So...could you be persuaded to do the binomial math for us?

Sure. It's complicated by the 6 mutual games left between the Yankees and Red Sox. I'm ignoring the fact that Toronto and Tampa aren't technically eliminated yet either, and assuming each remaining game is a coin flip. This is ignoring the tie case as well.

Here's a MATLAB script that calculates it:

mutual = sym(6); % 6 mutual Red Sox/Yankee games

totalgames = sym(78); % 78 total games left between the Red Sox and Yankees

nonmutual = totalgames - mutual; %Number of nonmutual games left

probweight = sym(1/2)^totalgames; % Assume a 50% probability of each game outcome

clinchnum = sym(33); % Current clinch number is 33 total Red Sox wins + Yankees losses

p = sym(0);

% Separate out non mutual games and Yankees/Red Sox games

% rwin is the number of Red Sox wins in mutual Red Sox/Yankees games.

% The clinch number drops by 2 for every Red Sox win in mutual games

for rwin = sym(0):mutual

p = p + sum(arrayfun(@(x)nchoosek(nonmutual, sym(x)),(clinchnum-sym(2)*rwin):nonmutual))*nchoosek(mutual,rwin)*probweight;

end

eval(p)*100

97.4%