El Duque's no sure thing

What are the odds of the Red Sox winning four out of five against the Yankees … without their best pitcher?

Updated: October 15, 2004, 3:23 PM ET
By Rob Neyer | ESPN Insider
Forget about The Curse. And for the moment, forget about the first two games in the American League Championship Series. They're history now and have little if no impact on what comes next. Instead, answer this simple question …
What are the odds against the Red Sox winning four out of five games against the Yankees … without their best pitcher?
If the Red Sox and Yankees were evenly matched – which a week ago they apparently were – the Red Sox would have roughly a 19 percent chance of winning. One in five.

In case you're wondering how I arrived at 19 percent, it's actually pretty simple. There are five "ways" the Red Sox can still win four games in this series. If we assume the teams are equal, the Red Sox have a .50 chance of winning any particular game. If you add up the probabilities of all five ways they can win the series, you wind up with .1875 (which I rounded up to 19 because it's somebody else's turn to win the pennant).

Make sense? If so, skip this section. If not, here's the "work": the five ways the Red Sox can win – again, assuming they've got a 50 percent chance of winning each game – followed by the probability of that way ...

W W W W    .0625 (.5*.5*.5*.5)
L W W W W  .03125 (.5*.5*.5*.5*.5)
W L W W W  .03125 (ditto)
W W L W W  .03125
W W W L W  .03125
Add all those probabilities together, and you get 18.75 percent.