be white, and in two-thirds black. In one-third of those urns of which the first ball was white, and also in one-third of those in which the first ball was black, the second ball would be white. In this way, we should have a distribution like that shown in the following table, where w stands for a white ball and b for a black one. The reader can, if he chooses, verify the table for himself.
| wwww. |
| wwwb.wwbw.wbww.bwww. |
| wwwb.wwbw.wbww.bwww. |
| wwbb.wbwb.bwwb.wbbw.bwbw.bbww. |
| wwbb.wbwb.bwwb.wbbw.bwbw.bbww. |
| wwbb.wbwb.bwwb.wbbw.bwbw.bbww. |
| wwbb.wbwb.bwwb.wbbw.bwbw.bbww. |
| wbbb.bwbb.bbwb.bbbw. |
| wbbb.bwbb.bbwb.bbbw. |
| wbbb.bwbb.bbwb.bbbw. |
| wbbb.bwbb.bbwb.bbbw. |
| wbbb.bwbb.bbwb.bbbw. |
| wbbb.bwbb.bbwb.bbbw. |
| wbbb.bwbb.bbwb.bbbw. |
| wbbb.bwbb.bbwb.bbbw. |
bbbb.
bbbb.
bbbb.
bbbb.
bbbb.
bbbb.
bbbb.
bbbb.
bbbb.
bbbb.
bbbb.
bbbb.
bbbb.
bbbb.
bbbb.
bbbb.
bbbb.
In the second group, where there is one b, there are two sets just alike; in the third there are 4, in the fourth 8, and in the fifth 16, doubling every time. This is because we have supposed twice as many black balls in the granary as white ones; had we supposed 10 times as many, instead of
1,2,4,8,16
sets we should have had
1,10,100,1000,10000
sets; on the other hand, had the numbers of black and white balls in the granary been even, there would have been but one set in each group. Now suppose two balls were drawn from one of these urns and were found to be both white, what would be the probability of the next one being white? If the two drawn out were the first two put into the urns, and the next to be drawn out were the third put in, then the probability of this third being white would be the same whatever the colors of the first two, for it has been supposed that just the same proportion of urns has the third ball white among those which have the first two white-white, white-black, black-white, and black-black. Thus, in this case, the chance
