Assuming the outcomes are equally likely recovers the counting rule

0:00You have a table of numbers, one against each outcome. How do you get from that to the value of an event? You add up the numbers sitting on the outcomes the event holds. That is the whole rule, and it sounds like a new one. It is not. It is the third condition, applied to the smallest events there are. An event is the union of the one-outcome events of its own members.

0:28No two of those can both happen, because the outcomes are different. So the third condition applies to them, one at a time, all the way along. Adding is not a convenience here. It is the only thing the conditions allow you to do, and they allow it every time. That deserves to be tested rather than believed. Take a function that assigns a number to every event, and ask whether it is the total of its own one-outcome values.

1:02Build one by totalling a table, and it is. On four outcomes it agrees on all sixteen events. On eight outcomes it agrees on all two hundred and fifty-six. Now take a function that is not additive. Give every event the square of its size, over the square of the list's size. It assigns a number to every event, so it looks like the same kind of object. But it is not the total of its own one-outcome values.

1:33It differs from that total on eleven of the sixteen events. The five it gets right are the empty event and the four one-outcome ones, which is to say the five where there is nothing to add. So totalling is not a fact about functions. It is a fact about functions that satisfy the third condition, and that is exactly what makes it legitimate. Three pens come off a machine, and each is recorded good or bad.

2:05Write out what can happen and there are eight outcomes. Weight them evenly, an eighth each. Now ask for exactly one bad pen. Walk the list and keep the outcomes that answer to that. Three of them do. Three eighths. Now ask for at least two bad pens. Four outcomes answer to that, so a half. Those two events cannot both happen, so their values add. Three eighths and a half is seven eighths, and not one.

2:36Which tells you they do not cover the list. The outcome they leave out is the one with no bad pen at all. Nothing here needed a formula. It needed the list, the table, and the willingness to add. Now the same machinery on an experiment that is not even. A coin is tossed twice. The outcome list is the familiar four. But the numbers on them are a quarter, a seventh, two sevenths, and nine twenty-eighths.

3:08Put those over a common denominator and they are seven, four, eight, and nine twenty-eighths. They total twenty-eight twenty-eighths, which is one. None of them is negative. So this table passes both tests, and it is a perfectly legitimate probability. Look at the two middle outcomes. One head and one tail, either way round. The same description, and different numbers. One of them is worth exactly twice the other. Nothing in the three conditions forbids that.

3:43The table is not uniform, and it does not have to be. Ask this table for the event that both tosses agree. That holds two heads and two tails. Seven twenty-eighths plus nine twenty-eighths. Sixteen twenty-eighths, which is four sevenths. Now ask the rule you had before. Two outcomes suit you, out of four possible ones. A half. Four sevenths is not a half. They differ by a fourteenth, and that is not a rounding error.

4:17Nothing was miscalculated. Run the same two routes on the even table and they agree exactly. The counting rule did not make an arithmetic mistake here. It made an assumption, and the assumption was false of this coin. Before deciding what that means, one warning. Ask the same weighted table for the event holding two heads on its own. Its value is a quarter, because that is the number sitting on it.

4:49And counting says one outcome out of four, which is also a quarter. Here the two agree. So a single agreement proves nothing whatever. Sweep the whole domain and counting is right on four of the sixteen events of this table. But two of those four are the empty event and the sure event. Every admissible table gets those two right, because the conditions force them. So the informative score is two right out of the other fourteen.

5:20Counting is not right on this table. It is right about a quarter of it, mostly by accident. How special is the case where counting works? Restrict the four entries to whole twenty-eighths, just to make the question answerable. Hand the twenty-eight units out among the four outcomes in every possible way. There are four thousand four hundred and ninety-five admissible tables. Exactly one of them is uniform. Now count the ones counting gets right.

5:56On the event that both tosses agree, two hundred and twenty-five tables give the same answer counting does. On the event holding two heads, two hundred and fifty-three do. Those are large numbers, and they mean almost nothing. Because the number of tables counting gets right on every single event is one. And it is the uniform one. A rule can agree with you on hundreds of tables at one question and still be the wrong rule.

6:29So make the assumption, out loud. Suppose every outcome carries one and the same number. Call it p. How much freedom does that leave? Do not assert the answer, search for it. Offer a list of candidate values, hand each one to every outcome at once, and ask the resulting table the two conditions. On two outcomes exactly one candidate survives, and it is a half. On three, one survives, a third.

7:02On four, six and eight, the same, one each time. Uniformity does not suggest the value. It determines it, and there is no second possibility. And it is worth knowing which condition does that work. Withhold the totalling condition and every candidate in range survives. The range condition alone rejects only the two candidates that were never numbers between nought and one. It is the totalling that pins p down.

7:34N equal numbers adding to one leaves each of them at one over n. Now the rule comes back. An event of size m is the union of m one-outcome events. Each of them is worth one over n. So the event is worth m over n. Favourable outcomes over possible outcomes. Test it rather than believing it. On outcome lists of every size from two to eight, the uniform table and the counting rule agree on every event, with no exceptions.

8:10That is the rule you already had. It has not been contradicted, and it has not been replaced. It has been given a hypothesis. Before, it was what probability meant. Now it is a theorem, and it is true exactly when the outcomes are interchangeable. Which is a claim about the experiment in front of you, and something you can be wrong about. Most of the time you will be right.

8:40Build a deck as a rank against a suit and there are fifty-two cards. Nothing distinguishes one from another as far as the draw is concerned, so the hypothesis holds. Thirteen of the fifty-two are diamonds, so a diamond is worth a quarter. Twenty-six are black, so a black card is worth a half. A bag holds four red discs, three blue and two yellow. Nine discs. Four ninths, a third, and two ninths.

9:12Those three values total one, and they had better. The three colours split the list. Every disc is one colour, and no disc is two. That is not a coincidence to be noticed afterwards. It is the check that the outcome list was built properly in the first place. Two men and two women, and a committee of two is chosen from them. The outcome list is the selections, and there are six.

9:42Now sort them by how many men they hold. No man, one of the six. One man, four of them. Two men, one. One, four, one, which totals six, so nothing has been double counted and nothing lost. A sixth, two thirds, a sixth. And those add to one, as a family that splits the list must. Notice what the outcome list is here. It is not the four people.

10:10It is the six committees, and it had to be built before anything could be counted. Sometimes the sizes are not obvious, and have to be counted themselves. Four cities are visited in some order. There are twenty-four orders, and they are interchangeable. A before B. Twelve of the twenty-four, so a half, which you could have argued by symmetry. A before B before C. Four of them, so a sixth.

10:42A first with B last. Two, so a twelfth. A in the first or second place. Twelve, so a half. A immediately before B. Six, so a quarter. That last one sits inside A before B, and not the other way round. Two different events that can both happen, and the smaller one is worth half the larger. Now a list nobody writes out. Seven cards are drawn from fifty-two.

11:13The number of hands is a hundred and thirty-three million, seven hundred and eighty-four thousand, five hundred and sixty. You do not list those. You count them, with the selection arithmetic from earlier in the course. How many hold all four kings? Choose the four kings, then three cards from the forty-eight that are not kings. One in seven thousand seven hundred and thirty-five. How many hold exactly three kings?

11:44Nine in one thousand five hundred and forty-seven. Put both over seven thousand seven hundred and thirty-five and the second is forty-five. Exactly three kings and all four kings cannot both happen. So they add, and at least three kings is forty-six in seven thousand seven hundred and thirty-five. The adding is licensed by the third condition, not by convenience. One thing the denominator is not. Three of five relay teams are placed first, second and third.

12:21Sixty orders. One named order is one of them, a sixtieth. Those same three teams in any order is six of them, a tenth. In every one of these the denominator counted the outcomes. Never the events. A list of four outcomes carries sixteen events. Four and sixteen are not the same number, and sixteen is never the denominator of anything. The function is asked about events. The counting rule divides by outcomes.

12:54Those are different collections, and confusing them is the easiest way to get a wrong answer that looks right. Here is where the hypothesis usually breaks. A fair coin is tossed four times. One point for a head, one and a half points off for a tail. Sixteen outcomes, and they are genuinely interchangeable. But how many different amounts can you finish with? Five. Minus six, minus three and a half, minus one, one and a half, and four.

13:28Four heads pays four, and four tails costs six. So the outcome list has sixteen members and the amounts have five. And the amounts are not interchangeable at all. They come out one sixteenth, a quarter, three eighths, a quarter, one sixteenth. Treat the five amounts as your outcome list and apply the counting rule and you get a fifth for each of them. That is wrong about all five.

13:57The coin was fair. The outcomes were interchangeable. The list you chose was not the list they were interchangeable on. One more, and it is the sharpest. A die is made with two faces carrying a one, three carrying a two, and one carrying a three. What is the probability of a two? There are two answers, and the difference between them is entirely a choice of list. Take the six faces as the outcome list.

14:29They are physically identical, so the hypothesis holds of them. Three of the six carry a two, so a half. The other three carry a one or a three, also a half. Take the three values as the outcome list instead. Three of them, so a two would be worth a third. A half and a third are different numbers, and only one of them is the answer. You can see which, because the faces settle what the values inherit.

14:59Under the faces, the three values come out a third, a half, and a sixth. That is not uniform, so the second hypothesis is false. At most one of the two can hold, and the die tells you which. Build the die differently, with two faces each carrying a one, a two and a three, and both readings give a third. So the clash was never about lists in general.

15:26It was about that particular die. So what was actually done here. The rule you started with was taken away and given back. It came back weaker in one specific sense. It now carries a condition it did not carry before. The outcomes have to be interchangeable, and that has to be true of the experiment and not merely convenient. When it is true, and it very often is, nothing changes.

15:57Cards, discs, committees, orders, hands of seven. Count what suits you, count what is possible, divide. When it is false, the counting rule does not become approximate. It becomes wrong, and it gives you no warning at all. The weighted coin was off by a fourteenth. The die read off its values instead of its faces was off by a sixth. Neither of those announced itself. The only defence is the one thing the axioms made you do.

16:29Write the hypothesis down, and then check it.