If A has £0, then he cannot play, so will have $0 for ever after.
If A has £1, then after 1 toss he will have either £0 or £2, each with probability 1/2, and £1, £3 or £4 with probability 0..
If A has £2, then after 1 toss he will have either £1 or £3, each with probability 1/2, and £0, £2 or £4 with probability 0..
If A has £4, then player B has £0, so cannot play and player A has £4 for ever after.
We can construct the table:
| A has\A has After 1 Toss | £0 | £1 | £2 | £3 | £4 |
| £0 | 1 | 0 | 0 | 0 | 0 |
| £1 | 1/2 | 0 | 1/2 | 0 | 0 |
| £2 | 0 | 1/2 | 0 | 1/2 | 0 |
| £3 | 0 | 0 | 1/2 | 0 | 1/2 |
| £4 | 0 | 0 | 0 | 0 | 1 |
\[P= \left( \begin{array}{ccccc} 1 & 0 & 0 & 0 & 0 \\ 1/2 & 0 & 1/2 & 0 & 0 \\ 0 & 1/2 & 0 & 1/2 & 0 \\ 0 & 0 & 1/2 & 0 & 1/2 \\ 0 & 0 & 0 & 0 & 1 \end{array} \right)\]
.The four step transition matrix
\[P^4\]
will give us the probabilities of each player having an amount of money after 4 tosses. It is \[P^4= \left( \begin{array}{ccccc} 1 & 0 & 0 & 0 & 0 \\ 5/8 & 1/8 & 0 & 1/8 & 1/8 \\ 3/8 & 0 & 1/4 & 0 & 3/8 \\ 1/8 & 1/8 & 0 & 1/8 & 5/8 \\ 0 & 0 & 0 & 0 & 1 \end{array} \right)\]
.The answer is the entry in the third row (starting with £2) and fits column (ending up with £0) and is equal to 3/8.