Nope.
All that bafflegab about somebody playing Monty Hall just cements the point. First off, note that Ross didn't go to such trouble (and was wrong). You're trying to add to the problem in ways that somehow make him right, that have nothing to do with the simple coin toss he used to introduce his discussion.
Ross was, statistically speaking, wrong. And yet by creating a false problem you have, I think, pegged exactly what Ross was trying to do.
The math of the Delaware electorate is quite simply ugly for a Republican state-wide candidate, and Ross didn't really address that. His only serious "data" referred to an "intensity gap," and his implication was that this intensity gap would make up for the significant numerical superiority of Democrats in that state.
You will note that Ross never actually bothered to do the math on that situation -- much less did he discuss ways by which O'Donnell could overcome the Democrats' natural advantage. No, he merely spun a web of hopes and assumptions about a presumed "intensity gap." Partly that's because Ross evidently just wants her to win, but has no concrete reasons for why she should.
Back to his example: if it's a coin-toss, it will be with a coin that gets Donkeys far more than half the time. In that case, the "Expected Value" of the coin toss favors the Democrat. That, sir, is how the statistics really work. (For a fuller explanation, see Bernoulli trial.)
From your link:
E[X] = 1*p + 0*(1-p) = p
If we restate the formula with a different payout for p and 1-p ($5 for heads and $0 for tails), it would look like this:
E[X] = $5*p + $0*(1-p) = $5p
If p = the probability of a coin toss landing heads, and we substitute .5 for the probability of a fair coin toss, then the expected value E[X] of the coin toss is $5*.5, or $2.50.
-PJ