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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 512 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 ſirſt 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*

- ↑ 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.