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 Playing With Marbles (Posted on 2003-10-13)
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?

 See The Solution Submitted by Ravi Raja Rating: 3.0000 (10 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 re: Extension to the problem | Comment 14 of 22 |
(In reply to Extension to the problem by Trevor Leitch)

The probability you have the majority-black box, by Bayes' rule is:

(1/2)(3/4)^(n-1) / ((1/2)(3/4)^(n-1) + (1/2)(1/4)^(n-1))

which reduces to 3^(n-1)/(3^(n-1)+1)

So the probability of getting the nth ball as black would be

3^(n-1)/(3^(n-1)+1) (3/4) + (1 - 3^(n-1)/(3^(n-1)+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

```

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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 2003-10-14 10:37:33

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