3 football clubs run an ad campaign
A supporter of club A has 10% chance switching B and 1% switching C
A supporter of club B has a 5% chance of switching to C and 1% of switching to A
A supported of club C has a 6% chance of switching to A and 2% of switching to B
10000 potential supporters
Find Long Run average expected supporters for each club
0.89 O.O1 0.06 = 0.96 = 3200
0.1 0.94 0.02 = 1.06 = 3533
0.01 0.05 0.92 = 0.98 = 3267
I don't think this is correct, 3x3 matrix, tried formatting but won't work am I on the right track with a 3x3 matrix?
We can construct a system of linear equations to figure out the number each of the teams has after a long time:
$$A = 0.89A+0.01B+0.06C\\ B = 0.1A + 0.94B + 0.02C\\ C = 0.01A + 0.05B + 0.92C$$
This is how the numbers of supporters change in one "iteration". If we subtract the left side from both sides, we end up with a nice system in terms of $A, B,$ and $C$: $$\begin{bmatrix}-0.11 & 0.01 & 0.06\\ 0.1 &-0.06 & 0.02\\ 0.01 & 0.05 & -0.08\end{bmatrix}\begin{bmatrix}A\\B\\C\end{bmatrix} = \begin{bmatrix}0\\0\\0\end{bmatrix}.$$
Solving this yields $B$ and $C$: $B = \frac{41A}{19}$ and $C = \frac{28A}{19}$, however leaves $A$ free to be any number.
To figure out exactly how many fans each team ends up with, we must set $A+B+C$ equal to $10000$. This will be easily solved for $A$, given that $B$ and $C$ are defined as fractions of it.
$$A+B+C = \frac{19A+41A+28A}{19} = \frac{88}{19}A=10000\\ \implies A = \frac{190000}{88}\approx2159.$$
Thus, after a long time, club $A$ will have about $2159$ fans, $B$ will have $4659$ fans, and club $C$ will have $3182$ fans.