Picking a speaker at random

We will consider a more general case; suppose there are $n$ candidates and $2m+1$ voting members in the party. We write $2m+1$ since there should be an odd number of party members to break ties. Define the following probability:
\[
p(n,m) = \left\{
\begin{array}{l}
\text{Probability that one of the $n$ candidates} \\
\text{wins a majority of the $2m+1$ votes.}
\end{array}
\right\}
\]Let’s find a formula for $p(n,m)$. For a particular candidate to win $k$ votes, $k$ members must vote for the candidate and the remaining members must vote for one of the other $n-1$ candidates. This occurs with probability $\binom{2m+1}{k}\left(\frac{1}{n}\right)^k \left(1-\frac{1}{n}\right)^{2m+1-k}$. The candidate wins a majority if they garner at least $m+1$ of the $2m+1$ votes. We multiply the final result by $n$ since we are counting the probability that any of the $n$ candidates wins a majority. Therefore,
\[
p(n,m) = n\sum_{k=m+1}^{2m+1} \binom{2m+1}{k}\left(\frac{1}{n}\right)^k \left(1-\frac{1}{n}\right)^{2m+1-k}
\]Note: We can add the probabilities of the different candidates winning as we did because the events are disjoint; it is impossible for two candidates to simultaneously win a majority of the votes. If the criteria allowed for overlap, e.g., one candidate winning at least 40% of the vote, then two candidates could win simultaneously, and we would have to account for this case to avoid double-counting.

Making the change of variables $k \mapsto 2m+1-k$, we obtain

$\displaystyle
p(n,m) = n \sum_{k=0}^m \binom{2m+1}{k} \left(1-\frac{1}{n}\right)^{k}\left(\frac{1}{n}\right)^{2m+1-k}
$

We recognize this as $n$ times the CDF of a binomial distribution with $2m+1$ trials and probability $1-\frac{1}{n}$.

What we’re really after is the expected number of rounds required for a candidate to secure a majority of the vote.
\[
f(n,m) = \left\{
\begin{array}{l}
\text{Expected number of rounds until one of the } \\
\text{that one of the $n$ candidates is elected.}
\end{array}
\right\}
\]This will happen in one round with probability $p(n,m)$. It will happen in $k$ rounds if it does not happen in the first $k-1$ rounds, and happens in the $k^\text{th}$ round. Putting this together, we get:

$\displaystyle
f(n,m) = \sum_{k=1}^{n-1} k\cdot p(n-k+1,m) \prod_{i=0}^{k-1} \bigl( 1-p(n-k+1+i,m) \bigr)
$

Unfortunately, these formulas do not simplify in any meaningful way, so we must evaluate them numerically. The first question asks us to compute $f(3,110)$ and the extra credit asks for $f(10,110)$. Calculating these, we obtain:
\begin{align}
f(3,110) &\approx 1.9999995110943353682 \\
f(10,110) &\approx 8.9999995110943296464
\end{align}These are rational numbers, of course, but $f(3,110)$ and $f(10,110)$ have 105-digit and 1357-digit denominators, respectively, when expressed as irreducible fractions, so I will not write them out here! One thing for sure is that it is very unlikely that the voting will end early. It is a virtual certainty that it will go the full $n-1$ rounds until there are only two candidates remaining.

Bonus: Changing the number of voters

The reason it is so difficult to elect a Speaker early is that there are so many voting members, and therefore it is unlikely that one candidate will obtain a majority of the votes by random chance. We can see how this probability would scale if we changed the number of voting members. Here is a plot of $f(n,m)$ for different values of $n$ (number of candidates), as a function of $2m+1$ (number of voting members). In all cases, we see that the expected number of rounds rapidly approaches $n-1$.