of ways to get a full house is

[Note the outcome of a “full house” can equivalently be broken into the following separate tasks: choose a 2-of-a-kind rank, then choose 2 cards from that rank (unordered); choose a 3-of-a-kind rank, then choose 3 cards from that rank (unordered). This gives the same answer: ]

c. How many ways are there to get a “three-of-a-kind” (i.e., exactly three cards of the same rank)? Note that a full house does not count as a three-of-a-kind.

The outcome of a “three-of-a-kind” can be broken into the following separate tasks: choose a 3-of-a-kind rank, then choose 3 cards from that rank (unordered); choose 2 other ranks (unordered), then choose 1 card from each of those ranks (the ranks must be different, as a “full house” does not count as a “three of a kind”). Hence, by the basic principle of counting, the number of ways to get a 3-of-a-kind is

Another method of solving this problem is to first count the total number of hands with 3 cards of the same rank and then subtract out the number of “full house” hands. The outcome of “3 cards of the same rank” can be broken into the following separate tasks: choose a 3-of-a-kind rank, then choose 3 cards from that rank; choose any other two cards (unordered and that are not of the 3-of-a-kind rank). By the basic principle of counting, this can be done in ways. We already know the number of ways to get a full house (answer b), so the number of ways to get a 3-of-a-kind is [Note in this case it was nice to have the actual numerical solution, as we can compare with the previous method and see they give the same answer.]

d. In regard to part c, explain why the following answers are incorrect (and how you could change them to make them correct):

i.

The first combination in this solution represents the selection of 3 different, unordered ranks. In fact, there is an implicit ordering on the ranks in a three-of-a-kind hand, in that one of them must be denoted the 3-of-a-kind rank. [For example, this method counts the hands (2 hearts, 2 diamonds, 2 clubs, 8 spades, 10 spades) and (2 spades, 8 hearts, 8 diamonds, 8 clubs, 10 spades) and (2 spades, 8 spades, 10 hearts, 10 diamonds, 10 clubs) as the same hand, but these are actually different poker hands.]

To make this solution correct, we simply need to additionally choose a 3-of-a-kind rank (as a separate task in the overall outcome):

[Practical Tip: If you are trying to determine why a counting-method solution is incorrect and you happen to know the correct answer, then an easy first step is to see if the incorrect answer is over- or under-counting in a particular way. In this case, the correct numerical answer is 3 times the incorrect numerical answer, so that gives a clue about how the incorrect answer is under-counting.]

ii.

Using the practical tip from above, this incorrect answer double counts certain hands (you can verify this by doing numerical calculations). In this case, the solution method imposes an order on the non-3-of-kind cards, when those cards should actually be considered unordered. [For example, this method counts the hands (2 hearts, 2 diamonds, 2 clubs, 8 spades, 10 spades) and (2 hearts, 2 diamonds, 2 clubs, 10 spades, 8 spades) as different, when in fact they are the same poker hand.]

Notice that in the solution is actually the number of permutations of 12 items taken 2 at a time. Since the order of the additional two cards does not matter in a poker hand, we should count the combinations, not the permutations: This is equivalent to dividing the incorrect answer by 2.

2. (Problem 11 in the textbook) In how many ways can 3 (distinct) novels, 2 (distinct) mathematics books, and 1 chemistry book be arranged on a bookshelf if

a. the books can be arranged in any order?

There are 6 total books (which are distinct). Since the question asks about ordered arrangements, the number of permutations is

b. the mathematics books must be together and the novels must be together?

A nice trick for solving problems like this is to initially think of the math books as one entity and the novels as one entity (since they must be placed together). This is what I call the “macro-level order” (there are 3 entities, so 3! macro-level orderings). Then consider the “micro-level order” within each set of books: 3! for the math books, 2! for the novels, and 1! for the chemistry book. So the outcome can be broken into macro-level and micro-level tasks, and by the basic principle of counting there are arrangements of the books where the math books are together and the novels are together. (I include the 1! for completeness sake only. This is not a necessary part of the solution.)

c. the novels must be together, but the other books can be arranged in any order?

Using the same reasoning as in part b, there are 4! macro-level arrangements of the books (considering the novels as one entity) and 3! micro-level arrangements of the novels. Hence, there are arrangements of the books where the novels are together.

Another solution method is to notice there are only 4 starting places for the novels (if you consider only the 6 slots on the shelf for the books). Then there are 3! arrangements of the novels (after you’ve chosen their starting spot) and 3! arrangements of the other books (around the novels). So by the basic principle of counting there are arrangements of the books where the novels are together.

3. (Problem 20 in the textbook) A person has 8 friends, of whom 5 will be invited to a party.

a. How many choices are there if 2 of the friends are feuding and will not attend together?

In this type of situation, it’s cleanest and easiest to break the outcome into distinct events: either neither of the friends is invited or only one of the friends is invited. Since these are distinct events, we can determine the number of ways each can happen and then add these numbers together.

Furthermore, this is a situation where we have two types of people (fighting friends and non-fighting friends) and where order is unimportant. Hence, by the basic principle of counting, there are ways to invite neither of the fighting friends and ways to invite only one of the fighting friends. Therefore, there are choices if 2 of the friends won’t attend together.

Note for the two fighting friends there are only 3 possibilities: neither is invited, only one is invited, or both are invited. So another solution method is to determine the total number of choices of friends and then subtract out the choices where both friends are invited: choices if 2 of the friends won’t attend together.

b. In regard to part a, explain why the following answer is incorrect (and how you could change it to make it correct):

i.

In this solution method, the outcome was not broken into distinct events (note the danger of trying to take-care-of-everything-at-once in these types of problems). The reasoning of this method seems to be: count the possibilities where fighting-friend-A is not invited and add that to the number of possibilities where fighting-friend-B is not invited. This method does count the ways in which neither friend is invited and in which only one friend is invited, but it double counts the cases where 5 non-fighting friends are invited.

Hence, we can correct this solution by subtracting out the number of ways that 5 non-fighting friends can be invited: choices if 2 of the friends won’t attend together.

c. How many choices are there if 2 of the friends will only attend together?

If 2 friends will only attend together, then the choices are limited to either both best-friends attend or neither best-friends attend. By the same reasoning as in part a, there are choices if 2 of the friends will only attend together.

Again, as in part a, an alternative solution method is to subtract the number of choices when only one friend attends from all the possible choices: choices if 2 of the friends will only attend together.

4. (Theoretical Exercise 12) Consider the following combinatorial identity:

a. Present a combinatorial argument for this identity by considering a set of n people and determining, in two ways, the number of possible selections of a committee of any size and a chairperson for the committee. Hint:

i. How many possible selections are there of a committee of size k and its chairperson?

Because committees are considered unordered (doesn’t matter how you arrange the people on the committee, they are still the same committee), there are ways to select a committee of size k from a group of n people. Once the committee is chosen, there are k possible choices for the chairperson. So for a given size k, there are possible selections of a committee and a chairperson of the committee. Since we’re interested in a committee of any (non-zero) size, k = 1, ..., n, we must add together the number of choices for all possible committee sizes. Hence, is the number of possible selections of a committee of any (non-zero) size and a chairperson for the committee.

ii. How many possible selections are there of a chairperson and the other committee members?

There are initially n possible choices for the chairperson. Once the chairperson is determined, there are choices for the other committee members (for a committee of size k). As in part i, we must add together the number of choices for all possible committee sizes: . We can re-index this sum by letting j = k – 1, so the answer becomes , where the last equality holds by using the binomial theorem with

[Note: We could also have gotten this answer by first selecting a chairperson and then for each of the remaining people, either selecting them or not selecting them for the committee. That is, the represents the number of ways, sampling with replacement from the options on-committee or off-committee for each person, to select a committee of any size, 0 through (in addition to the chairperson).]

Now since the answers to parts i and ii are both the number of possible selections of a committee of any size and a chairperson for the committee, they must be equal to one another. Hence, we obtain the identity