42. Suppose hypothetically that a certain, very corrupt political entity in a universe holds an election with two candidates, say $A$ and $B$. A total of $5,825,043$ votes are cast, but, in a sudden rainstorm, all the ballots get soaked. Undaunted, the election officials decide to guess what the ballots say. Each ballot has a $51 \%$ chance of being deemed a vote for $A$, and a $49 \%$ chance of being deemed a vote for $B$. The probability that $B$ will win is $10^{-X}$. What is $X$ rounded to the nearest 10 ?
Answer: 510
Solution: Let $N=2912521$, so that the number of ballots cast is $2 N+1$. Let $P$ be the probability that $B$ wins, and let $\alpha=51 \%$ and $\beta=49 \%$ and $\gamma=\beta / \alpha<1$. We have
$$
10^{-X}=P=\sum_{i=0}^{N}\binom{2 N+1}{N-i} \alpha^{N-i} \beta^{N+1+i}=\alpha^{N} \beta^{N+1} \sum_{i=0}^{N}\binom{2 N+1}{N-i} \gamma^{i}
$$
(think of $2 i+1$ as representing $B$ 's margin of victory). Now
$$
\frac{2^{2 N+1}}{2 N+1}<\binom{2 N+1}{N}<\sum_{i=0}^{N}\binom{2 N+1}{N-i} \gamma^{i}<2^{2 N+1},
$$

So
$$
-X=\log P=N \log \alpha+(N+1) \log \beta+(2 N+1) \log 2-\epsilon=N \log (2 \alpha)+(N+1) \log (2 \beta)-\epsilon,
$$
where $0<\epsilon<\log (2 N+1)<7$. With a calculator, we find that
$$
-X \approx 25048.2-25554.2-\epsilon=-506.0-\epsilon,
$$
so $X \approx 510$.