### Probability and Combinations

Just as with permutations, combinations crop up frequently in probability. In many card games the objective is to acquire a winning hand. In order to do that, it’s useful for players to know how likely a given hand is to arise, and also the probability that another player has a better hand. Mathematicians have been studying such games of chance for centuries.

#### Real-World Application: Word Games

A word game requires players to select 4 tiles from a bag containing 26 tiles, each with one of the letters A through Z written on it. If each letter appears once and only once, what is the probability that a player will be able to spell CATS with his tiles?

Since a player needs a \begin{align*}C\end{align*}, an \begin{align*}A\end{align*}, a \begin{align*}T\end{align*} and an \begin{align*}S\end{align*} in **any order**, we are looking at a **combination** calculation. We first need to determine how many combinations there are, choosing 5 letters from a total of 26:

Choosing 4 from 26:

\begin{align*}_{26}C_4 &= \frac{26!}{(26-4)! 4!} = \frac{26!}{22! 4! }\\ &= \frac{26 \times 25 \times 24 \times 23 }{ 4 \times 3 \times 2 \times 1}\\ &= 59,800 \ \text{combinations.}\end{align*}

Since only one combination allows a player to spell CATS, the probability of getting that combination is \begin{align*}\frac{1}{59,800}\end{align*}.

#### Real-World Application: Funfair Games

A funfair game consists of pulling numbered chips from a bag. The game starts with nine chips numbered 1 through 9, and players are allowed to pull out three chips. A player wins by drawing the number 7 chip. What is the probability that a player will win?

To find the probability for winning this game we need to know two pieces of information: 1) the total number of combinations for the game and 2) the number of combinations that contain a 7.

To find the total number of combinations for the game, use the formula \begin{align*}_nC_r\end{align*} with \begin{align*}n = 9\end{align*} and \begin{align*}r =3\end{align*}:

\begin{align*}{_9}C_2 = \frac{9!}{(9-3)! 3!} = \frac{9!}{6! 3! } = \frac{9 \times 8 \times 7 }{ 3 \times 2 \times 1} = 84 \ \text{combinations}\end{align*}

Now we need to determine how many combinations contain a 7. We can figure this out by reasoning as follows: given that there MUST be a 7 in the list, the number of combinations that contain a seven is the same as the number of combinations of choosing any *two* numbers out of the eight chips that *don’t* include the 7. In other words, if we imagine that we got to pick the 7 chip on purpose, how many ways would there be of picking the other two chips?

To find that number, we use the formula \begin{align*}_nC_r\end{align*} with \begin{align*}n =8\end{align*} and \begin{align*}r =2\end{align*}:

\begin{align*}{_8}C_2 = \frac{8!}{(8-2)! 2!} = \frac{8!}{6! 2! } = \frac{8 \times 7 }{ 2 \times 1} = 28 \ \text{combinations}\end{align*}

So the probability is given by:

\begin{align*}P(\text{getting a} \ 7) = \frac{84}{28} = \frac{1}{3} = \text{or one in three}.\end{align*}

#### Calculating Probability

Calculate the probability of being dealt four aces in a five card poker hand.

The first thing we need to know to solve this problem is the total number of unique hands. Since players can arrange their cards however they wish, the order of the cards is unimportant. So we will calculate, using the formula, the number of combinations for choosing 5 cards from a 52 card deck.

Choosing 5 from 52: \begin{align*}_{52}C_5 = \frac{52!}{(52-5)! 5!} = \frac{52!}{47! 5! } = \frac{52 \times 51 \times 50 \times 49 \times 48 }{ 5 \times 4 \times 3 \times 2 \times 1} = 2,598,960\end{align*} unique hands

Next we need to calculate how many hands there are that contain four aces. This sounds difficult, but we can think about it like this:

- If a hand contains four aces, it must also contain exactly ONE other card.
- Since the four aces are accounted for, there are 48 (that’s 52 – 4) cards left in the deck.

Since a unique hand is independent of the order in which the cards are dealt, there must be 48 unique hands that contain four aces (one unique hand for every non-ace card in the deck).

There are 48 possible hands that contain four aces. So the probability of being dealt four aces in poker is:

\begin{align*}P(\text{four aces})= \frac{48}{2,598,690} = \frac{1}{54,145} \end{align*}

### Review

For 1-3, calculate the number of combinations:

- \begin{align*}_8C_4\end{align*}
- \begin{align*}_{11}C_5\end{align*}
- \begin{align*}_{20}C_2\end{align*}

For 4-8, a town lottery requires players to choose three different numbers from the numbers 1 through 36.

- How many different combinations are there?
- What is the probability that a player’s numbers match all three numbers chosen by the computer?
- What is the probability that two of a player’s numbers match the numbers chosen by the computer?
- What is the probability that one of a player’s numbers matches the numbers chosen by the computer?
- What is the probability that none of a player’s numbers match the numbers chosen by the computer?

- A bag contains 13 dominoes. Each domino has a different number of dots on it, and the numbers of dots are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12. Peter selects 2 dominos at random from the bag. What is the probability that the total number of dots on the two dominoes he selects is 7?
- Looking at the odds that you came up with in question 4, devise a sensible payout plan for the lottery—in other words, how big should the prizes be for players who match 1, 2, or all 3 numbers? Assume that tickets cost $1. Don’t forget to take into account the following:
- The town uses the lottery to raise money for schools and sports clubs.
- Selling tickets costs the town a certain amount of money.
- If payouts are too low, nobody will play!

### Review (Answers)

To view the Review answers, open this PDF file and look for section 13.6.