Page:Philosophical Transactions - Volume 053.djvu/479

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Suppose, further, that he has heard 40 blanks drawn and 4 prizes; what will the before-mentioned chances be?

The answer here is .1525, for the former of these chances; and .527, for the latter. There will, therefore, now be an odds of only 5⁠1/2⁠ to 1 against the proportion of blanks to prizes lying between 9 to 1 and 11 to 1; and but little more than an equal chance that it is less than 9 to 1.

Once more. Suppose he has heard 100 blanks drawn and 10 prizes.

The answer here may still be found by the first rule; and the chance for a proportion of blanks to prizes less than 9 to 1 will be .44109, and for a proportion greater than 11 to 1 .3082. It would therefore be likely that there were not fewer than 9 or more than 11 blanks to a prize. But at the same time it will remain unlikely[1] that the true proportion should lie between 9 to 1 and 11 to 1, the chance for this being .2506 &c. There will therefore be still an odds of near 3 to 1 against this.

From these calculations it appears that, in the circumstances I have supposed, the chance for being right in guessing the proportion of blanks to prizes to be nearly the same with that of the number of blanks

  1. ↑ I suppose no attentive person will find any difficulty in this. It is only saying that, supposing the interval between nothing and certainty divided into a hundred equal chances, there will be 44 of them for a less proportion of blanks to prizes than 9 to 1, 31 for a greater than 11 to 1, and 25 for some proportion between 9 to 1 and 11 to 1; in which it is obvious that, though one of these suppositions must be true, yet, having each of them more chances against them than for them, they are all separately unlikely.
Vol. LIII
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