Question

One purse contains $5$ shillings and $1$ sovereign: a second purse contains $6$ shillings. Two coins are taken from the first and placed in the second; then $2$ are taken from the second and placed in the first: find the probable value of the contents of each purse.

Answer 1

The chance that the sovereign is in the first purse is equal to the sum of the chances that it has moved twice and that it has not moved at all;
That is, the chance $=\frac{1}{3}\cdot \frac{1}{4}+\frac{2}{3}\cdot 1=\frac{3}{4}$.
$\therefore$ the chance that the sovereign is in the second purse $=\frac{1}{4}$.
Hence the probable value of the first purse $=\frac{3}{4}$ of $25s.+\frac{1}{4}$ of $6s.=1.0s.3d.$
$\therefore$ the probable value of the second purse $=31s.20\frac{1}{4}s.=10s.9d.$
Or the problem may be solved as follows:
The probable value of the coins removed $=\frac{1}{3}$ of $25s.=8\frac{1}{3}s.;$
the probable value of the coins brought back $=\frac{1}{4}$ of $(6s.+8\frac{1}{3}s.)=3\frac{7}{12}s.;$
$\therefore$ the probable value of the first purse$=(25-8\frac{1}{3}+3\frac{7}{12})$ shillings$=1.0s.3d.,$ as before.