Why is best-of-N contest better than single-round for better player? Consider a game where team-better has probability of winning $>50\%$ against team-worse. Intuitively, why should team-better prefer to play a best-of-N contest versus a single game?
Is there an easy way to prove that formally? Is there a closed form way to calculate the probability?
I can compute a recursion:
       $$ P(N,N) = x\cdot P(N-1,N) + (1-x)\cdot P(N,N-1) $$
but I don't know how to obtain a closed form equation from that.
 A: Intuitively, the better player wants to average over many games, since this reduces the variance in the proportion of games won and the better player only loses becauses of deviations from the mean proportion (which is her winning probability). In the limit of a best-of-$\infty$ match, the better player will win with probability $1$, as the variance in the proportion of games won goes to zero.
For a proof, think of Pascal's triangle and consider the probabilities for crossing the symmetry axis. Denote the better player's winning probability by $p$ and a state with $k$ wins for the better player and $l$ wins for the other player by $(k,l)$. The probability of reaching state $(m,m-1)$ and then crossing the centre to reach $(m,m+1)$ is $\binom{2m-1}mp^m(1-p)^{m-1}(1-p)^2=\binom{2m-1}mp^m(1-p)^{m+1}$, whereas the probability for the opposite transition, reaching state $(m-1,m)$ and then crossing over the symmetry line to reach $(m+1,m)$ is $\binom{2m-1}mp^{m-1}(1-p)^mp^2=\binom{2m-1}mp^{m+1}(1-p)^m$. Since $p\gt1-p$, the latter transition probability is greater than the former. Since $m$ was arbitrary, in each step from $2m-1$ games played to $2m+1$ games played, the net flow of probability across the symmetry axis is in favour of the better player. Thus, the longer the set lasts, the more probability accumulates on her side.
