Why is the probability of fixation of an allele equal to its frequency? **Introduction** In a panmictic population, the probability of fixation of an allele at a neutral locus is equal to its frequency at that time. I will refer to this probability of fixation as calculated at time $t$ as $P_{fix,t}$. If $p_t$ is the frequency of the allele $A_1$ at time $t$, then the probability of the allele $A_1$ to reach fixation (rather than disappearing) is $P_{fix,t}=p_t$. Typically, the generation when the mutation occurred, the probability of the new allele to get fixed is $P_{fix,0}=p_0=\frac{1}{2N}$, where $N$ is the population size. **Question** This simple and classic result makes very good intuitive sense to me. However, I would fail to provide a **mathematical proof**. Can you please demonstrate that $P_{fix,t}=p_t$?

Several proofs are given here (p. 9). My favorite comes from the genealogical argument:

Consider the situation where there are $2N$ alleles: $A_1$, $A_2$, $A_3$ ... $A_{2N}$.

By the genealogical argument, we may state that at $t = \infty$, all alleles at this locus will be direct descendants of one particular allele present at $t = 0$.

Allelic variants at this locus are selectively neutral, so $Pr(A_{1fix})$ $=$ $Pr(A_{2fix})$ $=$ $Pr(A_{3fix})$ $= ... =$ $Pr(A_{2Nfix})$. For any given allele present at $t = 0$, the probability of fixation is therefore $\frac{1}{2N}$.

Now define allelic variants $A$ and a as complementary, non-overlapping groups of the initial alleles, such that $n_A$ + $n_a$ $=$ $2N$. From the above, the probability of fixation of an allele within group $A$ is $n_A * \frac{1}{2N} = \frac{n_A}{2N} = p_0$.