Suppose there are two horses (amongst others) engaged in a race, and that the odds are 2 to 1 against one, and 4 to 1 against the other-what are the odds that one of the two horses will win the race? This case will doubtless remind my readers of an amusing sketch by Leech, called—if I remember rightly—‘Signs of the Commission.’ Three or four undergraduates are at a ‘wine,’ discussing matters equine. One propounds to his neighbour the following question: I say, Charley, if the odds are 2 to 1 against _Rataplan_, and 4 to 1 against _Quick March_, what’s the betting about the pair?’—‘Don’t know, I’m sure,’ replies Charley; ‘but I’ll give you 6 to 1 against them.’ The absurdity of the reply is, of course, very obvious; we see at once that the odds cannot be heavier against a pair of horses than against either singly. Still, there are many who would not find it easy to give a correct reply to the question. What has been said above, however, will enable us at once to determine the just odds in this or any similar case. Thus-the odds against one horse being 2 to 1, his chance of winning is equal to that of drawing one white ball out of a bag of _three_, one only of which is white. In like manner, the chance of the second horse is equal to that of drawing one white ball out of a bag of _five_, one only of which is white. Now we have to find a number which is a multiple of both the numbers three and five. Fifteen is such a number. The chance of the first horse, modified according to the principle explained above, is equal to that of drawing a white ball out of a bag of fifteen of which _five_ are white. In like manner, the chance of the second is equal to that of drawing a white ball out of a bag of fifteen of which _three_ are white. Therefore the chance that _one of the two_ will win is equal to that of drawing a white ball out of a bag of fifteen balls of which _eight_ (_five_ added to _three_) are white. There remain _seven_ black balls, and therefore the odds are 8 to 7 _on_ the pair. To impress the method of treating such cases on the mind of the reader, let us take the betting about three horses—say 3 to 1, 7 to 2, and 9 to 1 _against_ the three horses respectively. Then their respective chances are equal to the chance of drawing (1) one white ball out of _four_, one only of which is white; (2) a white ball out of _nine_, of which two only are white; and (3) one white ball out of _ten_, one only of which is white. The least number which contains four, nine, and ten is 180; and the above chances, modified according to the principle explained above, become equal to the chance of drawing a white ball out of a bag containing 180 balls, when 45, 40, and 18 (respectively) are white. Therefore, the chance that one of the three will win is equal to that of drawing a white ball out of a bag containing 180 balls, of which 103 (the sum of 45, 40, and 18) are white. Therefore, the odds are 103 to 77 _on_ the three.