Adding two probabilities double-counts the overlap, so the overlap comes back off

0:00You have three conditions, and one of them does the work. Two events that cannot both happen have their values added. Read that restriction again, because it is doing something. It does not say adding is usually right. It says adding is licensed, and only on a pair that shares no outcome. Hand it any other pair and it is not wrong about them. It is silent about them. Three tossed coins give eight outcomes, and those eight outcomes carry two hundred and fifty-six events.

0:35Pair those events up, and there are thirty-two thousand six hundred and forty pairs to consider. Three thousand two hundred and eighty of them share nothing. The condition speaks about those. About the other twenty-nine thousand three hundred and sixty it says nothing at all. That is most of them, and that is the gap this video closes. Take two events on those eight outcomes. The first holds three of them.

1:06Heads, heads, tails. Heads, tails, heads. Tails, heads, heads. The second also holds three. Heads, tails, heads. Tails, heads, heads. Heads, heads, heads. Notice something before going on. The first event has a name. It is exactly two heads, and nothing else on the list qualifies. The second one does not. Write out twenty plain descriptions of a three-coin result, the kind anyone would reach for. At least one head, exactly one head, all three alike, the first is a head, an even number of heads.

1:47Exactly one of the twenty names the first event. None of the twenty names the second. That is not a defect in the example. An event is a collection of outcomes, and a collection needs no phrase to justify it. This one is handed over as a list, and a list is enough. Every outcome here is worth an eighth. So each event, holding three of them, totals three eighths.

2:16Add those two together and you get six eighths. Now count the union instead. Which outcomes lie in one event, or the other, or both? Heads heads tails, from the first. Heads tails heads and tails heads heads, from both. Heads heads heads, from the second. Four outcomes. The union is worth four eighths, which is one half. Six eighths is not four eighths. The gap is not a rounding error and it is not small.

2:47The total overshoots the truth by two eighths, on an answer whose true value is four eighths. It is out by half of the thing it was trying to measure. Where did two eighths of probability come from? Look at the two lists side by side and find what they have in common. Heads tails heads sits in both. Tails heads heads sits in both. Now follow the arithmetic rather than the intuition.

3:17Totalling over the first event picked up those two outcomes once. Totalling over the second event picked up the same two outcomes again. So in the sum of six eighths, those two are in there twice. The union holds them once, because a union holds each outcome once however many events claim it. The shared part holds two outcomes and is worth two eighths. The excess was two eighths. Those are the same number, and now you can see why they had to be.

3:51One example is a guess. So put the guess to a census before trying to prove it. Take every one of those thirty-two thousand six hundred and forty pairs of events. For each one, work out the two totals, subtract the union's value, and compare what is left against the shared part. They match on every single pair. Do it again with the outcomes weighted unevenly instead of equally, and they still match on every pair.

4:23Do it with a table that gives one outcome the whole of the probability and the other seven nothing, and they match there too. But a comparison that cannot come out false is not evidence. So run the same test on something that is not built by adding. Give every event the square of its size over the square of the list's size. That assigns a number to every event, gives the sure event one and the empty event nothing, and looks respectable.

4:55On a four-outcome list it has a hundred and twenty pairs to answer for, and the excess is not the shared part on fifty-five of them. So the test can fail. It just does not fail for anything built by totalling. Now the proof, and it starts by cutting. The union of two events divides into exactly three pieces. What lies in the first event and not the second. What lies in both.

5:26What lies in the second and not the first. On the coin example that is one outcome, then two, then one. One and two and one is four, which is the union's size, so nothing has been lost and nothing counted twice. And no two of those three pieces share an outcome, which is the condition the axiom was waiting for. Check that as sets rather than trusting it. Over every pair of events on lists of one outcome up to six, the three pieces put the union back together every time, and no two of them ever share.

6:04The cut is genuinely three pieces most of the time, too. On the three-coin list all three are occupied for twenty-three thousand three hundred and ten of the pairs. With the union cut into three, the axiom applies untouched. The first event is the first piece together with the middle piece, and those two share nothing, so its value is the sum of theirs. The second event is the middle piece together with the third piece, and those two share nothing either.

6:38The union is all three, no two sharing, so its value is the sum of all three. Add the first event's value to the second event's value. You get the first piece, plus the middle piece, plus the middle piece again, plus the third piece. That is the union, plus one extra copy of the middle. Take the middle off once and you have the union exactly. In eighths, on the example: one eighth, plus two eighths, plus one eighth is four eighths.

7:10The subtraction is not a correction bolted on after the fact. It is what is left when you write the union the only way the axiom will accept. Draw it once and it stops being algebra. A rectangle for everything that can happen, and two circles inside it that overlap. The left circle is the first event, the right circle the second. Three regions, not two. A crescent on the left, holding heads heads tails.

7:43A lens in the middle, holding heads tails heads and tails heads heads. A crescent on the right, holding heads heads heads. One eighth, two eighths, one eighth. The two crescents are the parts nobody argues about. The lens is the part that was counted twice. Everything outside both circles is the remaining four outcomes, which the union never claimed. The picture is not the proof, but it is the proof's shape, and the shape is why there are three terms and not two.

8:19There is a shorter route, and it is worth having because it works differently. Split the union a different way. It is the whole first event, together with whatever the second event holds that the first does not. Call that leftover the second event's own part. Those two share nothing, so the union's value is the first event's value plus the leftover's. That is one equation. Now split the second event, using the same leftover.

8:51It is the shared part, together with that leftover. Those two share nothing either, so the second event's value is the shared part plus the leftover. That is a second equation. Both were handed to you by the same axiom, and both contain the leftover. Subtract the second from the first and the leftover cancels. The union, less the second event, is the first event, less the shared part. Rearrange, and there is the rule again.

9:24On the coin example the leftover is heads heads heads alone, worth one eighth. Four eighths is three eighths plus one eighth. Three eighths is two eighths plus one eighth. Two equations, one subtraction, done. Here is the mistake that feels principled. The shared part was in both events, so surely it should come off twice. It should not, and the reason is worth saying carefully. It was counted twice, so it must be removed once, which leaves it counted once.

9:59Remove it twice and it is gone altogether. On the example, six eighths minus two eighths minus two eighths is two eighths. The union is four eighths. You are now short by exactly the piece you deleted. Run that across all thirty-two thousand six hundred and forty pairs and removing it twice is right three thousand two hundred and eighty times. Those are exactly the pairs that share nothing in the first place.

10:30In other words, taking it off twice is right only when there was nothing there to take off. This next point is the reason the rule subtracts a value rather than asking a yes-or-no question. With the outcomes equally weighted, plain adding gets the union right on three thousand two hundred and eighty pairs, and those are exactly the pairs that share nothing. The licence and the correct answers coincide, so nothing is learned.

11:03Now change the table. Give one outcome the whole of the probability and the other seven nothing at all. That table is perfectly legal. Its entries are never negative and they total one. Under it, plain adding gets the union right on twenty-four thousand five hundred and twelve pairs. Twenty-one thousand two hundred and thirty-two of those pairs do share an outcome. Adding was not licensed and it was right anyway.

11:34The eight thousand one hundred and twenty-eight where it still breaks are exactly the pairs both of whose events hold the outcome carrying the whole value. So a shared part can be occupied and still be worth nothing. The correction is the shared part's value, not the shared part's existence. Which tells you what happens at the other end. Take two events that genuinely cannot both occur. Their shared part is the empty event, and the empty event is worth nothing.

12:09So the rule subtracts nothing, and reads exactly as the axiom you started with. Across all three thousand two hundred and eighty pairs that share nothing, the rule and plain adding agree every time. The axiom has not been replaced. It has been extended, and the extension agrees with it wherever the axiom had anything to say. So a pair worth three fifths and one fifth that cannot both occur is worth four fifths together, and you do not need the general rule to say so.

12:45You just need to have checked that they cannot both occur. Two people sit an entrance test. Ana's chance of qualifying is nought point nought five. Ben's is nought point one. The chance they both qualify is nought point nought two. Four questions, and only the first is the rule. At least one qualifies. Nought point nought five plus nought point one, less nought point nought two, is nought point one three.

13:17Neither qualifies is everything else, so nought point eight seven. At least one fails is everything except both qualifying, so nought point nine eight. And exactly one qualifies. Ana alone is nought point nought three, Ben alone is nought point nought eight, and together that is nought point one one. There is a second route to that last one. The union less the shared part is precisely the outcomes belonging to one event and not the other, so nought point one three less nought point nought two is nought point one one.

13:56Two routes, same answer, free check. And notice that nought point one one and nought point nine eight are answers to different questions. Exactly one qualifying is not the same as at least one failing, and they are not close. Can any three numbers serve as two events and their shared part? Suppose someone offers you a half and seven tenths, with a shared part of six tenths. The shared part sits inside both events, so it cannot be worth more than either of them.

14:31Six tenths is more than a half. There is no experiment, no table of outcomes, nothing whatever, that carries those three numbers. They are refused. Compare a half and four tenths with a union of eight tenths. That forces the shared part to be one tenth, which sits below both, and it is not negative. Build the table and it exists. So there is a real test here, and it is easy to state.

15:01The shared part is at most the smaller of the two. That test is half the test, and the missing half is the more interesting one. Two examinations, one passed by eight tenths of the candidates and the other by seven tenths. Nothing wrong with that. Two probabilities are allowed to total more than one, because only a single event is capped. Now suppose the shared part is one tenth.

15:31That sits inside both, so it passes the test we just wrote down. But eight tenths plus seven tenths less one tenth is one point four, and no union is worth one point four. The shared part has a floor as well as a ceiling. It has to be at least the two values totalled, less one. Here that floor is a half, and at exactly a half the union comes out at one, which is the tightest it can be.

16:00Count how much this matters. Take every triple of numbers in tenths, one thousand three hundred and thirty-one of them. Five hundred and six pass the test that the shared part sits inside both events. Only two hundred and eighty-six of them describe an actual table. So two hundred and twenty triples pass the test that gets taught and describe nothing at all, and every one of them fails on the union instead.

16:31Nothing goes the other way. And that count of two hundred and eighty-six is reached twice over, once from the two bounds and once by building every table there is and reading off what it says. The rule has four quantities in it, and any three give you the fourth. A third and a fifth with a shared part of one fifteenth gives a union of seven fifteenths. Nought point three five and a shared part of nought point two five with a union of nought point six gives a second event of a half.

17:09A half and nought point three five with a union of nought point seven gives a shared part of nought point one five. Run it backwards on the two examinations. Eight tenths and seven tenths, with nought point nine five for passing at least one. Then passing both is nought point five five, which sits just above that floor of a half. Or start from the other end. Passing both is nought point five, passing neither is nought point one, so the union is nought point nine.

17:45If one subject is nought point seven five, the other is nought point six five. A quarter and a half with a shared part of an eighth give five eighths, so neither happening is three eighths. Nought point four two and nought point four eight with a shared part of nought point one six give nought point seven four for either, and nought point two six for neither. The rule earns its keep hardest where the overlap has to be spotted rather than handed to you.

18:19A class of sixty. Thirty play chess. Thirty-two sing in the choir. Twenty-four do both. Thirty plus thirty-two is sixty-two, which is more than the class holds, and that alone tells you something has been double counted. Take the twenty-four off once and you get thirty-eight, which is nineteen thirtieths. So twenty-two of them do neither, and the choir without the chess is eight. Or a shorter list. Five people, and you want the chance that a name drawn at random is a man, or over thirty-five.

18:57Two of the five are men, aged twenty-eight and fifty-one. Three are over thirty-five, aged thirty-six, forty-two and fifty-one. Two fifths and three fifths. Add them and you get one, which would say it is certain, and it is not. The fifty-one-year-old man is in both lists, and he is one fifth. So the answer is four fifths, and counting the four names directly agrees. The overlap did not announce itself.

19:28It had to be found. One last question, and it needs no new machinery. What about three events? Group the second and third together, treat that as a single event, and apply the two-event rule once. Then apply the two-event rule again inside, to the second and third themselves. One term is still awkward: the first event meeting the union of the other two. But meeting a union is the union of the meetings, so that becomes the first-with-second and the first-with-third, and those two overlap in exactly the part shared by all three.

20:10Apply the rule a third time to those, and everything unfolds. The three values, less the three pairwise shared parts, plus the part shared by all three. No fourth condition was introduced. That whole formula is the two-event rule, used repeatedly, on top of one fact about sets. Test it. On a four-outcome list there are four thousand and ninety-six triples of events, and the assembled route agrees with the union's own total on every one.

20:45Leave out the part shared by all three and it fails on one thousand six hundred and ninety-five of them. On the three coins, take the first two events again and add a third holding all heads and all tails. Three eighths, three eighths, two eighths. The pairwise shared parts are two eighths, nothing, and one eighth. Nothing at all lies in all three. So the union is three plus three plus two, less two, less nothing, less one, plus nothing.

21:17Five eighths, and the five outcomes are there on the board to count. The axiom on its own answers exactly one of those three pairs. Cut the union the only way the axiom accepts, and it answers all of them.