Problem:
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:
Seventeentwentysevenths.
Solution:
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 ×
+ ¼ ×
: i.e.
.
∴ the chance of drawing a
B is
After transferring this (unseen) counter to bag K, the chance that it shall contain
3W,   is × ¼;  i.e. .

2W, and  1B,  is × ½ + × ¼;  i.e. .

1W,  2B,  is × ¼ + × ½;  i.e. .

3B,   is × ¼;  i.e. ;

∴ the chance of drawing a W from it is
∴ the chance of drawing a
B is
.
Before transferring this to bag H, the chance that bag H shall contain
2W,  is  ¼ × 1 + ½ × ;
 i.e. .

1W, 1B,   ½ × + ¼ × ;
 i.e. ½

2B,   ¼ ×;  i.e. .

∴, after transferring it, the chance that bag H shall contain
3W,  is 
×
;  i.e.

2W, 1B,  
×
+ ½ ×
;  i.e.

1W, 2B,   ½ ×
+
×
;  i.e.

3B,  
×
;  i.e.
.

Hence the chance of drawing a W is
× { 25 × 1 + 50 ×
+ 29 ×
}; i.e.
.
Q.E.F.