10/05/08 7:10 AM ET

# Math gives hope to teams down 0-2

## Crunching numbers shows 42.6 percent shot for one underdog

By Matthew Leach / MLB.com

*one*of those teams will pull off the miracle.

The number has been repeated incessantly the past couple of days. In the history of baseball, 54 teams have fallen behind, 2-0, in a best-of-five series, with seven of those teams coming back to win.

Assuming that number is a true reflection of the chances -- and frankly, it's a small enough sample size that it's probably not a true reflection -- then there's more than a 40 percent chance that one of the four underdogs will win. Forty-two-point-six percent, to be exact.

Here's the math. If the true likelihood is 47 out of 54, that's an 87.04 percent chance that the team with the 2-0 lead will win the series.

As an aside, that's virtually the same likelihood the leading team would have if you just flipped a coin three straight times to determine a winner. If the Cubs or Brewers or Angels or White Sox had to win three straight coin flips to advance to the next round, each one would be eliminated 87.5 percent of the time and advance 12.5 percent of the time.

Back to the big picture, though. The calculation that you're looking for is the likelihood that all four leading teams will win their series. So you take the first percentage, the 87.04 number, and multiply it by itself three times, i.e. raise it to the fourth power. That is, (47 divided by 54) times (47 divided by 54) times (47 divided by 54) times (47 divided by 54).

Or (47/54)^4.

And when you calculate that out, you get a 57.39 percent chance that all four leading teams will advance. By deduction, then, it's determined that 42.61 percent of the time, at least one of those four teams trailing 0-2 will win.

**Matthew Leach** is a reporter for MLB.com. This story was not subject to the approval of Major League Baseball or its clubs.