Here's yet another question from my probability textbook:
In a company of sixty members, each member votes for one of the members to fill an office. If the votes be regarded as given at random, what is the chance that some member shall get a majority of the whole number of votes?
Assuming majority (i.e. at least $31$ votes) and not plurality (i.e. more votes than the candidate with the second most votes), then I ended up getting$${{\binom{60}{31} 59^{29} + \binom{60}{32} 59^{28} + \ldots + \binom{60}{59}59 + \binom{60}{60}}\over{60^{60}}}.$$The numerator recognizably looks like a fraction of $(59 + 1)^{60}$. I've messed around with trying to simplify it, not to any success. Could anyone help?
EDIT: Robert Shore writes helpfully in the comments:
I think your fraction gives the probability that a specific member gets a majority of votes, so you need to multiply it by $60$.
So the expression I want to simplify is actually$$60\left({{\binom{60}{31} 59^{29} + \binom{60}{32} 59^{28} + \ldots + \binom{60}{59}59 + \binom{60}{60}}\over{60^{60}}}\right) = {{\binom{60}{31} 59^{29} + \binom{60}{32} 59^{28} + \ldots + \binom{60}{59}59 + \binom{60}{60}}\over{60^{59}}}.$$
Let $X$ be a random variable with the $\text{Binomial}(60, \frac1{60})$ distribution: we do $60$ trials, each with probability of success $\frac1{60}$, and we want to know the number of successes. In particular, this is the distribution of the number of votes that any specific member gets. So $\Pr[X \ge 31]$ is the probability that a specific person gets a majority of the votes, and $60 \cdot \Pr[X \ge 31]$ is the probability that this happens to anyone.
Since we cannot exactly compute $\Pr[X \ge 31]$ except with the very long sum $\sum_{k=31}^{60} \binom{60}{k} (\frac1{60})^k (\frac{59}{60})^{n-k}$, we want to find approximations.
First, we can observe that if $\Pr[X \ge 31]$, then the likeliest outcome by far is that $\Pr[X=31]$ exactly. (And we can be more precise about how insignificant the other terms are.)
How does $\Pr[X=31]$ compare to $\Pr[X=32]$? The first is $\binom{60}{31} (\frac1{60})^{31} (\frac{59}{60})^{29}$. The second is $\binom{60}{32} (\frac1{60})^{32} (\frac{59}{60})^{28}$. We can say $\binom{60}{31} > \binom{60}{32}$, but there's actually not a huge relative difference there: $\binom{60}{32} = \frac{29}{32} \binom{60}{31}$. On the other hand, $ (\frac1{60})^{31} (\frac{59}{60})^{29}$ is bigger than $(\frac1{60})^{32} (\frac{59}{60})^{28}$ by a factor of $59$.
So we can say that $\Pr[X=32] = \frac{29}{32\cdot 59} \Pr[X=31] < \frac1{65} \Pr[X=31]$. In future steps, the decrease will be even steeper, because the binomial coefficients will begin to fall off even faster! So we can say that $\Pr[X=31+k] < (\frac1{65})^k \Pr[X=31]$. And in conclusion $$ \Pr[X=31] \le \Pr[X \ge 31] \le \Pr[X=31]\left(1 + \frac1{65} + \frac1{65^2}+ \dots \right) = \frac{65}{64} \Pr[X \ge 31]. $$ So if we estimate $\Pr[X \ge 31]$ just by the first term $\binom{60}{31} \frac{59^{29}}{60^{60}}$, and then multiply by $60$ because we want $60 \cdot \Pr[X \ge 31]$, we'll get the answer to within a relative error of $\frac1{64}$.
If we take the first two terms into account, the error becomes even better than that.
It might still be annoying to compute $60 \cdot \binom{60}{31} \frac{59^{29}}{60^{60}}$. So we can approximate this further.
We can estimate $\binom{n}{n/2} \approx \frac{2^n}{\sqrt{\pi n/2}}$ in two ways:
Very imprecisely, $\binom{60}{31}$ is about equal to $\binom{60}{30}$, so it is approximately $\frac{2^{60}}{\sqrt{30\pi}}$. This gives us a final approximation of $$ 60 \cdot \frac{2^{60} \cdot 59^{29}}{\sqrt{30\pi} \cdot 60^{60}} \approx 3.2995 \cdot 10^{-37}. $$ We can be a little more precise and add that $\binom{60}{31} = \frac{30}{31} \binom{60}{30}$, which gives $$ \frac{30}{31} \cdot 60 \cdot \frac{2^{60} \cdot 59^{29}}{\sqrt{30\pi} \cdot 60^{60}} \approx 3.1931 \cdot 10^{-37}. $$ For comparison, the exact value is approximately $3.2294 \cdot 10^{-37}$.