[ 413 ]
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 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 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.