27 October 2007

Red Sox rolling

In my last post I claimed that 4-2 is the single most likely World Series result if one team has between a 50% and 60% probability of winning each game.  Well, now the Red Sox are up 2-0 in the Series.  What's the single most likely result now, assuming that the Red Sox are only slightly favored to win each individual game?

Surprisingly, it's a 4-0 sweep.

Let's conservatively assume that the Red Sox have only a 51% chance of winning each game.  We can then compute the probability of each possible result by looking at every possible win/loss sequence.  For example, consider 4-2.  If we take "CCBB" to mean Colorado wins games 3 and 4 and Boston wins games 5 and 6, then there are only three possibilities that lead to a 4-2 result: CCBB, CBCB and BCCB.  The probability of each is 49% × 49% × 51% × 51% ≈ 6.245%, and multiplying by 3 gives the probability of a 4-2 result: 18.735%.  Doing the same for the other possible results, we get

Red Sox 4-026.01%
Red Sox 4-125.49%
Red Sox 4-218.735%
Red Sox 4-312.24%
Rockies 3-411.76%
Rockies 2-45.765%

So, unless the Rockies are truly the better team, a sweep is now more likely than any other result, and the Red Sox have at least a 81.25% chance of winning it.  (For the Rockies to have a 50% chance of winning the Series, they'd need to have a 68.62% chance of winning each game.)


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