In ellections between two candidates candidate A got $a$ number of votes and B got $b$ number of votes. It is given that $a>b$, assuming the votes were counted uniformly show that the probability that candidate A led over candidate B across the counting process equals to: $\frac{a-b}{a+b}$.
This is what I did so far: 1) define $S_k$ which will be the difference between "a" and "b" votes after "k" counted votes. 2) show that $X_k = \frac{S_a+b-k}{a+b-k}$ is martingale. 3) use the SOT and show that $E(X_0)=E(X_T)$ hence equal to $\frac{a-b}{a+b}$
I'm facing difficulties with showing that $X_k = \frac{S_a+b-k}{a+b-k}$ is martingale.
Here is what I tried:
$$E(X_{k+1} \mid X_0,\dots,X_k) = E(X_{k} + B_{k+1} \mid X_0,\dots,X_k)$$
where:
$B_{k+1}$ is if "a"/"b" vote was counted at the k+1 step
Thank you for helping :)
Let $a > b$ and $n = a + b$ with $S_k = $ the number of votes A is leading by after $k$ votes counted. Then $S_n = a - b$ and $X_k = \frac{S_{n - k}}{n - k}$ for $0 \leq k \leq n - 1$. To show $X_k$ is a martingale we have to find
$$E[X_k|X_0,...,X_{k-1}] = E\bigg[\frac{S_{n - k}}{n - k}|S_n,...,S_{n-k+1}\bigg]$$
First let $a_k$ = number of votes for A after $k$ votes are counted and $b_k$ = number of votes for B after $k$ votes are counted. We have that
$$a_{n-k+1} = \frac{n - k + 1 + S_{n - k + 1}}{2}$$ $$b_{n-k+1} = \frac{n - k + 1 - S_{n - k + 1}}{2}$$ and $$ S_{n-k} = \begin{cases} S_{n - k + 1} + 1 \quad \text{if n - k + 1 vote is for B} \\ S_{n - k + 1} - 1 \quad \text{if n - k + 1 vote is for A} \end{cases} $$
This leads to $$E[S_{n - k}|S_{n-k+1}] = (S_{n-k+1} + 1)\frac{a_{n - k + 1}}{(n - k + 1)} + (S_{n-k+1} - 1)\frac{b_{n - k + 1}}{(n - k + 1)}$$
$$E[S_{n - k}|S_{n-k+1}] = S_{n-k+1}\frac{n - k}{n - k + 1}$$
and back to the original expectation we have $$E[X_k|X_0,...,X_{k-1}] = E\bigg[\frac{S_{n - k}}{n - k}|S_n,...,S_{n-k+1}\bigg] = \frac{S_{n - k + 1}}{n - k + 1} = X_{k-1}$$
We have that $X_k$ is a martingale. By the martingale stopping theorem we have $$E[X_T] = E[X_0] = \frac{E[S_n]}{n}$$