Breaking Things Down

1. Introduction.

By itself, the method of obstacles only works in a narrow range of cases. For example, suppose two tickets will be drawn, one at a time, at random from the box shown below: A A Z Z If you want to find the chance the same letter shows on both tickets, the method of obstacles can handle this easily. The first ticket can be any ticket but the second must have the same letter as the first one: any ticket same letter as first

4/4 1/3 The answer is 1/3. Now suppose you want to find the same chance with a slightly different box: A A Z Z Z You can’t use an obstacle course, because the chance the second let- ter is the same as the first could be 1/4 or 1/2, depending on whether

Page 1 the letter on the first ticket is A or Z. Something else is required. But before reading on, you might try to guess which box gives a better chance of getting the same letter twice: the four ticket box or the five ticket box. Here is the method. You break down the possibility: same letter shows on both tickets into different ways it can happen: get A on both tickets or get Z on both tickets Next, you find the chances for these (using obstacles): get A on both tickets: 2/5 × 1/4 = 2/20 or get Z on both tickets: 3/5 × 2/4 = 6/20 And the last step is to add the chances of the different ways: 2/20 + 6/20 = 8/20 = 2/5. That ends the method. It leads to a 2/5 chance the same letter will come up twice. This is bigger than the chance for the four ticket box, which is 1/3. (You might have guessed the chance would go up by imagining an extreme case: that the box contained 2 As and 100 Zs, say. Then the chance would be almost certain both letters would be the same, because they would both be very likely to be Zs. So even adding a single Z to the box will put the chance up.)

Example 1. Someone shuffles a deck of cards and deals out, one at a time, five cards. Find the chance there is not more than one ace in the five cards. Do you think the chance will turn out to be closest to 50%, 75%, or 100%? Answer. First, break down the possibility: not more than one ace Ann wins Bob wins as follows. There won’t be more than one ace if either: there are 0 aces in the five cards 3/5 or there is 1 ace in the five cards There are two chances to find. The first one is straightforward, be-

Page 2 ace not an ace not an ace not an ace not an ace

4/52 48/51 47/50 46/49 45/48

cause 0 aces in five cards can be written as an obstacle course:

1st not ace 2nd not ace 3rd not ace 4th not ace 5th not ace

48/52 47/51 46/50 45/49 44/48 The chance there are 0 aces in the five cards: 48/52 × 47/51 × 46/50 × 45/49 × 44/48 ≈ 0.6588 That does the first line of the breakdown: not more than one ace in the five cards if either: there are 0 aces in the five cards: 0.6588 or there is 1 ace in the five cards

The second line takes most of the work. The possibility of 1 ace (in the five cards) can’t be represented as an obstacle course, because the first card dealt does not have to meet any condition, and nor do any of the other four cards. The possibility has to broken down further. A natural way to do it is according to the posi- tion of the ace in the five cards:

there is 1 ace in the five cards if either: there is 1 ace in the five cards and it is the 1st card dealt or there is 1 ace in the five cards and it is the 2nd card dealt or there is 1 ace in the five cards and it is the 3rd card dealt or there is 1 ace in the five cards and it is the 4th card dealt or there is 1 ace in the five cards and it is the 5th card dealt

More informally:

the 1 ace in the five cards could be: the 1st card dealt or the 2nd card dealt

Page 3 or the 3rd card dealt or the 4th card dealt or the 5th card dealt

There are five possibilities in the breakdown, so five chances to find. Start with the first one, but proceed with caution. Sometimes begin- ners misread the first possibility as saying only “the 1st card dealt is an ace”, and conclude the chance is 4/52. The longer version of the first possibility: there is 1 ace in the five cards and it is the 1st card dealt helps you avoid the mistake. It reminds you that when the first card is the ace, the other four cards cannot be aces, and that, in turn, points you at the right obstacle course: ace not an ace not an ace not an ace not an ace

4/52 48/51 47/50 46/49 45/48 The chance for the first line is: 4/52 × 48/51 × 47/50 × 46/49 × 45/48 ≈ 0.0599 Writing that in:

there is 1 ace in the five cards if either: 1 ace in the five cards and it is the 1st card dealt: 0.0599 or 1 ace in the five cards and it is the 2nd card dealt or 1 ace in the five cards and it is the 3rd card dealt or 1 ace in the five cards and it is the 4th card dealt or 1 ace in the five cards and it is the 5th card dealt

The next line of the breakdown: there is 1 ace in the five cards and it is the 2nd card dealt

Page 4 leads to the obstacle course: not an ace ace not an ace not an ace not an ace

48/52 4/51 47/50 46/49 45/48 The chance is: 48/52 × 4/51 × 47/50 × 46/49 × 45/48 ≈ 0.0599, which is equal to the chance for the first line. This is not a fluke. The five possibilities in the breakdown have the same chance, because the numerators in the fractions under the obstacles are always 4, 48, 47, 46, 45, although not necessarily in that order, and the denominators are always 52, 51, 50, 49, 48. Putting in these chances:

there is 1 ace in the five cards if either: 1 ace in the five cards and it is the 1st card dealt: 0.0599 or 1 ace in the five cards and it is the 2nd card dealt: 0.0599 or 1 ace in the five cards and it is the 3rd card dealt: 0.0599 or 1 ace in the five cards and it is the 4th card dealt: 0.0599 or 1 ace in the five cards and it is the 5th card dealt: 0.0599

To get the chance of 1 ace in the five cards, add the chances for the possibilities in the breakdown: 0.0599 + 0.0599 + 0.0599 + 0.0599 + 0.0599 = 0.2995

The rest is routine.

Now you have the chances for original breakdown: not more than one ace in the five cards if either: 0 aces in the five cards: 0.6588 or 1 ace in the five cards: 0.2995 The sum is 0.9583. The chance is nearly 96%. A poker hand contains 5 cards, so around 96% of poker hands will have only one ace or none.

Page 5 That does Example 1.

If you take a moment to look back at the answer, you will see a fair amount of reasoning expressed in ordinary English. As you might guess, people don’t usually write it all out like this. The next section presents a way to express the reasoning briefly.

2. A key notation.

No new principles for finding chances are presented in this section. The only goal here is to show you a very helpful notation for listing the possibilities in a breakdown. Example 2. A tiny town contains 20 voters: 12 Democrats, 6 Republicans, and 2 independent voters. A polling organization plans to interview a sam- ple of 4 people from the town. The organization puts twenty tickets in a box, one for each voter:

twelve D s six R s two I s 20 tickets in box To get the voters it will interview, the organization draws out 4 tick- ets, one at at time, at random, from the box. More than half of the voters (12 out of 20) in the town are Democrats. Find the chance more than half the sample will be Democrats.

Answer. More than half the voter in the sample will be Democrats if either: there are 3 Democrats in the sample or there are 4 Democrats in the sample The 3 Democrats possibility is going to have to be broken down fur- ther. When faced with something you want to break down, a good first step is to pause a little and think of one way it could happen. For 3 Democrats, you might think of something like the following: the 1st voter in the sample is a Democrat, the 2nd is also, the 3rd is a Republican, and the 4th a Democrat. This is clear enough. You can see how to calculate its chance. But you wouldn’t want write it out for each possibility in the breakdown. To get t