# Pillow-Problems: Problem #50

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Problem #47 | Problem #52
For more information on this collection, see Pillow-Problems by Charles L. Dodgson (Lewis Carroll).

## Problem:

50.

There are 2 bags, H and K, each containing 2 counters: and it is known that each counter is either black or white. A white counter is added to bag H, the bag is shaken up, and one counter transferred (without looking at it) to bag K, where the process is repeated, a counter being transferred to bag H. What is now the chance of drawing a white counter from bag H?

50.

Seventeen-twentysevenths.

## Solution:

50.

At first, the chance that bag H shall contain

 2W counters, is ¼; 1W and 1B, is ½; 2B, is ¼.
∴ after adding a W, the chance that it shall contain
 3W, is ¼; 2W, 1B, is ½; 1W, 2B, is ¼.
hence the chance of drawing a W from it is
¼ × 1 + ½ ×
 (2)/ (3)
+ ¼ ×
 (1)/ (3)
: i.e.
 (2)/ (3)
.
∴ the chance of drawing a B is
 (1)/ (3)

After transferring this (unseen) counter to bag K, the chance that it shall contain

3W,is
 (2)/ (3)
× ¼;
i.e.
 (1)/ (6)
.
2W, and 1B, is
 (2)/ (3)
× ½ +
 (1)/ (3)
× ¼;
i.e.
 (5)/ (12)
.
1W, 2B, is
 (2)/ (3)
× ¼ +
 (1)/ (3)
× ½;
i.e.
 (1)/ (3)
.
3B, is
 (1)/ (3)
× ¼;
i.e.
 (1)/ (12)
;
∴ the chance of drawing a W from it is
 (1)/ (12)
× 1 + ½ ×
 (2)/ (3)
+
 (1)/ (3)
×
 (1)/ (3)
;     i.e.
 (5)/ (9)
.
∴ the chance of drawing a B is
 (4)/ (9)
.

Before transferring this to bag H, the chance that bag H shall contain

2W, is ¼ × 1 + ½ ×
 (1)/ (3)
;
i.e.
 (5)/ (12)
.
1W, 1B, ½ ×
 (2)/ (3)
+ ¼ ×
 (2)/ (3)
;
i.e. ½
2B, ¼ ×
 (1)/ (3)
;
i.e.
 (1)/ (12)
.
∴, after transferring it, the chance that bag H shall contain
3W,is
 (5)/ (12)
×
 (5)/ (9)
;
i.e.
 (25)/ (108)
2W, 1B,
 (5)/ (12)
×
 (4)/ (9)
+ ½ ×
 (5)/ (9)
;
i.e.
 (50)/ (108)
1W, 2B, ½ ×
 (4)/ (9)
+
 (1)/ (12)
×
 (5)/ (9)
;
i.e.
 (29)/ (108)
3B,
 (1)/ (12)
×
 (4)/ (9)
;
i.e.
 (4)/ (108)
.

Hence the chance of drawing a W is

 (1)/ (108)
× { 25 × 1 + 50 ×
 (2)/ (3)
+ 29 ×
 (1)/ (3)
}; i.e.
 (17)/ (27)
.

Q.E.F.