Theoretical probability: counting favourable outcomes when all are equally likely

0:00Everyone learns the same formula for a chance: favourable outcomes over possible outcomes. Ask for a four on a die and it answers one in six, without your rolling anything. That is theoretical probability, and it is genuinely powerful. But look at where the six came from. It is a count of faces, and counting faces says nothing about chance. So the formula is not a definition. It is a result, true under one condition, and the whole of the mathematics is hiding inside that condition.

0:38The condition is that the things being counted are equally likely. This video is about where that permission comes from, and about the harder question underneath: what you should be counting at all. Start with the die, and ask the awkward question. Why should any face be as likely as any other? The answer is not about probability but about shape. Pick a cube up and set it down again, any way you like.

1:09There are exactly twenty-four ways to do it, each one a rearrangement of the six faces. Take any face and any other, and exactly four of those turns carry the first onto the second. Four, for every single one of the thirty-six pairs. No face is special: nothing you could say about one is not equally true of the rest. That is what interchangeable means, and it is what forces the six chances to be equal.

1:41The denominator six is that argument, written down as a number. Now glue a small weight to one face, and watch the argument fall apart. Only the turns that leave the weight underneath still apply. Four survive out of twenty-four. And under those four the faces no longer mix. The weighted face stays put, the one opposite stays put, and the four around the side swap only among themselves. Three separate families where there was one.

2:15The up face can no longer be carried onto the front face, so nothing makes them equal. Notice what did not change. There are still six faces, and six is still the right count. What was lost was the reason for dividing by it. One over six is a fact about a fair cube, not about the number six. So write the formula with its condition attached, and read the whole thing every time.

2:45If, and only if, the outcomes are equally likely, the probability of an event is the number of favourable outcomes divided by the total. Favourable sounds like approval. It means only that the outcome counts as the event you asked about. Ask for a four on a fair die: one outcome qualifies out of six, so one sixth. As a decimal that is nought point one six seven, to three places.

3:16As a percentage, sixteen point seven. And the six answers, one per face, add to exactly one: the check that you counted outcomes and not something else. Now the harder half, and it has nothing to do with fairness. It is about what you decide to count. Write out the word probability, and take one letter from it at random. The word runs to eleven letters. Nine are different, because two letters appear twice: the B and the I.

3:50So what is the chance of drawing a B? Two reasonable answers come back, and both use the same formula. One says: there are two B positions among eleven positions, so two elevenths. The other says: there are nine different letters and B is one of them, so one ninth. Both divide favourable by possible. They cannot both be right. The formula does not settle it, because the formula was never the disagreement.

4:23The disagreement is about what an outcome is. Test each collection against the condition. The eleven positions are a fair thing to count: no slot has anything the others lack. The nine different letters are not. Work out what each is worth: seven of the nine come out at one eleventh, and two at two elevenths. Two different weights in one collection. Dividing by nine treats them as though they were the same, and they are not.

4:54So the answer is two elevenths, about eighteen point two per cent. One ninth is not bad arithmetic. It is correct arithmetic on the wrong collection. That example is one case of a rule worth carrying around. Take any set of equally likely outcomes and bundle them into groups. The groups are themselves equally likely exactly when every group holds the same number of outcomes. Three sweets, one of each flavour: groups of size one all round, so a third each, and here the objects and the categories coincide.

5:35The letters of that long word: group sizes of one and two mixed together, so no. The odd and even faces of a die: three each, so yes, a half. Whenever a chance problem feels slippery, this is usually why: somebody has grouped the outcomes and then counted the groups. Here is a question that deserves a longer answer than it usually gets. A bag holds three purple marbles and seven blue ones, and you draw one without looking.

6:09Ten marbles, the same size, reached for the same way. Equally likely, and the rule agrees: ten groups of one. So each individual marble is one tenth. But nobody asks about an individual marble. They ask about colour, and colour is a grouping: three in one group and seven in the other. Different sizes, so the two colours are not equally likely. Purple is three tenths and blue is seven tenths.

6:40Those still add to one, and neither is a half. The marbles are equally likely; the colours are not; both sentences are about the same bag. Some questions have two outcomes and no symmetry at all. Turn a key in a car: it starts, or it does not. Two outcomes. Is it a half? Obviously not, and there is nothing in the car resembling the twenty-four turns of a cube.

7:10The same trap catches a subtler case. A baby is born; the two possibilities are often treated as an even split. They are close to even but not exactly even, and nothing here could tell you the true figure. Say the outcomes are not equally likely, so the formula does not apply, and finding the number would take data rather than counting. Two outcomes never make a half on their own.

7:39Symmetry does that, or evidence does, and if you have neither you have no number. When the condition genuinely holds, the counting is quick. An even number on a fair die: three faces qualify out of six, so a half. Ten cards numbered one to ten, face down: five are even, so a half again, the same number by a different route. A number above four on a die: five and six, two of six, so a third.

8:11A bag with three purple, two blue and one green: three of the six are not purple, so a half. And the short word peace, five letters with the E doubled: a P, an E or a C is four fifths, and not an E is three fifths. Check that the outcomes are equally likely, then count. A spinning wheel divided into eight sectors, numbered one to eight, and one detail worth noticing first.

8:42Nobody measured those sectors. You are told they are equal, and being told is a good reason, as long as you know that is what happened. A drawing is not evidence about a real wheel. Any one number is one eighth. An odd number: four of the eight qualify, so a half. A number greater than two: six of the eight, so three quarters. A number less than nine: all eight, so the answer is one, an event that cannot fail.

9:13A multiple of three: three and six, so two eighths, a quarter. Sometimes the condition is easy and the counting is the work. Roll two dice and ask for a total that is prime and greater than five. Only seven and eleven qualify, but you cannot count totals: totals are groups, and they come in different sizes. Count the pairs instead: thirty-six, equally likely, and eight of them total seven or eleven.

9:45Eight thirty-sixths, which is two ninths. Or take the digits one to four and arrange all four, no digit repeated. There are twenty-four such numbers, and twelve end in an even digit, so an even number is a half. In each case the formula did nothing. The counting did everything, and it had to be done on the right collection. Ask what the outcomes are, ask what makes them equally likely, and only then divide.