Coins, dice, bags and a deck of 52: getting the denominator right

0:00Every probability here is one number over another. The top is the easy half. Count the results that go your way. The bottom is where the marks are lost. Because the bottom number is not something you read off the question. It is something you decide. It has to be the number of separate results the experiment can produce, with nothing about the experiment favouring any one of them over the rest.

0:29And a description is not a result. That one sentence is the whole of this video. Almost every wrong answer in this material comes from letting a description stand in for a list. So before anything gets divided, ask one question. What exactly is the bottom number counting? When the list is short, write it out. When it is long, count it in one piece. Either way you must know what is in it.

1:00Start with the containers, because they make the habit obvious. A bag holds three blue marbles, two white and four red. One is taken out without looking. The bottom number is nine. And nine is nowhere in that question. You had to add three, two and four to get it. That is the move for every container. Empty it out, line up what came out, count. Two of the nine are white, so white is two ninths. Blue is three of nine, a third. Red is four of nine.

1:34A money box: a hundred five-cent coins, fifty ten-cent, twenty twenty-cent and ten fifty-cent. The bottom number is a hundred and eighty. A five-cent coin is a hundred over a hundred and eighty, five ninths. Anything but a fifty-cent coin is a hundred and seventy over a hundred and eighty, seventeen eighteenths. Twelve faulty pens are mixed into a hundred and thirty-two good ones. The bottom number is a hundred and forty-four, and a good pen is eleven twelfths.

2:13Now turn it round. One experiment, several questions, and the bottom number never moves. An ordinary die. Six faces, six results, and six is the denominator for every question you can ask about one throw. The chance of a prime. Two, three and five are prime, so three of six. A half. The chance of a number strictly between two and six. Three, four and five. Three of six again. A half.

2:45The chance of an odd number. One, three, five. A half. Three different questions, three different lists on top, one answer. The numerator moved; the denominator did not. One caution on the middle one. Strictly between leaves two and six out. Read it as between and including, and the list becomes two, three, four, five, six - five of the six faces, five sixths. Different reading, different answer. Say which one you are using.

3:17And here is a die whose faces read A, B, C, D, E, and A again. Six faces, but only five different labels. An A is two faces of six, a third. A D is one face of six, a sixth. The denominator is still six, because there are still six faces. It never counts labels. A pack of playing cards is the neatest structured denominator there is, and worth knowing cold.

3:48Fifty-two cards. Four suits - spades, hearts, diamonds and clubs - and each suit holds thirteen cards. Four thirteens, fifty-two. Two of the suits are red, hearts and diamonds. Two are black, spades and clubs. So twenty-six cards of each colour. Exactly half the pack. Inside a suit the ranks run ace, king, queen, jack, then ten down to two. Thirteen ranks, and every rank turns up once in each suit, so there are four of any given rank.

4:23The face cards are the kings, the queens and the jacks. Three per suit, four suits, so twelve face cards. Notice what that leaves out. The aces are not face cards. Twelve, not sixteen. That single confusion costs more marks than anything else in this topic. Half the face cards are red, so six red face cards. Now every card question is the same question. The denominator is fifty-two and it stays fifty-two. All that changes is which cards the description picks out.

5:01A red king. There are four kings, one per suit, and two of those suits are red. Two cards. Two over fifty-two, one twenty-sixth. A face card. Twelve over fifty-two, three thirteenths. A red face card. Six over fifty-two, three twenty-sixths. The jack of hearts. One card. One over fifty-two. A spade. Thirteen over fifty-two, a quarter. The queen of diamonds. One over fifty-two again. An ace. Four cards, four over fifty-two, one thirteenth.

5:38Not an ace. Forty-eight over fifty-two, twelve thirteenths. Eight descriptions, eight numerators, one denominator. The work was never the dividing. The work was knowing how the pack is put together. Sometimes the list is people. A class of forty. Twenty-five girls and fifteen boys. Every name is written on a card, the forty cards are identical, they are shuffled, and one is drawn. A girl's name is twenty-five over forty. Five eighths.

6:13A boy's name is fifteen over forty. Three eighths. The word carrying the argument there is identical. If one card were bigger, or bent, or stickier, the forty results would not be interchangeable, and forty would not be the denominator. The same job is done by at random, by well shuffled, and by marbles of the same size. Those phrases are not scene setting. They are the guarantee that the results you are about to count are the kind you are allowed to count.

6:49Now the case that catches everybody. Two coins are tossed together. How many results are there? The tempting answer is three. Two heads, two tails, one of each. That does sound complete, and it is complete - as a set of descriptions. Here is the defence against it. Make the two coins different. One is a five-cent piece and the other is a ten. Now you can tell them apart, so you have to record which is which.

7:20Five-cent head, ten-cent head. Five-cent head, ten-cent tail. Five-cent tail, ten-cent head. Five-cent tail, ten-cent tail. Four results. H H, H T, T H, T T. And they are interchangeable in the way a denominator needs: nothing about either coin favours any one of the four. So the chance of at least one head is three of the four. Three quarters. Now set the three descriptions against the four results and watch exactly where the merge happens.

7:56Two heads is one result, H H. A quarter. Two tails is one result, T T. A quarter. One of each is two results, H T and T H. A half. A quarter, a half, a quarter. Not a third, a third, a third. And look how respectable the three descriptions are. Every result is described, none is described twice, and the three chances add to one. That is exactly why the mistake feels safe.

8:28What fails is size. One of the three descriptions holds two results and the other two hold one each. The coins did not have to be different for that to be true. H T and T H were always two results. Giving the coins different values is an argument, not a rule of the game. It makes visible what was there anyway. The same argument, one size up. Two dice thrown together. Make them different colours - one blue, one grey.

9:01Record every throw as an ordered pair, blue first. Blue one, grey one. Blue one, grey two. All the way to blue six, grey six. Six choices for the blue die, and for each of those, six for the grey. Thirty-six ordered pairs, laid out as a six by six grid. Blue down the side, grey along the top. Thirty-six is the denominator for anything you can ask about one throw of two dice.

9:31Now the question people argue about. Blue one with grey four, and blue four with grey one. One throw or two? Two. They are different arrangements of two physical objects, and you can see they are different because the dice are different colours. Merge them and every mixed pair collapses, and thirty-six becomes twenty-one. The colouring is not decoration. It is the argument that thirty-six is the honest count. With the grid drawn, a numerator is something you look at rather than work out.

10:09The chance the two dice total eight. Walk the grid and shade every cell whose two numbers add to eight. Two and six. Three and five. Four and four. Five and three. Six and two. Five cells. Five over thirty-six. Look at where they sit. They run as a diagonal across the grid, not as a block. And it is why five and three has to be shaded as well as three and five. Drop one of them because the numbers look the same, and the shading falls to four cells, and the answer is wrong.

10:46Any description you can state, you can shade. Here is one that is not about totals: the blue die showing five or more. Twelve cells, all of them in the last two rows. A third. And the blue die beating the grey: fifteen cells, five twelfths. The grey beating the blue is fifteen cells too - but different cells, on the other side of the diagonal. Six cells are drawn.

11:15Now the trap this whole video has been walking towards. The total of two dice runs from two up to twelve. That is eleven possible totals. Eleven results, so one eleventh each? No. And the reason is exactly the reason two coins are not three results. Count the cells for each total, off the grid. Two: one cell. Three: two. Four: three. Five: four. Six: five. Seven: six. Then back down. Eight: five. Nine: four. Ten: three. Eleven: two. Twelve: one.

11:53One, two, three, four, five, six, five, four, three, two, one. Those add to thirty-six, as they must. So the chances run one thirty-sixth, one eighteenth, one twelfth, one ninth, five thirty-sixths, one sixth, and then back down the same way. A seven is six times as likely as a two. And the eleven totals really do cover the grid with no gaps and no overlaps, and their chances really do add to one.

12:25Everything reassuring about them is true. They are simply not the same size, and not one of the eleven is one eleventh. That deserves a fair question, because the shortcut does work sometimes. Six faces of a die, six descriptions, one sixth each. Correct. Four suits of a pack, a quarter each. Correct. So how often is one over the number of descriptions safe? Here is a measurement. Take every way of grouping up to six results into named descriptions. Two hundred and seventy-eight schemes in all.

13:05In thirty-nine of them the shortcut gets every description right. In the other two hundred and thirty-nine it is wrong about at least one. Counting descriptions rather than schemes: eight hundred and seventy-six descriptions across the sweep. The shortcut is right about a hundred and fifty-seven of them, and wrong about seven hundred and nineteen. Wrong more than four times as often as it is right. And when it is right, it is right for one reason only. The descriptions came out all the same size.

13:42A fact about the grouping, not about how many names there are. One last way the denominator moves, and this time it moves mid-question. Five cards face down: the ten, the jack, the queen, the king and the ace of one suit. One is drawn. The chance it is the queen. One of five. Now suppose it was the queen, and it is put aside rather than returned. A second card is drawn.

14:13The list is different now. Four cards: the ten, the jack, the king and the ace. The chance the second is an ace. One of four, not one of five. And the chance the second is a queen. Zero. Not small. Zero. There is no queen left to draw. A lot of twenty bulbs, four of them faulty. A bulb taken at random is faulty with chance four over twenty, one fifth.

14:42Now say that first bulb was sound, and it is not put back. Nineteen bulbs remain, fifteen of them sound, so the next one is sound with chance fifteen over nineteen. Both numbers changed, and they changed because the list did. So, the whole thing, as something to do rather than something to know. One. Read the description and ask what a single result of this experiment actually is. A card. An ordered pair of dice. A string of heads and tails, one letter per coin.

15:18Two. Get the size of that list. If the question hands you contents rather than a total, add them up first. The denominator is very often a number the question never prints. Three. Check the results are interchangeable. Identical cards, at random, well shuffled, marbles of the same size. Those phrases are the licence, and without something like them the count is not a probability. Four. Only now go looking for the results that fit the description, and count those.

15:55Five. If the experiment moves - something taken out and not put back - go back to step two. Get the list right and the arithmetic is one division. Get the list wrong, and every step after it is flawless.