Multiplying along a chain of dependent draws, and extending it past two events

0:00Almost every real probability question is not about one thing happening. It is about several things all happening. A king and then a queen, drawn one after the other. Two black balls out of a bag. Three sound oranges out of fifteen. And the direct way to answer those is horrible. You would have to count the set of all the ways the whole run could come out the way you want, and that set is enormous. For three cards from a full deck there are a hundred and thirty two thousand six hundred ordered triples to sort through.

0:38There is one move that replaces the counting with arithmetic, and it turns out to be a move you have already seen, rearranged. Here is the whole idea, and it takes ten seconds. Last time you had a definition: the chance of one thing given another is the chance both happen, divided by the chance of the one you were told. Multiply both sides by the chance of the one you were told. The denominator on the right cancels.

1:12And what is left is the chance that both happen, written as a product: the chance of the first, times the chance of the second given the first. That is it. That is the entire result. No new idea entered the room — the definition was simply turned round so that the intersection is on the left instead of buried in a numerator. But turning it round changes what it is good for. As a definition it tells you how to work out a conditional. As a product it tells you how to build an intersection you could never count.

1:52Now do the same thing with the roles swapped. Start from the chance of the second given the first. Multiply through. You get the chance that both happen as the chance of the second, times the chance of the first given the second. Two products, and they are equal, because both of them are the chance that both events happen — and an intersection does not care which order you name its two events in.

2:20I checked that over every ordered pair of events you can build out of a two-draw experiment where both events are possible. Two hundred and twenty five pairs, both orders, landing on the directly counted answer every time. But equal value does not mean interchangeable. They are different computations. One of them needs the chance of the first event and a conditional going forwards. The other needs the chance of the second event and a conditional going backwards, which is often a thing nobody has told you.

2:59So you get two ways in, and in any real question one of the two is handed to you and the other is not. Let us do one properly. A bag holds fifteen balls. Ten are black, five are white. You draw two, one after the other, and you do not put the first one back. What is the chance that both are black? First factor. The chance the first ball is black. Ten black out of fifteen balls.

3:29And I am going to leave that as ten over fifteen rather than tidying it to two thirds, for a reason that matters in about twenty seconds: fifteen is the number the next factor has to be built out of, and I do not want it hidden. Second factor. The chance the second ball is black, given that the first one was. This is where the whole topic lives, so say the state of the bag out loud before you write anything.

4:01One black ball has gone. It did not come back. So the bag no longer holds fifteen balls, and it no longer holds ten black ones. It holds fourteen balls, nine of them black. That sentence is the answer. Nine out of fourteen. Read straight off the bag as it now is. Ten fifteenths times nine fourteenths. The fives cancel, the sevens cancel, and you are left with three sevenths.

4:31And the direct count agrees. There are two hundred and ten ordered ways to draw two balls out of fifteen; ninety of them have black in both places; ninety over two hundred and ten is three sevenths. Same number, much more work. Now the mistake, which is the single most common wrong answer in this whole topic. Using ten over fifteen again for the second factor, because black balls are two thirds of the bag. That gives four ninths, and four ninths is not three sevenths.

5:07What makes it seductive is that two thirds is a real number about this experiment. The chance that the second ball is black, asked with no information at all, genuinely is two thirds — the same as the first. It is just not the factor the product wants. Three events now, and there is exactly one thing to get right. The chance all three happen is the chance of the first, times the chance of the second given the first, times the chance of the third given the first and the second.

5:45Both of them. Not just the one immediately before it. That is the error worth spending a minute on, because it looks harmless. Conditioning the third factor on only the second stage feels like it should be close enough. It is not close at all. I built every three-stage pattern of a deck draw — sixty four of them — as a running product, twice: once conditioning each factor on everything before it, once conditioning only on the stage just before.

6:20The first way is right at all sixty four. The second way is right at none of them. Not a few. None. And the same thing happens with a bag drawn four times: sixteen patterns, all sixteen right the proper way, zero right the shortsighted way. Three cards from a full deck, none of them put back. You want a king, then another king, then an ace. Watch the denominators. Fifty two cards, then fifty one, then fifty. Each draw takes one card out of the world and the next factor is worked out in the world that is left.

7:01First factor: four kings out of fifty two. Second factor: one king has gone, so three kings out of fifty one. Third factor: two cards have gone and neither of them was an ace, so all four aces are still there — but only fifty cards are. Four out of fifty. Multiply. Forty eight over one hundred and thirty two thousand six hundred. That is not an answer anybody wants to look at, so reduce it: both parts divide by twenty four, and what is left is two over five thousand five hundred and twenty five.

7:41I want that division on screen rather than hidden, because an answer that appears without the cancellation looks like something that was looked up. And if you had conditioned the ace on only the second king, you would have written four over fifty one instead of four over fifty, and got four over eleven thousand two hundred and seventy one. Which looks every bit as respectable as the right answer.

8:10Now change one word and watch what happens. A bag of ten black and eight red. Draw two — but this time put the first one back before drawing the second. The rule does not change. It is the same product, the same two factors, in the same order. What changes is that the second factor stops depending on the first. The bag at the second draw holds eighteen balls, eight of them red, whatever happened first. Four ninths either way.

8:42So both red is eight eighteenths squared, sixteen over eighty one. Black then red is twenty over eighty one. One of each in either order is forty over eighty one. And there is a sharper way to see what putting it back did. I took the bag with three draws, and worked out every pattern twice — once properly, and once reading every factor off the bag as it started, ignoring everything that had happened.

9:13Without replacement, the lazy reading is wrong at every pattern. With replacement, it is right at every pattern. That is not a different rule. That is the same rule, in a situation where the conditional factor happens to equal the unconditional one. Which is a thought worth holding on to. Here is the picture I would keep. Draw the run as a path. Each edge is one stage. Write on the edge the chance of that stage, worked out in the world the earlier edges have already produced. Write at each node the product so far.

9:55Then the number at any node is the chance of getting that far, and the number at the end is the answer. Black all the way out of that bag of fifteen: two thirds, then three sevenths, then twenty four ninety firsts, then two thirteenths, then twelve over a hundred and forty three. Each of those I counted directly off the three hundred and sixty thousand ordered draws, and each one matches the running product exactly.

10:26One warning about trees, because you will meet both conventions and they look identical. On the picture I have just described, an edge label is a conditional chance and a node label is a whole outcome's probability. You will also see trees drawn with outcome probabilities on the edges. On those, multiplying along the path gives you nonsense. So every time a tree appears, decide which kind it is before you multiply anything.

11:01Back to the question I opened with, because it deserves an answer rather than an atmosphere. A king and then a queen, from a full deck, with nothing put back. Four kings out of fifty two, then four queens out of fifty one — taking a king out does not remove any queens. That is four over six hundred and sixty three. But notice the wording is doing something quietly. Did you want the king first? Or just one of each, in whichever order they came?

11:34If either order counts, add the other path: queen first, then king, which is the same four over six hundred and sixty three, and the two share no outcome at all, so it is a plain sum. Eight over six hundred and sixty three. Two different answers, and the words alone do not settle which. So say which reading you have taken, every time. That habit is worth more marks than any formula on this board.

12:07Two to try, and they are the same move with different furniture. Two cards from a full deck, nothing put back, both black. Twenty six over fifty two, times twenty five over fifty one. Twenty five over a hundred and two. That is the bag of balls again with a deck instead of a bag. Fifteen oranges, twelve of them sound, three drawn, all three wanted sound. Twelve fifteenths, then eleven fourteenths, then ten thirteenths. Forty four over ninety one.

12:43Notice what you did not do in either. You did not count a single intersection. You wrote down what was in the container, took one thing out, wrote down what was in it now, and multiplied. And one last thing about running out. A chain can stop being possible. Take a container of two black balls and one white and draw three times: of the eight patterns you might write down, only three actually exist, and a chain asked for one of the others simply has nowhere to stand at the second or third step.

13:20The answer there is not a small number. It is that the question has no ground to be asked on. What to keep. The chance that several things all happen is a product, not a count. You never have to enumerate the intersection, and that is the entire point of the result. Each factor is worked out in the world the earlier factors have already created. Say what the container holds before you write the next fraction down.

13:53The condition on a factor carries every earlier stage, not just the one before it. Conditioning on only the previous stage was wrong at every single pattern I tested, in every experiment I tested it in. Both orders are legal and they give the same value, because an intersection is symmetric. They need different inputs, and in practice only one of the two will be available. Put it back and the conditional factor stops depending on anything. Same rule, special case — and that special case has a name and a topic of its own coming next.

14:35And when the wording is ambiguous about order, answer it twice and say so.