Being told one event happened shrinks the sample space you count against

0:00Here is the only question this whole topic is built on. You are about to bet on something. Somebody tells you that a different thing has already happened. Should that change your bet? Sometimes it should, sometimes it should not, and the point of the next twenty minutes is that there is a single mechanical answer to when and by how much. And I want to say at the start what the mechanism is not. Nothing physical changes. The coins have already landed. Nobody moves them.

0:34What changes is the list of outcomes you are entitled to count. That is a statement about your information, not about the world, and every formula in this topic is bookkeeping on top of it. Three fair coins on the table. Eight outcomes, and because the coins are fair each one carries a chance of one eighth. Two events matter here. The first is that at least two heads show. That holds four of the eight outcomes: three heads, or heads heads tails, or heads tails heads, or tails heads heads.

1:13The second is that the first coin shows tails. That also holds four: tails heads heads, tails heads tails, tails tails heads, tails tails tails. Four out of eight each, so each has a chance of one half. And they share exactly one outcome between them — tails, then heads, then heads — so the chance that both happen is one eighth. Those three numbers are the whole of the arithmetic. Everything from here is about which of them goes where.

1:49Now somebody tells you the first coin came up tails. Four of the eight outcomes contradict that, and they are gone. Not unlikely. Gone. Heads heads heads cannot be the truth if the first coin is tails. What is left is four outcomes, and you are allowed to treat that shorter list as a fresh experiment in its own right. That sentence is the entire idea of this topic, and everything after it is arithmetic.

2:20I am going to call the shorter list the reduced sample space. That is my name for it, not a standard one, but it is worth having a name for the thing that does all the work. So: among those four survivors, how many still give you at least two heads? One. Tails heads heads. The other three survivors have at most one head between them. And look at which outcome that is. It is the one both events already contained. That is not a coincidence about coins.

2:55To be favourable now, an outcome has to do two things. It has to agree with what you were told, or it is not on the list at all. And it has to be in the event you are asking about, or it is not favourable. Both conditions at once is exactly what the intersection of the two events means. So the favourable survivors are the shared outcomes, always, with no further argument.

3:26Which gives the answer straight away, as a ratio of two head counts. One favourable survivor. Four survivors. One in four. Compare that with the chance of at least two heads before anybody told you anything, which was one half. The information has halved it. Notice what is underneath that fraction. It is the number of outcomes in the event you were told about. You are standing inside that event and counting; there is nowhere else to count from.

3:58If you remember nothing else from this video, remember the shape of that fraction, because in a minute it will be written with probabilities instead of counts and the denominator will look arbitrary. It is not. It is the list you are standing in. Now one small piece of bookkeeping, and it takes three seconds. Divide the top of that fraction by eight, and the bottom by eight. The value cannot change; that is what dividing both parts means.

4:32The top was one outcome. One over eight is the chance of both events happening — the shared eighth we worked out earlier. The bottom was four outcomes. Four over eight is one half, which is the chance of the event you were told about. So the same answer, one quarter, is now written as a chance of both, divided by a chance of the one you were told about. And you watched the division happen, so you never have to memorise which one goes underneath.

5:07That is the definition, and it is the first thing here that gets set down formally. The probability of one event given another is the probability that both happen, divided by the probability of the one you were given. There is a condition attached, and it is not decoration. The probability underneath must not be nought. Two things go wrong at once if it is. The obvious one is that you would be dividing by nought. The real one is that an event of probability nought leaves you no reduced list at all. There is nothing to stand inside and count.

5:48So the answer there is not nought. It is not defined. There is a standard exercise item whose entire purpose is to make you write those two words instead of a number, and people still write nought. A word about the notation, because it is read wrongly more often than any other symbol here. The vertical bar has two sides and they do completely different jobs. On the right of the bar is the ground — the thing you were told, the event that shrank the list. On the left is what you are measuring on that ground.

6:27Read it aloud as: the chance of the left-hand thing, given the right-hand thing. Given. Told. Standing on. And the rule that follows is the one everybody gets backwards. You divide by the probability of the event on the right. Not the one you are asking about. The one you were told. If you ever lose it, go back to the counts. In the count form the denominator is visibly the list you are standing in, and no amount of forgetting can move it.

7:01So what happens if you turn the bar around? You get a different number. In general, a completely different number. Here is a pair that makes it unmissable. One event has probability six tenths, the other three tenths, and the chance that both happen is two tenths. Two tenths over three tenths is two thirds. Two tenths over six tenths is one third. Same shared part on top; different ground underneath; twice the answer one way round.

7:36They agree only when the two events are equally likely, because then the two denominators are the same number. And that is not a guess. I ran every pair of events you can build out of the three-coin experiment — nearly fifty nine thousand ordered pairs that share at least one outcome — and the two conditionals agree at eleven thousand seven hundred and sixty three of them. The pairs where they agree and the pairs that are equally likely are the same pairs. Not one exception.

8:09Five shapes of the same move, quickly, because the mechanism never changes and only the list does. Two children. Four ordered outcomes: boy boy, boy girl, girl boy, girl girl. Told that at least one is a boy, three survive, and one of those three is two boys. One third. Most people say one half before they see the list, because they think of boy girl and girl boy as one case. They are two outcomes, and both survive, which is why the answer is a third.

8:45Ten numbered cards. Told the number is over three, seven survive. Four of those seven are even. Four sevenths — and notice that is bigger than the plain chance of an even card, which was one half. Conditioning does not always shrink things. A school of a thousand with four hundred and thirty girls, a tenth of whom are in the twelfth year. The chance of a girl is nought point four three, the chance of a twelfth year girl is nought point nought four three, and dividing gives nought point one. Which is the tenth you were handed at the start, arriving back unchanged. That makes it an excellent check and a terrible puzzle.

9:31A die thrown three times, two hundred and sixteen outcomes. Told the first two throws were a six and a five, six outcomes survive, and one of them has a four on the third throw. One sixth — which is exactly the chance of a four on one throw. The information changed nothing at all, and that is the first hint of independence. A die thrown twice, told the two faces add to six. Five outcomes survive; two of them contain a four. Two fifths. Without the information, the chance a four shows at all was eleven thirty sixths.

10:14Now the case that separates the two forms of the formula, and it is the reason the definition is written in probabilities rather than in counts. Toss a coin. If it comes up heads, toss it again. If it comes up tails, throw a die instead. Eight outcomes. But they are not equally likely. The two head outcomes carry a quarter each. The six tail outcomes carry a twelfth each. A quarter and a quarter and six twelfths is one, so it is a proper sample space.

10:49Now: at least one tail appears — that is three quarters. The die shows a number bigger than four — that is two outcomes, a twelfth each, one sixth. And those two outcomes are also the only ones the two events share. So the conditional is one sixth divided by three quarters, which is two ninths. Try it by counting instead. Seven outcomes survive, two of them are favourable, two sevenths. That is wrong. Two sevenths is not two ninths.

11:23The head count assumed every survivor was as likely as every other, and here they are not. The definition never assumed it. That is what the probability form buys you, and it is why it is the one that gets written down. That two stage experiment is usually drawn as a tree. One branching at the first stage, then a branching at the second, and eight leaves at the bottom.

11:54A tree is worth drawing, because it turns a staged experiment back into a flat list of outcomes, which is the only form the definition knows how to eat. But there is a trap in how the numbers get written on it, and I have to flag it because nothing warns you. You will see this tree drawn with a half on each first level branch, and then a quarter on each head branch and a twelfth on each tail branch at the second level.

12:25Those second level numbers are the probabilities of the whole outcomes. They are not the chances along those branches. The chance along a head branch is a half, and a half of a half is a quarter. The chance along a tail branch is a sixth, and a half of a sixth is a twelfth. So if you do the natural thing and multiply the two labels along a path, you get a half times a quarter, which is an eighth. The outcome is a quarter. You are out by a factor of two, on every single path.

13:01The reading I have just given you is mine, and a careful reader should check it against whatever they are looking at. Multiply branch chances, never leaf probabilities — and if a label is already the answer, stop. What to keep. Being told something does not change the experiment. It changes the list of outcomes you are allowed to count against. Cross out everything that contradicts what you were told. Among what is left, the favourable ones are the outcomes both events contained.

13:39That gives a ratio of two counts. Divide both by the size of the original list and it becomes the chance of both, over the chance of the one you were told. Divide by the event on the right of the bar. Always. And if that event has probability nought, the answer is not defined rather than nought. Turning the bar around gives a different number unless the two events are equally likely.

14:09And the conditional can be bigger, smaller or exactly the same as the plain chance. Two boys given at least one boy went up. A four on the third throw given the first two went nowhere at all. Conditioning is not shrinking; it is recounting.