Election between $2$ candidates ends in a tie: probability one candidate leads until the penultimate vote

Let's shift gears and use the ballot theorem instead of reinventing it for the case of ties. Other than the one appeal to the ballot theorem, this will be a conditional probability problem. ^_^

Our sample space will be all cases where each candidate received $p$ votes. Let $A$ be the event that $C_1$ lead all the way until the moment before the final vote was read, and let $B$ be the event that $C_2$ received the final vote.

We know that

• $P(B)=P(\overline B)=\frac12$

• $P(A\mid B)=\frac{p-(p-1)}{p+(p-1)}=\frac1{2p-1}\quad$ This is where we are using the ballot theorem.

• $P(A\mid \overline B)=0\quad$ Obviously, $C_1$ could not have lead throughout the counting and received the final vote, because the count ended in a tie.

Using all this and the law of total probability,

$$P(A)=P(A\mid B)\cdot P(B)+P(A\mid \overline B)\cdot P(\overline B)\\=\frac{1}{2p-1}\cdot\frac12+0\cdot\frac12=\frac1{4p-2}$$

Note that this formula agrees with almagest's calculation that the probability was $\frac1{10}$ when $p=3$.

HINT

Why not just use Bertrand's ballot theorem?

A feasible $2p$-steps path in the OP problem consists of a $(2p-1)$-long front segment where $C_1$ leads throughout, and then a last vote for $C_2$. If you consider just the front segment, this fits exactly the ballot theorem.

• $M = {2p-1 \choose p} =$ no. of possible front segments.

• The ballot theorem gives the probability, i.e. the fraction $f$, of such front segments with $C_1$ leading throughout. So the no. of such front segments $= X = ???$

• The no. of $2p$-long paths where $C_1$ leads until the very end $= Y = ???$

• The total no. of $2p$-long paths is of course ${2p \choose p}$, so $P = ???$

Can you finish now?

By the ballot theorem, the fraction of such $(2p-1)$-long segments is $$f={p - (p-1) \over p + (p-1)} = {1 \over 2p-1}$$ among all ${2p-1 \choose p}$ ways to arrange the first $2p-1$ votes. Thus the no. of paths feasible for OP is $$Y = X = {1 \over 2p-1} {2p-1 \choose p} = {(2p-2)! \over p! (p-1)!}$$ The required probability is: $$P = Y \big/ {2p \choose p} = {(2p-2)! \over p! (p-1)!} \big/ {2p \choose p} = {p \over (2p) (2p-1)} = {1 \over 2(2p-1)}$$ E.g. when $p=3$ this gives $P={1 \over 10}$ agreeing with the comment by @almagest