You choose one of two identical looking bags at random. One bag has three black marbles and one white marble. The other has three white marbles and one black marble.
After choosing a bag you draw one marble out at random. You notice it is black. You then put it back and draw another marble out of the same bag at random.
What is the probability that the second marble drawn is black?
(In reply to
Extension to the problem by Trevor Leitch)
The probability you have the majorityblack box, by Bayes' rule is:
(1/2)(3/4)^(n1) / ((1/2)(3/4)^(n1) + (1/2)(1/4)^(n1))
which reduces to 3^(n1)/(3^(n1)+1)
So the probability of getting the nth ball as black would be
3^(n1)/(3^(n1)+1) (3/4) + (1  3^(n1)/(3^(n1)+1)) (1/4)
As n gets larger, this approaches 3/4, as it becomes a near certainty you have the first bag.
2 0.750000 0.250000 0.625000
3 0.900000 0.100000 0.700000
4 0.964286 0.035714 0.732143
5 0.987805 0.012195 0.743902
6 0.995902 0.004098 0.747951
7 0.998630 0.001370 0.749315
8 0.999543 0.000457 0.749771
9 0.999848 0.000152 0.749924
10 0.999949 0.000051 0.749975
11 0.999983 0.000017 0.749992
12 0.999994 0.000006 0.749997
13 0.999998 0.000002 0.749999
14 0.999999 0.000001 0.750000
15 1.000000 0.000000 0.750000

The above from a spreadsheet, showing the probability that you have bag 1, the probability you have bag 2 and the probability that the next ball is black.

Posted by Charlie
on 20031014 10:37:33 