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ds against a horse are 7 to 1, we infer that the cognoscenti consider his chance equal to that of drawing one particular ball out of a bag of eight. A similar treatment applies when the odds are not given as so many to one. Thus, if the. odds against a horse are as 5 to 2, we infer that the horse's chance is equal to that of drawing a white ball out of a bag containing five black and two white balls--or seven in all. We must notice also that the number of balls may be increased to any extent, provided the proportion between the total number and the number of a specified color remains unchanged. Thus, if the odds are 5 to 1 against a horse, his chance is assumed to be equivalent to that of drawing one white ball out of a bag containing six balls, only one of which is white; or to that of drawing a white ball out of a bag containing sixty balls, of which ten are white--and so on. This is a very important principle, as we shall now see.
Suppose there are two horses (amongst others) engaged in a race, and that the odds are 2 to 1 against one, and 4 to I against the other-what are the odds
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 neighbor the following question :--' I say, Charley, if the odds are 2 to I against Rat plan, 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.
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