The Boston Red Sox are tied 1-1 with the New York Yankees in the ALDS. What does history tell us about how pivotal Game 3 is in this scenario?
The Boston Red Sox and New York Yankees are tied at one game apiece in the American League Division Series. After tonight’s game, one of these two teams will be a single win from advancing to the American League Championship Series. This is probably not news to anyone reading this column. What I set out to discover today is how much of an advantage does tonight’s winner hold for the rest of the series.
Intuitively, we would think the winner of tonight’s game would win about 3/4ths of the time. The easiest way to picture this is that if we imagine each game as a coin flip and tails as a loss, the winner of tonight’s game wins the series so long as we don’t flip tails twice in a row (which happens 1/4th of the time). But this assumes that the two teams are evenly matched. It could be that, more often than not, winning two out of the first three games correlates to being the better team and thus closing out the series successfully at an even higher rate.
I decided to go through all division series since 1995 that started off tied at one game apiece and record the outcomes. Below are the results.
|Year/League||Series Winner||Series Loser||Game three winner||Length of series (games)|
|AL 1998||Indians||Red Sox||Indians||4|
So these results appear to confirm what we would have expected. The team that won Game 3 won the series in 28 out of 37 tries or about 76% of the time. Game 5 was forced 19 out of 37 tries or just over 50% of the time.
Nothing here is earth-shattering news. My hot take is that the Red Sox would be better off if they win tonight’s game than if they lose. Sometimes baseball results will surprise you. It is the unexplainable and counter-intuitive outcomes that keep me analyzing this sport. This is not one of those outcomes.