Listing every outcome: the sample space

0:00Before you can work out any probability, you need the list of things that could happen. That list has a name, a symbol and a size. The list is called the sample space and it gets the letter S, written inside braces with commas between the entries. One entry in it is called an element. How many elements it holds is written n of S. Roll a die and S is one, two, three, four, five, six, so n of S is six.

0:31That is the whole of the notation, and the rest of this video is about the part nobody tells you: the list is something you choose. Two rules come with the definition, and they are the obvious ones. First: nothing possible may be left out. Write the outcomes of one die as one, two, three, four and five and you have not made a smaller experiment; you have made a wrong list.

1:00Every probability you compute from it is then divided by five when it should be divided by six. The rule sounds like tidiness and it is really about the denominator. Second: nothing may be written twice. If six appears at both ends of your list, that same denominator goes up to seven and every answer shrinks. Both rules are guarding the same thing, which is the count. Keep them in mind, because in a few minutes you will meet a list that obeys both of them and is still not a sample space.

1:40Now toss two coins at the same time, and be careful here. It is tempting to say there are three results: two heads, two tails, or one of each. Set it out as a table instead, with a column for the first coin and a column for the second. Heads and heads. Heads and tails. Tails and heads. Tails and tails. Four rows, because the coins are two different coins, and the middle two are different outcomes even though they look alike from a distance.

2:13This is not pedantry. Fuse them into a single entry and you have a three-item list, and asking for one head and one tail gives one in three. Keep them apart and it is two of four, which is one half. One of those numbers is right, and it is the one that comes from the four-row table. Five quick sample spaces, to fix the notation. Whether it rains tomorrow: rain, no rain. Size two.

2:44How a team's match ends: win, lose, draw. Size three. One coin: heads, tails. Size two. One die: one through six. Size six. Two coins: the four rows we just built. Size four. Look at the second one for a moment, because something is quietly wrong with it and almost nobody says so. Win, lose and draw is a perfectly good sample space whose three entries are nothing like equally likely.

3:17Which brings us to the choice. Take tomorrow's weather again, with the list rain and no rain. By writing those two you have committed to caring only whether it rains at all. If you want to tell a drizzle from a downpour, that list cannot do it, and you need a wider one: no rain, drizzle, light rain, heavy rain. Same tomorrow. Two sample spaces, one of size two and one of size four.

3:46Neither is the right one in general, and neither is more accurate than the other. The list is fixed by the question you are asking, not by the weather. So a sample space is not something you discover about an experiment. It is something you decide. And that decision settles something the rules never mention. Here is a box holding twelve balls: five green and seven red, and you draw one without looking.

4:17One sample space is the twelve balls themselves, and every one of them carries the same weight, one twelfth. The other is just the two colours, green and red. Both lists leave nothing out and repeat nothing, so both are legal. But green is five of the twelve and red is seven of the twelve, so those two entries are not equally likely at all. Count in the two-colour list and you will say the chance of green is one half. It is five twelfths.

4:50The two-colour list is a sample space you must not divide into. So when is grouping outcomes together safe? There is an exact answer, and it is worth more than the example. Grouping keeps things equally likely precisely when the groups you make are all the same size. Take a die and group it into odd and even: three faces in each group, so the two groups are equally likely, one half each, and you may count in them.

5:23Take the same die and group it into over four and not over four: that is two faces against four, so it is two thirds against one third and you may not. The balls fail for the same reason, five against seven. Nothing about coarsening is wrong in itself. What matters is whether the groups came out even. One practical thing before the hard part. You often need how big a sample space is without needing to write it out.

5:56A die rolled together with a coin: each of six faces can meet each of two, so twelve. Three snacks with two drinks: six pairs. A coin tossed with one of six numbered cards: twelve again, the same shape as the first. Four numbered balls, one drawn and put back, then another drawn: four times four, sixteen. Do not put the first one back and the second draw has only three left, so four times three, twelve.

6:29Each of those sizes comes straight off the experiment, and none of them needed a list. Now the promised list that obeys both rules and is not a sample space. Toss three coins, and write down only how many heads you got. Which of these four lists is the sample space? Zero, one, two, three. Or one, two, three. Or zero, one, two. Or zero, one, two, three, four. Two of them are easy: one, two, three leaves out zero, and zero, one, two leaves out three, so both break the first rule.

7:08Now look hard at zero, one, two, three, four. It leaves nothing out, since every count that can happen is on it. It writes nothing twice. It passes both of the rules and it is still wrong, because three coins cannot give you four heads. There is a third requirement nobody writes down: everything on the list must be something the experiment can actually produce. That extra entry is not untidiness, and here is what it costs.

7:41Ask for the probability of exactly two heads. Count inside the five-item list and you get one out of five. Take the impossible four away and count inside the correct four-item list, and you get one out of four. Now count where the outcomes really are equally likely, which is the eight ways three coins can land, three of which show exactly two heads. Three out of eight. Three different answers, and only the last one is right.

8:14The impossible entry inflated the denominator, and the correct list was still the wrong thing to count in, because its four entries carry one eighth, three eighths, three eighths and one eighth. So put the two lists for three coins side by side and look at what you have. Eight outcomes that are equally likely and that you may count in, and four counts of heads that are a perfectly correct sample space you may not count in.

8:46One experiment, two lists, and the arithmetic only works in one of them. Which is the whole lesson, and it explains a small trap in the wording of questions too. Suppose you want a random integer between minus five and plus five: are the ends included? Leave them out and there are nine integers; put them in and there are eleven. Nothing in the words settles it, so the sample space has to be stated rather than assumed.

9:17Three things to carry away, then. Leave nothing out. Write nothing twice. Put nothing on the list that cannot happen. And one more, which is the one that matters: choose the list your question needs, and check whether its entries are equally likely before you divide by how many there are.