Take the case of three dice: As three numbers out of six must come up, it might be supposed that it was an even thing that one would be an ace. But the possible throws with three dice are 6 × 6 × 6 = 216; and those that do not contain an ace are 5 × 5 × 5 = 125; so that the odds against getting an ace in one throw with three dice, or three throws with one die, are 125/216, or 125 to 91 against it.
To find the probability of getting a given total on the faces of two or three dice we must find the number of ways that the desired number can come. In the 36 possible throws with two dice there are 6 which will show a total of seven pips. The probability of throwing seven is therefore 6/36, or 5 to 1 against it. A complete list of the combinations with two dice were given in connection with Craps.
Poker. In calculating the probability of certain conflicting events, both of which cannot occur, but either of which would be favourable, we must make the denominator of our fraction equal in both cases, which will, of course, necessitate a proportionate change in our numerator. Suppose a poker player has three of a kind, and intends to draw one card only, the odds against his getting a full hand are 1/16; against getting four of a kind, 1/48. To find the total probability of improvement, we must make the first fraction proportionate to the last, which we can do by multiplying it by 3. The result will be 3/48 + 1/48 = 4/48; showing that the total chance of improvement is 1 in 12, or 11 to 1 against it.
Whist. To calculate the probable positions of certain named cards is rather a difficult matter, but the process may be understood from a simple example. Suppose a suit so distributed that you have four to the King, and each of the other players has three cards; what are the probabilities that your partner has both Ace and Queen? The common solution is to put down all the possible positions of the two named cards, and finding only one out of nine to answer, to assume that the odds are 8 to 1 against partner having both cards. This is not correct, because the nine positions are not equally probable. We must first find the number of possible positions for the Ace and Queen separately, afterward multiplying them together, which will give us the denominator; and then the number of positions that are favourable, which will give us the numerator.
As there are nine unknown cards, and the Ace may be any one of them, it is obvious that the Queen may be any one of the remaining eight, which gives us 9 × 8 = 72 different ways for the two cards to lie. To find how many of these 72 will give us both cards in partner’s hand we must begin with the ace, which may be any one of his three cards. The Queen may be either of the other two, which gives us the numerator, 3 × 2 = 6; and the fraction of probability, 6/72, = 1/12; or 11 to 1 against both Ace and Queen.
If we wished to find the probability of his having the Ace, but not the Queen, our denominator would remain the same; but the numerator would be the three possible positions of the Ace, multiplied by the six possible positions of the Queen among the six other unknown cards, in the other hands, giving us the fraction 18/72. The same would be true of the Queen but not the Ace. To prove both these, we must find the probability that he has neither Ace nor Queen. There being six cards apart from his three, the Ace may be any one of them, and the Queen may be any one of the remaining five. This gives us 6 × 5 = 30, and the fraction 30/72. If we now add these four numerators together, we have:—for both cards in partner’s hand, 6; for Ace alone, 18; for Queen alone, 18; and for neither, 30; a total of 72, or unity, proving all the calculations correct.
In some of the problems connected with Whist, it is important to know the probability of the suits being distributed in various ways among the four players at the table; or, what is the same thing, the probable distribution of the four suits in any one hand. The author is indebted to Dr. Pole’s “Philosophy of Whist” for these calculations. As an example of the use of this table, suppose it was required to find the probability of any other player at the table holding four or more trumps if you had six. Take all the combinations in which the figure 6 appears, and add together the number of times they will probably occur. That will be your denominator, 166. The numerator will be the number of times that the combinations occur which contain a figure larger than 3, in addition to the 6. This will be found to be 74, and the probability will therefore be 74/166.
| DISTRIBUTIONS. | TIMES IN 1000. | |||
|---|---|---|---|---|
| 8 | 2 | 2 | 1 | 2 |
| 8 | 3 | 1 | 1 | 1½ |
| 8 | 3 | 2 | 0 | 1 |
| 8 | 4 | 1 | 0 | ½ |
| 8 | 5 | 0 | 0 | 0 |
| 7 | 2 | 2 | 2 | 5 |
| 7 | 3 | 2 | 1 | 19 |
| 7 | 3 | 3 | 0 | 3 |
| 7 | 4 | 1 | 1 | 4 |
| 7 | 4 | 2 | 0 | 3 |
| 7 | 5 | 1 | 0 | 1 |
| 7 | 6 | 0 | 0 | 0 |
| 6 | 3 | 2 | 2 | 57 |
| 6 | 3 | 3 | 1 | 35 |
| 6 | 4 | 2 | 1 | 47 |
| 6 | 4 | 3 | 0 | 13 |
| 6 | 5 | 1 | 1 | 7 |
| 6 | 5 | 2 | 0 | 6 |
| 6 | 6 | 1 | 0 | 1 |
| 5 | 3 | 3 | 2 | 155 |
| 5 | 4 | 2 | 2 | 106 |
| 5 | 4 | 3 | 1 | 130 |
| 5 | 4 | 4 | 0 | 12 |
| 5 | 5 | 2 | 1 | 32 |
| 5 | 5 | 3 | 0 | 9 |
| 4 | 3 | 3 | 3 | 105 |
| 4 | 4 | 3 | 2 | 215 |
| 4 | 4 | 4 | 1 | 30 |
MARTINGALES. Many gamblers believe that as the science of probabilities teaches us that events will equalise themselves in time, all that is necessary is to devise some system that will keep a person from guessing, so that he may catch the pendulum as it swings; and to add to it some system of betting, so that he will have the best of it in the long run. Some content themselves with playing a “system” against banking games, which is merely a guide to the placing of the bets, the simplest example of which would be to bet always on heads if a coin was tossed a thousand times, or to bet on nothing but red at Roulette. Others depend more on martingales, which are guides to the amount of the bets themselves, irrespective of what they are placed on.
The most common form of martingale is called doubling up, which proceeds upon the theory that if you lose the first time and bet double the amount the next time, and continue to double until you win, you must eventually win the original amount staked. If there was no end to your capital, and no betting limit to the game, this would be an easy way to make money; but all banking games have studied these systems, and have so arranged matters that they can extend their heartiest welcome to those who play them.