# Problem involving selfing (inbreeding it with itself) a plant to generate purebred lines

I am working on a past exam problem where the first bit is as follows

A plant is repeatably selfed to generate inbred lines. Let $\mathbb{P}(He|He)$ denote the probability that a heterozygous parent gives rise to a heterozygous offspring, and let $\mathbb{P}(Ho|He)$ denote the probability that a heterozygous parent gives rise to homozygous offspring.

I am asked to work out the two probabilities as well as an interpretation and state the value of $\mathbb{P}(Ho|Ho)$. The interpretation is obvious and I calculate the conditional probabilities to be

$$\mathbb{P}(He|He)=\frac12, \quad \mathbb{P}(Ho|He)=\frac12, \quad \mathbb{P}(Ho|Ho)=1$$

Next I am given the following

The inbreeding protocol is carried out $i$ times in an effort to obtain representatives of all alleles that can occur at a locus of interest. Suppose that there are 3 alleles, equally prevalent in the population at large and not under selective pressure during the inbreeding experiment. Let $p_j(i)$ denote the probability that, after $i$ repetitions the lines share $j$ distinct alleles between them. Let $p(i) \in [0,1]^3$ be a vector whose $j$th element is $p_j(i)$. Consider the following equation: $$p(i+1) = M p(i)$$

I would like to calculate the transition matrix $M$ and explain the boundary condition $p(1) = [ 1, 0 ,0 ]^T$.

I am lost on how to do this, I'm given the answer for $M$ to be $$\begin{pmatrix} \frac13 & 0 & 0 \\ \frac23 & \frac23 & 0 \\ 0 & \frac13 & 1 \end{pmatrix}$$

But for example looking along the first line I would interpret this as $$\mathbb{P}(\text{There is 1 distinct allele between all lines at time t}) = \frac{1}{3} \mathbb{P}(\text{There is 1 distinct allele between all lines at time t-1})$$

However if there is a single allele between the lines at time $t-1$ then this means every line is homozygous for a single allele so why would that change at the next time step?!

As for the boundary condition, I would of thought it would be $[0, 0, 1]^T$ since is the alleles occur with equal chance and you select a large enough number of lines, you should find all $3$ alleles between them. I was thinking perhaps there is a mistake in the question/solution and the vectors are defined the wrong way round, but even then I would be confused. Surely the number of lines we have makes a large difference, if every line is of genotype $\alpha_1 \alpha_2$ and we would like the probability that there is a single allele between lines at the next selfing step, then every single line would need to become homozygous for the same allele, which seems unlikely if we have a massive number of lines.

Can someone help me work out what is going on here?