Probability with arrangements and combinations; real-life events and games of chance

25 min

When the sample space is too big to list

In a lottery where 6 numbers are drawn from 1 to 45, there are more than eight million possible draws. Nobody can list them, but the counting techniques of Unit 2 can count them, and then $P(A) = \frac{n(A)}{n(S)}$ does the rest. This lesson combines the two units: count the equally likely outcomes with permutations and combinations, then divide. The same method lets you judge real games of chance and see why the house, or the lottery, always wins in the long run.

Definition

Chosen at random means every possible selection or arrangement is equally likely, so $P(A) = \frac{n(A)}{n(S)}$ applies.
Use ${}^{n}C_{r} = \frac{n!}{r!\,(n-r)!}$ to count selections (order does not matter) and ${}^{n}P_{r} = \frac{n!}{(n-r)!}$ or $n!$ to count arrangements (order matters). Count $n (A)$ and $n (S)$ in the same way.
In a game of chance, the expected gain is the average amount won or lost per game in the long run: $E = \sum x\,P(x)$ , where $x$ is each possible net gain.
A game is fair if the expected gain is 0.

Worked example

Worked example 1: a committee. A committee of 3 is chosen at random from 6 boys and 4 girls. Find the probability that it contains (a) only girls; (b) exactly 2 girls; (c) at least one girl.

All selections of 3 from 10 are equally likely: $n(S) = {}^{10}C_{3} = 120$ .

(a) $n(A) = {}^{4}C_{3} = 4$ , so $P = \frac{4}{120} = \frac{1}{30}$ .

(b) Choose 2 of the 4 girls and 1 of the 6 boys: $P = \frac{{}^{4}C_{2} \times {}^{6}C_{1}}{{}^{10}C_{3}} = \frac{6 \times 6}{120} = \frac{36}{120} = \frac{3}{10}$

Worked example

Worked example 2: arranging letters. The letters of RWANDA are arranged in a random order. Find the probability that (a) the arrangement begins with R; (b) the two A's are next to each other.

There are 6 letters with A repeated twice: $n(S) = \frac{6!}{2!} = 360$ .

(a) Fix R first and arrange W, A, N, D, A: $\frac{5!}{2!} = 60$ . $P = \frac{60}{360} = \frac{1}{6}$ . (This agrees with common sense: each of the 6 letters is equally likely to come first.)

(b) Glue the two A's into one block AA. Then R, W, N, D and the block make 5 different items: $5! = 120$ . $P = \frac{120}{360} = \frac{1}{3}$ .

Worked example

Worked example 3: a lottery. In a lottery (illustrative rules), a player chooses 6 different numbers from 1 to 45, and 6 winning numbers are drawn at random. Find the probability of (a) matching all 6; (b) matching exactly 5.

$n(S) = {}^{45}C_{6} = 8\,145\,060$ .

(a) Only one draw matches all six: $P = \frac{1}{8\,145\,060} \approx 1.23 \times 10^{-7}$ .

(b) Choose 5 of the player's 6 numbers and 1 of the 39 numbers the player did not choose:

P = \frac{{}^{6}C_{5} \times {}^{39}C_{1}}{{}^{45}C_{6}} = \frac{6 \times 39}{8\,145\,060} = \frac{234}{8\,145\,060} \approx 2.87 \times 10^{-5}

To put (a) in perspective: buying one ticket a week, a player would expect to wait more than 150,000 years for a jackpot.

Worked example

Worked example 4: is the game fair? At a school fun day, it costs 100 Frw to roll one die. If it shows a 6, the player receives 400 Frw. Find the expected gain per game.

The net gain is $400 - 100 = 300$ Frw with probability $\frac{1}{6}$ , and $- 100$ Frw with probability $\frac{5}{6}$ :

E = 300 \times \frac{1}{6} + (-100) \times \frac{5}{6} = \frac{300 - 500}{6} = -\frac{200}{6} \approx -33.3 \text{ Frw}

On average a player loses about 33 Frw per game, and the stall gains it. For the game to be fair, the prize $w$ would have to satisfy $\frac{w}{6} - 100 = 0$ , so $w = 600$ Frw.

A surprising real-life result: shared birthdays

How many people must be in a room before it is more likely than not that two share a birthday? Ignoring 29 February and assuming all 365 days are equally likely, the probability that $n$ people all have different birthdays is $\frac{{}^{365}P_{n}}{365^{n}}$ , because the first person may have any of 365 days, the second any of the other 364, and so on. So

P(\text{at least one shared birthday}) = 1 - \frac{365 \times 364 \times \cdots \times (366 - n)}{365^{n}}

With only 23 people the probability already passes one half (0.507), and in a class of 50 it is 0.970. Our intuition fails because we think of our own birthday; but 23 people form ${}^{23}C_{2} = 253$ different pairs, and any pair may match.

Warning

The gambler's fallacy. After five tails in a row, a fair coin is not "due" to land heads: the tosses are independent, and $P(\text{head})$ is still $\frac{1}{2}$ . Games of chance are designed so that the expected gain for the player is negative; playing more often makes a loss more certain, not less. Gambling is also restricted by age in law, and it can become an addiction.

Tip

Count the sample space and the event the same way. If you use combinations for $n (S)$ , use combinations for $n (A)$ ; mixing ordered and unordered counts gives wrong answers.

Khan Academy: "Probability using combinations". Notice that the event and the sample space are both counted with combinations, as in Worked example 1.

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Practice

Three counters are taken at random from a bag of 5 red and 3 blue. Find the probability of exactly 2 red and 1 blue.
Four boys and three girls sit in a row of 7 chairs in random order. Find the probability that the three girls sit together.
Six friends stand in a row in random order. Find the probability that Ishimwe and Nshuti stand next to each other.
A 4-digit PIN is chosen at random (digits 0 to 9, repeats allowed). Find the probability that all four digits are different.
Five cards are dealt from a pack of 52. Find the probability that all five are hearts, to 3 significant figures.
In the lottery of Worked example 3, find the probability of matching exactly 3 numbers, to 3 significant figures.
A game costs 50 Frw. Two dice are rolled, and a double wins 300 Frw (the stake is not returned). Is the game fair? Show your working.
Find the probability that in a group of 4 people at least two share a birthday (365 equally likely days), to 3 decimal places.

Show answersHide answers

$\frac{{}^{5}C_{2} \times {}^{3}C_{1}}{{}^{8}C_{3}} = \frac{10 \times 3}{56} = \frac{30}{56} = \frac{15}{28}$ .
Block the girls: $5! \times 3! = 720$ of $7! = 5040$ : $\frac{720}{5040} = \frac{1}{7}$ .
$\frac{5! \times 2}{6!} = \frac{240}{720} = \frac{1}{3}$ .
$\frac{10 \times 9 \times 8 \times 7}{10^4} = \frac{5040}{10000} = 0.504$ .
$\frac{{}^{13}C_{5}}{{}^{52}C_{5}} = \frac{1287}{2\,598\,960} \approx 0.000495$ .
$\frac{{}^{6}C_{3} \times {}^{39}C_{3}}{{}^{45}C_{6}} = \frac{20 \times 9139}{8\,145\,060} \approx 0.0224$ .
$P(\text{double}) = \frac{1}{6}$ . $E = 250 \times \frac{1}{6} + (-50) \times \frac{5}{6} = \frac{250 - 250}{6} = 0$ : the game is fair.
$1 - \frac{365 \times 364 \times 363 \times 362}{365^4} \approx 1 - 0.984 = 0.016$ .

Important

For random selections or arrangements, $P(A) = \frac{n(A)}{n(S)}$ with both counted by the same method ( ${}^{n}C_{r}$ , ${}^{n}P_{r}$ or $n!$ ). Expected gain $E = \sum x\,P(x)$ ; a fair game has $E = 0$ .

1When the sample space is too big to list

2A surprising real-life result: shared birthdays

When the sample space is too big to list

A surprising real-life result: shared birthdays