On the first page of Lionel Penrose's 1946 paper "The Elementary Statistics of Majority Voting," Penrose claims that "the amount by which [an individual voter's] chance of winning exceeds one half" (call this quantity $A$) is equal to "half the likelihood of a situation in which an individual vote can be decisive--that is to say, a situation in which the remaining votes are equally divided upon the issue at stake" (call this quantity $B$).
Suppose an individual Bob has a given preference on a vote between two options/candidates. Based on the discussion in the paragraph quoted above, it seems like what Penrose means by quantity $A$ is the difference between $\frac{1}{2}$ and the probability that enough other voters will agree with Bob for his choice to win a majority vote, assuming that the probability any other voter agrees with Bob is $\frac{1}{2}$. Hence, if the number of voters is $2k+1$, it seems that what Penrose means by what we call quantity $A$ is $\sum_{j=k}^{2k} {2k \choose j}\big(\frac{1}{2}\big)^{2k}-\frac{1}{2}$.
Similarly, it seems that what Penrose means by quantity $B$ is one half times the probability that exactly one half of all voters other than Bob will agree with him, assuming each other voter has a $\frac{1}{2}$ chance of agreeing with Bob. So if there are $2k+1$ voters, it seems that what Penrose has in mind for what we call quantity $B$ is $\frac{1}{2}\Big({2k \choose k}\big(\frac{1}{2}\big)^{2k}\Big)$.
So it seems that Penrose is claiming that for all integers $k\ge 1$ we have $$\sum_{j=k}^{2k} {2k \choose j}\big(\frac{1}{2}\big)^{2k}-\frac{1}{2}=\frac{1}{2}\Big({2k \choose k}\big(\frac{1}{2}\big)^{2k}\Big).$$
But I'm wondering, how can one prove this identity? And is it even true? It seems to hold for the cases $k=1$ and $k=2$, but I haven't yet checked any others.
Take out the factor $(1/2)^{2k}$ and repeat the terms in reverse (we will have the middle term twice) \begin{eqnarray*} \sum_{j=k}^{2k} \binom{2k}{j} \left( \frac{1}{2} \right)^{2k} &=& \left( \frac{1}{2} \right)^{2k+1} \left( \sum_{j=0}^{2k} \binom{2k}{j} + \binom{2k}{k} \right) \\ &=& \left( \frac{1}{2} \right)^{2k+1} \left( 2^{2k} + \binom{2k}{k}\right) \\ &=& \frac{1}{2} + \binom{2k}{k} \left(\frac{1}{2} \right)^{2k} \end{eqnarray*}