The rules conditional probability inherits, including addition on a restricted space

0:00Last time you built a new number. You took an event, you were told a second event had happened, and you counted the first one on the shorter list the second left standing. Here is the question nobody asks out loud. Is that new number still a probability? It is not a silly question. You built it by dividing one probability by another. Dividing does all sorts of things to numbers. Nothing so far guarantees that what came out still behaves the way probabilities behave.

0:36And a great deal is riding on the answer. Everything you already know about probability — adding, complements, the whole apparatus — you are about to use on these new numbers without thinking twice about it. So the next few minutes are a certificate. Three results, and together they say one thing: conditioning builds a new probability, it does not build a new kind of arithmetic. The first result is two statements, and they look too obvious to bother with.

1:11The whole space, given an event, is one. And the event, given itself, is one. Both proofs are a single substitution. The whole space intersected with your ground is the ground. The ground intersected with itself is the ground. So both fractions are the chance of the ground, over the chance of the ground. The same number over itself. One. Twice. I ran that over every ground a die offers — sixty three of them, every non-empty event — and over all two hundred and fifty five of a three-coin space, and over fifteen grounds of a space whose outcomes are deliberately not equally likely. Not one exception anywhere.

2:01So why prove it at all? Nothing I have seen says why, so let me tell you what I think it is for. A thing is a probability when three conditions hold. Every event gets a number that is not negative. The whole space gets one. And two events that share no outcome have their chances simply added. The first result is the second of those conditions. It is the one that says the new numbers total one, which is the only thing stopping them from being an arbitrary pile of fractions.

2:37And it is worth seeing that all three conditions are load bearing rather than decorative. Take the same construction and forget to divide, and what comes out still adds properly but only totals one on the certain ground — it passes on one ground out of sixty three. Take a version that gets the single outcomes right and returns nought for anything bigger, and it totals one everywhere and stops adding — it passes on no ground at all. The real thing passes on all sixty three, and on all fifteen of the unequal space too.

3:18Second result. The addition rule, with a ground held fixed. The chance that either of two events happens, given a third, is the first given the third, plus the second given the third, minus their overlap given the third. Look at the shape of that rather than the letters. It is exactly the addition rule you already had. Every single term simply carries the same fixed thing on the right of the bar.

3:48That is the claim in one line: the ground does not change the arithmetic. It comes along for the ride, on every term, unchanged. I swept it over every triple a die offers — sixty four events, by sixty four events, by sixty three grounds. Two hundred and fifty eight thousand and forty eight triples, and it fails at none of them. And it does not need the outcomes to be equally likely. On the lopsided space — a half, a quarter, an eighth, an eighth — all three thousand eight hundred and forty triples hold too.

4:30The proof is four lines, and three of them are bookkeeping. Line one. Write the left-hand side out using the definition. The top becomes the chance that the union and the ground both happen; the bottom is the chance of the ground. Line two. Rewrite that top. This is the only real step and I will come back to it. Line three. The top is now the chance of one thing or another thing, so use the ordinary addition rule on it. That leaves three chances on top, with the third one subtracted.

5:09Line four. Split the one fraction into three fractions over the same bottom, and read each of the three back as a conditional. That is the result. Three of those four lines move symbols around. The second line is the one that says something. So here is line two on its own. You have the chance that the union happens and the ground happens. You want it as two overlapping pieces you can feed the addition rule.

5:42The fact you need is about sets, not about probability. Take the union of two events and intersect the whole thing with a fixed third. What you get is the union of the two separate intersections. Shade it and the two pictures are the same shape. Cut the combined region down to the ground, or cut each piece down to the ground and then combine — same region, every time.

6:10That is intersection spreading across a union, and I checked it on all two hundred and sixty two thousand one hundred and forty four triples of events a die gives. It holds at every one. Once you have it, line three follows, because the overlap of those two pieces is the triple intersection, and that is exactly the term that ends up subtracted. A warning about that step, because it is very often labelled, and the label you will usually see has the two operations the wrong way round.

6:49There are two distributive facts about sets and they are different statements. Intersection spreading across a union is the one this proof uses. Union spreading across an intersection is the other one. Both are true. That is what makes the slip hard to spot — nothing you are told is false. But they are different objects, and I measured how different. Set the proof's own quantity against the other arrangement and they agree at only forty six thousand six hundred and fifty six of those two hundred and sixty two thousand triples, and disagree at two hundred and fifteen thousand four hundred and eighty eight.

7:33So if you go looking for the identity by that name, you are looking for the wrong one. Name it by what it does instead: cutting a union down to a fixed set. Now the special case, which is where a real mistake lives. If the two events share no outcome at all, the subtracted term is the chance of the empty event, on a ground. That is nought. So the rule shortens to a plain sum.

8:04Forty five thousand nine hundred and twenty seven triples of the die have no shared outcome, and the short form is right at every single one. Which is exactly why people remember the short form and then use it everywhere. Run it on all two hundred and fifty eight thousand triples and it is wrong at a hundred and forty four thousand four hundred and ninety five of them, overshooting by as much as a whole one.

8:34And there is a nicer way to say when the short form survives. It is right at a hundred and thirteen thousand five hundred and fifty three triples. Forty five thousand of those share nothing. The other sixty seven thousand six hundred and twenty six do share something — it just carries no chance once you are standing on the ground. What has to vanish is the term, not the overlap. That is a slightly bigger class, and it is the honest statement of the special case.

9:09Third result, and it is the shortest. An event's opposite, on a fixed ground, is one minus the event on that same ground. Four thousand and thirty two pairs on the die, two hundred and forty on the lopsided space, no failures. But the interesting thing is not that it holds. It is where it comes from. An event and its opposite share no outcome — that is what opposite means. And together they are the whole space.

9:41So: they are disjoint, which means the short form of the second result adds them. And their union is the whole space, which the first result says is one on any ground. One equals the event plus its opposite. Rearrange. Three lines, and both of them are results you already proved. That is the clearest picture in this topic of one result being used to buy the next. Now the part that matters more than any of the three, and it is about what they do not say.

10:18Every one of those results holds with a single fixed thing on the right of the bar. The events on the left move. The ground does not. So nothing here lets you combine two grounds. Nothing here lets you form a union or an intersection across the bar. The rule that looks like the addition rule applied to the right-hand side is not a rule at all. It is not even close. Run that shape on the right of the bar across the die and it disagrees at a hundred and nineteen thousand one hundred and sixty of two hundred and fifteen thousand usable triples.

11:00One case is enough to retire it. Ask for an even face. On the ground of the first three faces the answer is a third; on the last three faces, two thirds. On either of those two grounds it is a half — and the two grounds share nothing, so the term you would want to subtract cannot even be written down. And the ground genuinely carries the answer. Swap any ground for its opposite and the number changes at three thousand one hundred and twenty four of the die's three thousand nine hundred and sixty eight pairs.

11:39Let us use all of it on one die. Three events. The odd faces: one, three, five. A two-face event: two and three. And a four-face ground: two, three, four, five. Ask for the union of the first two, on that ground. Count it. The union is one, two, three and five — four faces. Three of them are in the ground of four. Three quarters. Now get the same number out of the rule instead. The odd faces on that ground: two of the four, a half. The two-face event on that ground: also two of the four, a half. Their overlap is the single face three, which is in the ground: a quarter.

12:24A half plus a half minus a quarter. Three quarters. The same number, and that is the addition rule checked on something you can count on your fingers. Drop the subtracted term and you get one. That would say a union of four faces is certain on a ground that only contains three of them. The short form does not just lose accuracy here, it says something impossible. One more thing, and it is about what gets remembered rather than what gets proved.

13:00Two facts tend to travel together in revision notes on this topic. One is that a conditional probability lies between nought and one. The other is that the first of the three results quietly goes missing. The bound is true. Four thousand and thirty two pairs on the die, two hundred and forty on the lopsided space, none outside nought and one. And the reason is a containment, not luck: what the top counts is part of what the bottom counts, so the fraction cannot pass one.

13:37But notice it is a real test rather than an empty one. Run the same check on the plain sum of two conditionals — the thing the short form would hand you — and eighty six thousand six hundred and seventy four of the die's triples come out above one. And the result that goes missing is the one everything else rests on. Without it the new numbers are not known to total one, and there is no certificate at all. If you only keep two of the three, keep that one.

14:12What to keep. Three results, one claim: fix a ground, and what sits on the left of the bar is an ordinary probability. That is why nothing later has to re-derive anything. The whole space on a ground is one. That is the result that makes the total come out right, and it is the one people drop. Two events on a fixed ground add with the overlap taken off once. The short form is the special case, not the rule, and using it everywhere overshoots — sometimes by a whole one.

14:50The step in the proof that is not bookkeeping is cutting a union down to a fixed set. If you meet it under a name, check the name — the two distributive facts are both true and only one of them is this. An event and its opposite on a ground still fill it, and that falls straight out of the other two. And the ground never moves. Everything here is licensed one fixed ground at a time. The moment something on the right of the bar starts combining, you have left the certificate behind.