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Probability with combinations

Probability · level 6, lesson 3 of 6

A question to start

In a classic 6/49 lottery, you choose 6 different numbers from 1 to 49. Six winning numbers are drawn. You win the jackpot if your six are exactly the six drawn, in any order.

The chance of that is 1 in 13 983 816. If you bought one ticket every week, you would expect to wait about 13 983 816 weeks for a win, which is more than 260 000 years.

Where does such a precise number come from? Nobody listed nearly 14 million tickets. In this lesson you use combinations to count huge numbers of equally likely choices, and turn the counts into probabilities: lottery prizes, poker hands and committees picked out of a hat.

The idea

Combinations, quickly

A combination is a choice of some objects from a group where the order does not matter. Choosing Ana and Ben for a team is the same choice as choosing Ben and Ana.

The number of ways to choose \(r\) objects from \(n\) is written \(\binom{n}{r}\), read "\(n\) choose \(r\)". (Some books and calculators write \(C(n, r)\) or \(^nC_r\).) Choosing 2 letters from A, B, C, D, E gives 10 combinations: each one is a line joining two of the five points below.

ABCDE AB AC AD AEBC BD BECD CEDE 10 lines: 5 choose 2 = 10

The formula is

\[\binom{n}{r} = \frac{n!}{r!\,(n - r)!},\]

where \(n!\) ("\(n\) factorial") means \(n \times (n - 1) \times \cdots \times 2 \times 1\).

Why? If order did matter, there would be \(n\) choices for the first object, \(n - 1\) for the second, and so on: \(n \times (n - 1) \times \cdots \times (n - r + 1)\) ordered choices. But each unordered group of \(r\) objects appears in \(r!\) different orders, so the ordered count is \(r!\) times too big. Dividing by \(r!\) fixes it. For 2 from 5: \(5 \times 4 = 20\) ordered pairs, each pair counted twice (AB and BA), so \(\tfrac{20}{2} = 10\). (The full story is in Counting.)

A handy shortcut: for small \(r\), \(\binom{n}{r} = \dfrac{n \times (n-1) \times \cdots}{r \times (r - 1) \times \cdots \times 1}\) with \(r\) factors on top and on the bottom. So \(\binom{10}{3} = \dfrac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120\).

The key idea: every group is equally likely

When objects are chosen at random (from a well-shuffled pack, from a hat, by a fair lottery machine), every possible group of \(r\) is equally likely. So the basic rule of level 3 applies, with groups as the outcomes:

\[P(\text{event}) = \frac{\text{number of groups that give the event}}{\binom{n}{r}}.\]

The bottom is "all the ways to choose". The top is "the ways that work".

Counting the groups that work

Usually the good groups are built in parts, and you multiply the number of ways to make each part (the product rule: choose this part and that part).

For example, to choose 3 people from 6 girls and 4 boys so that exactly 2 are girls: choose the 2 girls, \(\binom62 = 15\) ways, and choose the 1 boy, \(\binom41 = 4\) ways. That is \(15 \times 4 = 60\) groups.

Two methods, one answer

Many of these questions can also be done step by step with the multiplication rule from level 5 (one pick after another, without replacement). Both are right. The combination method counts unordered groups; the step-by-step method follows one order of picks. Doing a question both ways is an excellent check, and worked example 2 does exactly that.

Worked examples

1. A committee by lot. A club has 10 members: 6 girls and 4 boys. Three names are drawn from a hat to form a committee. Find the chance that (a) all three are girls, (b) exactly two are girls, (c) a particular member, Zoe, is on the committee.

  • All possible committees: \(\binom{10}{3} = 120\).
  • (a) All girls: choose 3 of the 6 girls, \(\binom63 = 20\). So \(P = \tfrac{20}{120} = \tfrac16\).
  • (b) Exactly 2 girls: \(\binom62 \times \binom41 = 15 \times 4 = 60\). So \(P = \tfrac{60}{120} = \tfrac12\).
  • (c) Zoe is on it: Zoe takes one place, and the other 2 places are filled from the other 9 members: \(\binom92 = 36\). So \(P = \tfrac{36}{120} = \tfrac{3}{10}\). That is \(\tfrac{3}{10}\) because 3 of the 10 members are chosen, and each member has the same chance.

2. The same question step by step. All three girls, one draw at a time:

\[\frac{6}{10} \times \frac59 \times \frac48 = \frac{120}{720} = \frac16. \checkmark\]

Exactly two girls, step by step, needs care: the boy can be drawn first, second or third (GGB, GBG, BGG), and each order has chance \(\tfrac{6 \times 5 \times 4}{10 \times 9 \times 8} = \tfrac{120}{720} = \tfrac16\). Three orders: \(3 \times \tfrac16 = \tfrac12\). ✓ The combination method counted all three orders at once.

3. Two aces. Two cards are dealt from a full pack. Find \(P(\text{both aces})\).

  • All 2-card hands: \(\binom{52}{2} = \tfrac{52 \times 51}{2} = 1326\).
  • Hands of 2 aces: \(\binom42 = 6\).
  • \(P = \tfrac{6}{1326} = \tfrac{1}{221}\), the same as the step-by-step answer in level 5.

4. A full house. In poker, a 5-card hand is a full house if it has three cards of one rank and two of another (like three 8s and two kings). Find its chance.

  • All 5-card hands: \(\binom{52}{5} = 2\,598\,960\).
  • Build a full house in four steps. Choose the rank for the three: 13 ways. Choose 3 of its 4 suits: \(\binom43 = 4\). Choose a different rank for the pair: 12 ways. Choose 2 of its 4 suits: \(\binom42 = 6\).
  • Number of full houses: \(13 \times 4 \times 12 \times 6 = 3744\).
  • \(P = \tfrac{3744}{2\,598\,960} \approx 0.00144\), about 1 in 694.

Why \(13 \times 12\) and not \(\binom{13}{2}\)? Because the two ranks play different roles: three 8s and two kings is a different hand from three kings and two 8s. Order matters between the roles, so we do not divide by 2.

5. The lottery. In a 6/49 lottery, all possible tickets: \(\binom{49}{6} = 13\,983\,816\).

  • Jackpot: only 1 ticket matches all six. \(P = \tfrac{1}{13\,983\,816}\).
  • Exactly 3 of your numbers drawn: your ticket has 3 of the 6 winning numbers, \(\binom63 = 20\) ways, and 3 of the 43 losing numbers, \(\binom{43}{3} = 12\,341\) ways. That is \(20 \times 12\,341 = 246\,820\) tickets. \(P = \tfrac{246\,820}{13\,983\,816} \approx 0.0177\), about 1 in 57.

Doing this for every number of matches gives the full picture:

Matches Tickets Chance
6 1 about 1 in 14 million
5 258 about 1 in 54 000
4 13 545 about 0.00097
3 246 820 about 0.018
2 1 851 150 about 0.132
1 5 775 588 about 0.413
0 6 096 454 about 0.436

The ticket counts add up to 13 983 816, as they must. Matching no numbers at all is the most likely result.

Try it

  1. Work out \(\binom62\), \(\binom{10}{3}\) and \(\binom{52}{2}\).
  2. A bag holds 5 red and 4 blue balls. Three are taken at random. Find \(P(\text{all red})\) and \(P(\text{exactly 2 red})\).
  3. A committee of 4 is chosen at random from 12 people, including Ana. Find the chance that Ana is on it.
  4. Find the chance that a 5-card hand is all hearts, and the chance it is a flush (all five cards the same suit).
  5. A four of a kind is a 5-card hand with all four cards of one rank plus one other card. How many are there, and what is the chance?
  6. In the 6/49 lottery, check the number 258 in the table for exactly 5 matches.
  7. A box holds 20 light bulbs, and 3 of them are faulty. An inspector tests 4 at random. Find the chance that none of the 4 is faulty, and the chance that at least one is.
  8. A small lottery: choose 3 numbers from 1 to 10. Find the chance of the jackpot, and the chance of matching exactly 2 of the 3.
Answers
  1. \(\binom62 = \dfrac{6 \times 5}{2} = 15\). \(\binom{10}{3} = 120\) (worked out above). \(\binom{52}{2} = \dfrac{52 \times 51}{2} = 1326\).
  2. All groups: \(\binom93 = 84\). All red: \(\binom53 = 10\), so \(\dfrac{10}{84} = \dfrac{5}{42}\). Exactly 2 red: \(\binom52 \times \binom41 = 10 \times 4 = 40\), so \(\dfrac{40}{84} = \dfrac{10}{21}\).
  3. Committees with Ana: \(\binom{11}{3} = 165\). All committees: \(\binom{12}{4} = 495\). \(P = \dfrac{165}{495} = \dfrac13\) (which is \(\dfrac{4}{12}\): 4 of 12 people are chosen).
  4. All hearts: \(\binom{13}{5} = 1287\), so \(P = \dfrac{1287}{2\,598\,960} \approx 0.000495\). A flush can be in any of the 4 suits: \(4 \times 1287 = 5148\), so \(P \approx 0.00198\), about 1 in 505.
  5. Choose the rank: 13 ways. All 4 cards of it: 1 way. The fifth card: any of the other 48. So \(13 \times 48 = 624\) hands, and \(P = \dfrac{624}{2\,598\,960} = \dfrac{1}{4165} \approx 0.00024\).
  6. Choose 5 of the 6 winning numbers, \(\binom65 = 6\), and 1 of the 43 others, 43 ways: \(6 \times 43 = 258\). ✓
  7. No faulty bulb: choose all 4 from the 17 good ones. \(\dfrac{\binom{17}{4}}{\binom{20}{4}} = \dfrac{2380}{4845} \approx 0.491\). At least one faulty: \(1 - 0.491 = 0.509\).
  8. All tickets: \(\binom{10}{3} = 120\), so the jackpot is \(\dfrac{1}{120}\). Exactly 2: choose 2 of the 3 winning numbers, \(\binom32 = 3\), and 1 of the 7 others, 7 ways: \(21\) tickets, \(P = \dfrac{21}{120} = \dfrac{7}{40}\).

Common mistake

Mixing ordered and unordered counts. If the bottom counts unordered groups (\(\binom{n}{r}\)), the top must count unordered groups too. If you count orders on top (like \(6 \times 5 \times 4\)) you must count orders on the bottom (\(10 \times 9 \times 8\)). Mixing them gives answers that are too big by a factor like \(3! = 6\).

Forgetting the "other" part. "Exactly 2 girls" in a group of 3 means 2 girls and 1 boy. Counting only \(\binom62\) forgets to choose the boy. Every place in the group must be filled from somewhere.

Where it leads