The Coin Tossing Game With Each Player Starting With Different Amounts of Money

Two players pay a coin tossing game for money. If the coin shows heads, A wins ad if it shows tails, B wins. Draws are not possible. If player A starts with £3 and player B starts with £2, and each bets £1 on each throw, construct a transition matrix and find how long it will take for each player to win or lose all their money.
Each player can have between £0 and £5.
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, £4 or £5 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, £4 or £5 each with probability 0..
If A has £4, then after 1 toss he will have either £3 or £5, each with probability 1/2, and £0, £1, £2 or £4 with probability 0.
If A has £5, then player B has £0, so cannot play and player A has £5 for ever after.
We can construct the table:

A has\A has After 1 Toss	£0	£1	£2	£3	£4	£5
£0	1	0	0	0	0	0
£1	1/2	0	1/2	0	0	0
£2	0	1/2	0	1/2	0	0
£3	0	0	1/2	0	1/2	0
£4	0	0	0	1/2	0	1/2
£5	0	0	0	0	0	1

The transition matrix is  .
An absorbing state of a Markov chain when the probability of that state in the transition matrix is 1. Rewrite the table as

A has\A has After 1 Toss	£0	£5	£1	£2	£3	£4
£0	1	0	0	0	0	0
£5	0	1	0	0	0	0
£1	1/2	0	0	1/2	0	0
£2	0	0	1/2	0	1/2	0
£3	0	0	0	1/2	0	1/2
£4	0	1/2	0	0	1/2	0

This matrix is of the form    where
 

If a player starts with £2 the expected length of the game is (from row 2)1.2+2.4+1.6+0.8=6. If a player starts with £3 the expected length of the game is (from row 3) 0.8+1.6+2.4+1.2=6.