That this is so may easily be demonstrated by putting down on a sheet of paper the total number of events that may happen if any agreed number of games are played, expressing wins by a stroke, and losses by a cipher. Take the case of two games only. There are four different events which may happen to Smith, as shown in the margin. He may win both games or lose both; or he may win one and lose the other, either first. Only one of these four equally probable events being favourable to his winning both games, and three being unfavourable, the odds are 3 to 1 that he does not win both; but these are the odds before he begins to play. Having won the first game, there are only two events possible, those which begin with a win, and he has an equal chance to win again.
| GAMES. | ||
|---|---|---|
| 1st | 2nd | 3rd |
| 1 | 1 | 1 |
| 1 | 1 | 0 |
| 1 | 0 | 1 |
| 1 | 0 | 0 |
| 0 | 0 | 0 |
| 0 | 0 | 1 |
| 0 | 1 | 0 |
| 0 | 1 | 1 |
If the agreement had been to play three games, there would have been eight possible events, one of which must happen but all of which were equally probable. These are shown in the margin. If Smith wins the first game, there are only four possible events remaining; those in which the first game was won. Of these, there are two in which he may win the second game, and two in which he may lose it, showing that it is still exactly an even thing that he will win the second game. If he wins the second game, there are only two possible events, the first two on the list in the margin, which begin with two wins for Smith. Of these he has one chance to win the third game, and one to lose it. No matter how far we continue a series of successive events it will always be found that having won a certain number of games, it is still exactly an even thing that he will win the next also. The odds of 1023 to 1 against his winning ten games in succession existed only before he began to play. After he has won the first game, the odds against his winning the remaining nine are only 511 to 1, and so on, until it is an even thing that he wins the tenth, even if he has won the nine preceding it.
In the statistics of 4000 coups at roulette at Monte Carlo it was found that if one colour had come five times in succession, it was an exactly even bet that it would come again; for in twenty runs of five times there were ten which went on to six. In the author’s examination of 500 consecutive deals of faro, there were 815 cards that either won or lost three times in succession, and of these 412 won or lost out. In a gambling house in Little Rock a roulette wheel with three zeros on it did not come up green for 115 rolls, and several gamblers lost all they had betting on the eagle and O’s. When the game closed the banker informed them that the green had come up more than twenty times earlier in the evening. They thought the maturity of the chances would compel the green to come; whereas the chances really were that it would not come, as it had over-run its average so much earlier in the evening. The pendulum swings as far one way as the other, but no method of catching it on the turn has ever yet been discovered.
Compound Events. In order to ascertain the probability of compound or concurrent events, we must find the product of their separate probability. For instance: The odds against your cutting an ace from a pack of 52 cards are 48 to 4, or 12 to 1; because there are 52 cards and only 4 of them are aces. The probability fraction is therefore 1/13. But the probabilities of drawing an ace from two separate packs are 1/13 × 1/13 = 1/169, or 168 to 1 against it.
Suppose a person bets that you will not cut a court card, K Q or J, from a pack of 52 cards, what are the odds against you? In this case there are three favourable events, but only one can happen, and as any of them will preclude the others, they are called conflicting events, and the probability of one of them is the sum of the probability of all of them. In this case the probability of any one event separately is 1/13, and the sum of the three is therefore 1/13 + 1/13 + 1/13 = 3/13; or ten to 3 against it.
In order to prove any calculation of this kind all that is necessary is to ascertain the number of remaining events, and if their sum, added to that already found, equals unity, the calculation must be correct. For instance: The probability of turning a black trump at whist is 13/52 + 13/52 = 26/52; because there are two black suits of 13 cards each. The only other event which can happen is a red trump, the probability of which is also 26/52, and the sum of these two probabilities is therefore 26/52 + 26/52 = 52/52, or unity.
Another fallacy in connection with the maturity of the chances is shown in betting against two successive events, both improbable, one of which has happened. The odds against drawing two aces in succession from a pack of 52 cards are 220 to 1; but after an ace has been drawn the odds against the second card being an ace also are only 16 to 1, although some persons would be mad enough to bet 1000 to 1 against it, on the principle that the first draw was a great piece of luck and the second ace was practically impossible. While the four aces were in the pack the probability of drawing one was 4/52. One ace having been drawn, 3 remain in 51 cards, so the probability of getting the second is 3/51, or 1/17. Before a card was drawn, the probability of getting two aces in succession was the product of these fractions; 1/13 × 1/17 = 1/221. On the same principle the odds against two players cutting cards that are a tie, such as two Fours, are not 220 to 1, unless it is specified that the first card shall be a Four. The first player having cut, the odds against the second cutting a card of equal value are only 16 to 1.
Dice. In calculating the probabilities of throws with two or more dice, we must multiply together the total number of throws possible with each die separately, and then find the number of throws that will give the result required. Suppose two dice are used. Six different throws may be made with each, therefore 6 × 6 = 36 different throws are possible with the two dice together. What are the odds against one of these dice being an ace? A person unfamiliar with the science of probabilities would say that as two numbers must come up, and there are only six numbers altogether, the probability is 2/6, or exactly 2 to 1 against an ace being thrown. But this is not correct, as will be immediately apparent if we write out all the 36 possible throws with two dice; for we shall find that only 11 of the 36 contain an ace, and 25 do not. The proper way to calculate this is to take the chances against the ace on each die separately, and then to multiply them together. There are five other numbers that might come up, and the fraction of their probability is ⅚ × ⅚ = 25/36, or 25 to 11 in their favour.