Will the Blue Jays clinch the World Series title? Q & A with statistics professor Jeffrey Rosenthal
Published: October 8, 2015
With the American League East title firmly in their back pocket, the Toronto Blue Jays are now focusing their attention on the World Series title, giving fans a reason to hope and rejoice.
There hasn’t been much reason for either since the Blue Jays last won the World Series in 1993.
But, the Jays’ fortune seems to be turning and U of T News turns once again to professor Jeffrey Rosenthal for his take on their odds of clinching the World Series title again this year.
Rosenthal is an award-winning professor in the department of statistics renowned for his ability to teach quantitiative reasoning to students who lack confidence in math and the bestselling author of Struck by Lightning: The Curious World of Probabilities.
What are the odds of the Blue Jays clinching the World Series title again after 22 years?
Well, at this point, to become World Series champions, they have to win three different series along the way. Each of the three series will be against another top-level MLB team. So, naively, you could say the Jays have a 50 per cent chance of winning each series, so their chance of becoming champion is 50 per cent x 50 per cent x 50 per cent, or 1/8, or 12.5 per cent.
Of course, you could try to nuance the probabilities a bit better, based on everything from regular-season records to pitcher rotations to home-field advantage and more, but it’s pretty tricky.
Suppose you wanted to take the regular-season records into account. How might you do that?
To take one possible approach, consider their first series, against the Texas Rangers. Now, the Blue Jays won 93 out of 162 games in the regular season, while the Rangers only won 88. So, you could approximate the probability of the Jays winning any one game against the Rangers as 93/(93+88), which is about 51.4 per cent. Assuming this, the chance of the Jays winning the best-of-five series then turns out (using the “binomial distribution”) to be slightly better, about 52.6 per cent. So, that’s a little bit better than 1/2, but not much.
How would the probabilities play out in subsequent rounds?
Well, in the next round they might face the Kansas City Royals, who won 95 regular-season games – two more than the Jays. This gives an estimate of about 49.5 per cent of the Jays beating the Royals in any one game, which turns out to give them (again using the “binomial distribution”) a probability of 48.8 per cent of winning the best-of-seven series there.
If they win both of those series, then in the World Series their worst case would be to face the St. Louis Cardinals who won 100 regular-season games. If so, then their chance of winning any one game is about 48.2 per cent and their chance of winning the best-of-seven series is about 46.0 per cent.
So according to this analysis, what is the probability that the Jays win the championship?
Putting all of the above together, the Jays’ chance of becoming world series champions is 52.6 per cent x 48.8 per cent x 46.0 per cent, which works out to 11.8 per cent – just slightly less than the naive estimate of 12.5 per cent.
In short, I wish the Jays all the best, but I wouldn’t bet my life savings on them going all the way.
