Pillow-Problems: Problem #50 (Math Lair)

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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?

Answer:

50.

Seventeen-twentysevenths.

Solution:

50.

At first, the chance that bag H shall contain

2`W` counters,	is	¼;
1`W` and 1`B`,	is	½;
2`B`,	is	¼.

∴ after adding a W, the chance that it shall contain

3`W`,	is	¼;
2`W`, 1`B`,	is	½;
1`W`, 2`B`,	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,

(2)/

(3)

× ¼;

i.e.

(1)/

(6)

2W, and

1B,

(2)/

(3)

× ½ +

(1)/

(3)

× ¼;

i.e.

(5)/

(12)

1W,

2B,

(2)/

(3)

× ¼ +

(1)/

(3)

× ½;

i.e.

(1)/

(3)

3B,

(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,

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,

(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.