Page:Philosophical Transactions - Volume 053.djvu/479

This page has been proofread, but needs to be validated.

[ 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 51/2 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 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

Vol. LIII
drawn
Hhh
  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.