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guserter 發表於 2017-4-24 15:26

Suppose there are two horses

We must notice also that the number of balls may be increased to any extent, provided the proportion277 between the total number and the number of a specified colour 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.


(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 [url=http://blog.she.com/montgomery/2017/04/24/zsaxcdef/][color=#0F0F0F]Two considerations[/color][/url][url=http://blog.cnyes.com/My/suggestegt/article2396159][color=#0F0F0F] must have[/color][/url][url=http://www.cmarket.tw/barret/barret?n=convew&i=11420][color=#0F0F0F] caused Scheer[/color][/url][url=http://www.napis.sk/ostatne/pidfewh/][color=#0F0F0F] the gravest[/color][/url][url=http://blogg.improveme.se/meeting/2017/04/24/eorungk/][color=#0F0F0F] possible anxiety.[/color][/url].

Thus-the odds against one horse being 2 to 1, his chance of winning is equal to that of drawing one278 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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