Pillow-Problems: Problem #41

Problem #40 | Problem #44
For more information on this collection, see Pillow-Problems by Charles L. Dodgson (Lewis Carroll).

Problem:

My friend brings me a bag containing four counters, each of which is either black or white. He bids me draw two, both of which prove to be white. He then says "I meant to tell you, before you began, that there was at least one white counter in the bag. However, you know it now, without my telling you. Draw again."

(1) What is now my chance of drawing white?

(2) What would it have been, if he had not spoken?

Answer:

(1) Seven-twelfths.   (2) One-half.

Solution:

(1) As there was certainly at least one W in the bag at first, the ‘a priori’ chances for the various states of the bag, ‘WWWW, WWWB, WWBB, WBBB,’ were ‘

(1)/

(8)

,

(3)/

(8)

,

(3)/

(8)

,

(1)/

(8)

’.

These would have given, to the observed event, the chances ‘1, ½,

(1)/

(6)

, 0’.

Hence the chances, after the event, for the various states, are proportional to ‘

(1)/

(8)

· 1,

(3)/

(8)

· ½,

(3)/

(8)

·

(1)/

(6)

’.; i.e .to ‘

(1)/

(8)

,

(3)/

(16)

,

(1)/

(16)

’; i.e. to ‘2, 3, 1’. Hence their actual values are ‘

(1)/

(3)

, ½,

(1)/

(6)

’.

Hence the chance, of now drawing W, is ‘

(1)/

(3)

· 1 + ½ · ½’; i.e. it is

(7)/

(12)

.

Q.E.F.

(2) If he had not spoken, the ‘a priori’ chances for the states ‘WWWW, WWWB, WWBB, WBBB,’, would have been

‘1, 4, 6, 4, 1’

16

.

These would have given, to the observed event, the chances ‘1, ½

1

6

, 0, 0’.

Hence the chances, after the event, for the various states, are proportional to ‘

1

16

· 1, ¼ · ½,

1

6

·

3

8

’; i.e. to ‘1, 2, 1’. Hence their actual values are ‘¼, ½, ¼’.

Hence the chance, of now drawing W, is ‘¼ · 1 + ½ · ½’; i.e. it is ½.

Q.E.F.