Math 207—Solutions to Counting-Method 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 e—of this solution.) Hence, the number of poker hands is simply .

b. What is the probability of 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

Then

[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. What is the probability of a straight (i.e., five cards in order—the ace can be used as a high card or a low card, but a straight can’t be, say Queen, King, Ace, Two, Three), but not a straight flush (i.e., five cards in order and all of the same suit)?

The outcome of a “straight” can be broken into the following separate tasks: choose the starting spot of the straight and then choose 1 card from each rank of the straight (as defined by the starting spot). Hence, by the basic principle of counting, the number of ways to get a straight is

Now we must subtract out the possible straight flushes. The outcome of a “straight flush” can be broken down in the following separate tasks: choose the starting spot of the straight and then choose the suit for the cards in the straight. Hence, by the basic principle of counting, the number of ways to get a straight flush is

So the probability of a straight (but not a straight flush) is

d. 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 (part b), so the number of ways to get a 3-of-a-kind is

So the probability of a three-of-a-kind is

e. In regard to part d, explain why the following answers are incorrect for the number of ways to get a three-of-a-kind (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. Suppose you have 3 distinct Green Bay Packers fans and 3 distinct Chicago Bears fans.

a. In how many ways can the fans sit in a row?

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

b. What is the probability of the fans sitting in a row such that the Packers fans are seated together and the Bears fans are each seated together?

A nice trick for solving problems like this is to initially think of the Packers fans as one entity and the Bears fans as one entity (since they must be placed together). This is what I call the “macro-level order” (there are 2 entities, so 2! macro-level orderings). Then consider the “micro-level order” within each set of fans: 3! for the Packers fans and 3! for the Bears fans. 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 fans where the Packers fans sit together and the Bears fans sit together.

Another solution method is to breakdown this event into 6 separate tasks—counting the number of ways to fill each seat in the row. Then there are 6 choices for the first spot, only 2 choices for the second spot (must be the same type of fan), and 1 choice for the third spot (the last of that type of fan); then there are 3 choices for the fourth spot (choose any of the other type of fan), 2 for the fifth spot, and 1 for the sixth spot. So by the basic principle of counting there are arrangements of the fans where the Packers fans sit together and the Bears fans sit together.

So the probability that the Packers fans are seated together and the Bears fans are seated together is

c. What is the probability of the fans sitting in a row such that the Packers fans are seated together?

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

Another solution method is to notice there are only 4 starting places for the Packers fans (if you consider only the 6 seats in the row for the fans). Then there are 3! arrangements of the Packers fans (after you’ve chosen their starting spot) and 3! arrangements of the other fans (around the Packer fans). So by the basic principle of counting there are arrangements of the the people where the Packer fans are together.

So the probability that the Packers fans are seated together is

d. What is the probability of the fans sitting in a row such that no two fans of the same team are seated together?

This event can be broken into the following separate tasks: select the type of fan that will be seated first (then the other spots, in terms of fan-type, are determined), order the Packers fans, and order the Bears fans. Then by the basic principle of counting there are arrangements of the people where no two fans of the same team are seated together.

Another solution method is to breakdown this event into 6 separate tasks—counting the number of ways to fill each seat in the row. Then there are 6 choices for the first spot, only 3 choices for the second spot (must be a different fan type), 2 choices for the third spot (must be one of the remaining fan-types as the first seat), 2 choices for the fourth spot, 1 for the fifth spot, and 1 for the sixth spot. So by the basic principle of counting there are arrangements of the people where no two fans of the same team are seated together.

So the probability of the fans sitting in a row such that no two fans of the same team are seated together is

3. A person has 8 friends, of whom only 5 will be (randomly) invited to a party (perhaps he only has 6 place settings).

a. How many choices are there if two of the friends are fighting and won’t attend together? (Here you are simply counting, not determining a probability.)

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 if two of the friends will only attend together (perhaps they just started dating and are currently inseparable)?

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.