GAMING, lairs of. These are found ed on the doctrine of chances. See CHANCE.
M. de Moivre, in a treatise "De Men. sura Sortis," has computed the variety of chances in several cases that occur in gaming, the laws of which may be under stood by what follows : Suppose p the number of cases in which an event may happen, and q the number of cases wherein it may not happen, both sides have the degree of probability, which is to each other as p to q.
If two gamesters, A and B, engage on this footing, that, if the cases p happen, A shall win; but if q happen, B shall win, and the stake be a; the chance of A will be and that of B 4 as ; con if , if they sell the expectancies, they should have that for them respec tively.
If A and B play with a single die, on this condition, that, if A throw two or more aces at eight throws, he shall win ; otherwise B shall win ; what is the ratio of their chances ? Since there is but one case wherein an ace may turn up, and five wherein it may not, let a = 1, and b = 5. And again, since there are eight throws of the die,let n = 8 and you will have a — bn—n a b — 1, to bn n a bn-1 : that is, the chance of A will be to that of B, as 663,991 to 10,156,525, or nearly as 2 to 3.
A and B are engaged at single quoits, and, after playing some time, A wants 4 of being up, and B 6; but B is so much the better gamester, that his chance against A upon a single throw would be as 3 to 2; what is the ratio of their chances ? Since A wants 4, and B 6, game will be ended at nine throws; there fore raise a-I-6 to the ninth power, and it will be b 36 a7 b 5+84 bi+ 126 as 54+126 a4 bi, to 84 a a +6 a 63+b, • call a 3, and b 2, and you will have the ratio of chances in numbers, -viz. 1,759,077 to 194,048.
A and B play at single quoits, and A is the best gamester, so that he can give B 2 in 3; what is the ratio of their chances at a single throw ? Suppose the chances as z to 1, and raise z + 1 to its cube, which will be :3 + 3 z' + 3 z + 1. Now since A could give B 2 out of 3, A might undertake to win three rows running ; and, consequently, the chances in this case will be as z3 to 3 z' + 3 z 1. Hence, z3 = 3 z=
3 z -F 1: or, 2 z3 = z3 + 3 z= 3 z ? 1. And, therefore, z p/ 2 = z 1; 1 and, consequently, z The „p/ 2— 1' 1 chances therefore, are — — and 1, respectively.
Again, suppose I have two wagers de pending, in the first of which I have 3 to 2 the best of the lay, and in the second, 7 to 4, what is the probability I win both wagers ? 1. The probability of winning the first that is, the number of chances I 7 have to win divided by the number of all the chances : the probability of winning the second is . therefore, multiplying these two fractions together, the product will be 3 2 ' 1 which is the probability of 3 winning both wagers. Now, this fraction being subtracted from 1, the remainder is s which is the probability I do not win both wagers: therefore the odds against me are 34 to 21.
2. HI would know what the probability is of winning the first, and losing the se cond, I argue thus : the probability of winning the first is 4, the probability of losing the second is a therefore, mul tiplying . by the product 41 will be the probability of my winning the first, and losing the second; which being sub tracted from 1, there will remain 44, which is the probability I do not win the first, and at the same time lose the second.
3. If I would know what the probability is of winning the second, and at the same time losing the first, I say thus: the pro bability of winning the second is ; the probability of losing the first is 4 ; therefore, multiplying these two fractions together, the product 44/ is the proba bility I win the second, and also lose the first.
4. If I would know what the proba bility is of losing both wagers, I say, the probability of losing the first is 4, and the probability of losing the second _4 - therefore, the probability of los ing them both is • which being tracted from l, there remains 4.-k ; there fore, the odds of losing both wagers is 47 to 8.