Events that cannot both happen, and events that between them must

0:00Here are two questions you can ask about a family of events. Can any two of them happen together? And between them, do they leave any outcome out? Those sound like one question. They are not, and the whole of this topic is the difference. Take one die and two descriptions: an odd face, and a face below four. The first names one, three and five. The second names one, two and three.

0:29Now the first question. What lies in both? One and three do. So a single throw can satisfy both descriptions at once, and the answer to the first question is no, they are not exclusive. And the second question. What lies in neither? Four and six. So the two of them do not cover the list either. Notice that the two questions were answered by looking at two different things.

0:58The first is a question about what the two events share. The second is a question about what they claim between them. Nothing so far says one answer has to follow from the other. The example almost everybody meets first is the odds against the evens. One, three and five against two, four and six. Ask the first question. Nothing lies in both, so one throw cannot produce both. They exclude.

1:31Ask the second question. Between them they hold all six faces, so nothing is left out. They cover. And that is exactly what makes this a poor example to generalise from. It satisfies both conditions at once. If it is the only case you have looked at, you cannot tell which condition is doing which piece of work, because here they arrive together. So we will keep it, and we will never use it alone.

2:03Before going further, be precise about the first condition. It is not a condition on how the two descriptions sound. It is a condition on one set: the outcomes lying in both. If that set is empty, the two events cannot both occur, and if it is not, they can. That is the whole test. And it is not a matter of degree. Here is a pair sharing exactly one outcome, and a pair that is the same event twice, sharing all three.

2:35Both fail. Not one of them fails a little and the other a lot. One shared outcome is a complete failure of the condition, which is why producing a single witness settles the question. You never have to list the whole overlap to answer it. Although if you are going to list it, list all of it. Which brings us back to the odd faces against the low faces. It is common to settle that pair by pointing at the three, which is odd and is also below four.

3:11That is a complete argument, and it is enough. But the overlap is not just the three. The one is odd as well, and the one is also below four. So the two events share two outcomes, not one. The single witness answers the question; the full overlap is what is actually there. And this pair fails the second condition too, leaving four and six unclaimed. Which makes it a strangely useless example for the point we are building towards.

3:44It fails both conditions, so it is no evidence at all about whether one of them forces the other. For that we need pairs that pass exactly one. Here is the easiest family to be sure about. An event holding exactly one outcome. A die of six faces carries six of those, one for each face. Take any two of them. Can they overlap? The first holds a single outcome, the second holds a single outcome, and the two outcomes are different, because the events are different.

4:21So there is nothing that could be in both. Notice what that argument did not use. It did not use the die. It did not use six. It used only the fact that the two events are distinct, so it holds on any experiment whatever, and it is checked here from two outcomes out to eight without a single exception. One-point events are never in each other's way. And keep this one in reserve: an event does not exclude itself, unless it is empty.

4:55That is why the condition is always stated for two different members of the family. Now we can settle one direction. Does excluding force covering? Take the one-point event holding just the one, and the one-point event holding just the two. They exclude, by the argument we have just made. What do they cover? Between them, two faces out of six. Three, four, five and six are claimed by neither.

5:26So no: events that cannot both happen may perfectly well leave most of the list unclaimed. That refutation is worth remembering because it is so cheap. Two different single outcomes, on any experiment with a third outcome in it, and you are done. Now the other direction, which is harder and more interesting. Does covering force excluding? Here is a family of three events on the same die. A face below four, which is one, two and three.

6:00A face above two but below five, which is three and four. And a face above four, which is five and six. Take the union of all three and walk the list. One is claimed. Two is claimed. Three is claimed. Four is claimed. Five and six are claimed. Nothing is left over. So this family covers. It is a perfectly ordinary family of three sentences about a die, and between them they account for every face.

6:34But look at the first two members again. One, two and three. Three and four. They both hold the three. So this family covers the list and is not exclusive. One direction settled, and the other settled: neither condition forces the other. There is a cheaper way to have spotted that overlap, and it is worth having. Add up the sizes. Three, then two, then two. That is seven. And the outcome list has six things in it.

7:06Seven claims spread over six outcomes, and every outcome is claimed at least once, so some outcome was claimed twice. The excess is one, and the outcome counted twice is the three. That is the whole demonstration, in one line of arithmetic, and it needs no picture. So does the size check settle the question? Not on its own, and it is worth seeing exactly why. Here is the accounting.

7:37Add up the sizes of a family. Every outcome claimed twice adds one to that total, and every outcome claimed by nobody takes one away. So the excess is the double-counting less the gap. That identity holds for every family of three on a five-outcome list with no exceptions, and it tells you immediately what the size check can and cannot see. If you already know the family covers, the gap is nought, and the excess is exactly the double-counting.

8:12Nought excess then means no overlap, and the check is exact. But if you do not know that, the two can cancel. Watch. One and two. One and three. One and four. Three events of two outcomes each, on a die. The sizes add to six, and the list has six faces in it. The size check says nothing is wrong. But the one is claimed three times over, and the five and the six are claimed by nobody at all.

8:43An overlap of two and a gap of two, cancelling in the total. So the check earns its keep only alongside one of the two conditions. Given covering, it settles exclusion. Given nothing, it settles nothing, and on a five-outcome list there are four hundred and ten families of three that pass it while failing both conditions. When both conditions hold together, the family cuts the outcome list into pieces with no seam and no gap.

9:17The overlapping family we built is one repair away from that. Take the three out of the middle member, so it holds only the four. Now the three is claimed once, by the first member, the seam is closed, and the cover is untouched. That is a clean split. But be careful which member you take it out of. Take the three out of both members instead and the seam closes and a gap opens in its place: nothing holds the three at all.

9:49The two conditions really do pull in opposite directions. Making the pieces smaller closes seams and opens gaps. Making them bigger closes gaps and opens seams. A clean split is the point where you have done exactly enough of each. We have now seen a pair in each of three situations, and it is worth asking whether that is a fair sample or three lucky examples. So take every event on a list of five outcomes, pair them up every way, and sort the pairs by the two answers.

10:27There are four hundred and ninety-six pairs. Sixteen of them both exclude and cover. A hundred and five exclude without covering. A hundred and five cover without excluding. And two hundred and seventy do neither. All four boxes are occupied, and they stay occupied on lists of three and four outcomes as well. That is what independence looks like when it is counted rather than asserted. And look at the sixteen in the first box.

10:59A pair that excludes and covers is a pair holding an event and everything it leaves out. There are thirty-two events on a five-outcome list, so there are sixteen such pairs, and they are the only ones. The two middle numbers are the same, and that is not a coincidence. Take any pair of events and replace each of them by everything it leaves out. What happens to the two answers?

11:28The pair covered the list exactly when nothing was outside both, which is exactly when the two replacements share nothing. So covering turns into excluding. And running the same sentence the other way, excluding turns into covering. That is the law that swaps a complemented union for an intersection of complements, arriving here as a symmetry of the whole picture. It matches up the pairs in one middle box with the pairs in the other, one for one, over all four hundred and ninety-six.

12:05Which is why those boxes hold a hundred and five each, and why the two corner boxes are the ones the swap leaves alone. Now a bigger experiment, because with two events there are only so many pairs to check. Throw a die twice and record both faces in order. Thirty-six outcomes, and every one of them is a single ordered pair. Add the two faces on each, and here are four descriptions of the total.

12:36An even total. A total that is a multiple of three. A total below four. And a total above eleven. Those name eighteen outcomes, twelve outcomes, three outcomes and one outcome. Eighteen is half of the thirty-six, which is worth pausing on: the even totals are not six equally likely numbers. They are the outcomes producing those totals, and there are eighteen of them. Four events give six pairs, and the question is which of them cannot both happen.

13:14Even, and a multiple of three. Both hold every outcome totalling six, and every outcome totalling twelve, which is six outcomes. They can happen together. Even, and below four. A total below four is two or three, and only the two is even, so they share the pair of ones. One outcome is enough. They can happen together. Even, and above eleven. Above eleven means twelve, which is even, so they share the pair of sixes.

13:48A multiple of three, and below four. Three is a multiple of three, so they share the two outcomes totalling three. A multiple of three, and above eleven. Twelve is a multiple of three, so again the pair of sixes. And finally, below four and above eleven. Nothing at all. No total is under four and over eleven at the same time, so those two cannot both happen. Exactly one of the six pairs excludes, and five of them do not.

14:22Which is the answer to a question people rarely ask out loud: is checking one pair enough? Here, checking one pair would have had five chances out of six of telling you the wrong thing. The condition is asked of every pair, and it means every pair. The smallest version of this question is worth doing too, because it catches a particular kind of error. One die. The first event is the face showing four.

14:53The second is an even face. Can they both happen? It is tempting to say no, because a four is one specific thing and an even face is a whole collection, and they feel like different kinds of statement. But throw a four. The first event has occurred, and the second event has occurred, because four is even. They share the four, so they are not exclusive. In fact one of them sits entirely inside the other, which is about as far from excluding as a pair can get while still being two different events.

15:30Containment and sharing are different questions, and only one of them is the one being asked. Finally, a family where every seam closes. Toss three coins. Eight outcomes. Three descriptions, by how many heads turn up. No head at all. Exactly one head. At least two heads. The first names one outcome, the all-tails one. The second names three. The third names four. One and three and four is eight, which is the size of the outcome list.

16:07And this time we know the family covers, because every outcome has some number of heads and that number is nought, or one, or at least two. So the size check is exact here, and it says the excess is nought, and nothing is claimed twice. Both conditions hold. It is a clean split. And notice the first member. It holds a single outcome. So a family that cuts the outcome list cleanly may perfectly well contain a one-point event, and the smallest clean split of all is the one made of nothing but one-point events, one for every outcome.

16:50So what were the two conditions for? They are genuinely two, and the counting says so: on a five-outcome list, a hundred and five pairs have one without the other, in each direction. They are not degrees of the same thing. The first is decided on what two events share, and one shared outcome fails it outright. The second is decided on what the family claims between them, and one unclaimed outcome fails it outright.

17:21The size check reads the difference between those two failures, and on its own it cannot tell them apart, because an overlap and a gap cancel. What is coming next needs only the first of them. The rule for adding up across two events asks that they cannot both happen, and asks nothing about what they leave out. So exclusion alone is the working condition. Covering is what turns a family of exclusive pieces into a family whose parts account for the whole thing, with nothing outside them.

17:57That is why the two are named together, and it is also why they had to be separated first.