Monday, November 02, 2015

More World Series Math

In a previous post, I (teasingly) criticized my brother-in-law about taking his kids to Game 3 of the World Series.  At a $1000 per ticket, and 0 chance of seeing them win even if they won, I argued for the statistically most probable 7th game, if he couldn't see all of them.  Turns out I was wrong - he managed to pick the one game the Mets won, and he is a die-hard Mets fan so I am sure he enjoyed it.  Kudos to him for picking right.

After I wrote that post, my sister asked how the probabilities would change game by game.  So here goes.  Now you might be thinking, why is this clown writing this now?  The Series is already over.  Well, maybe it is because I was too busy watching real sports, like football.  Or maybe I think some people will appreciate a break from all the how-could-the-Mets-have-blown-it-like-that?  posts.

Probabilities before game 1:

70 possible win/loss combinations, possible, and each team had a:
1% chance of a sweep
6% chance of winning in 5
14% chance of winning in 6
29% chance of winning in 7

After the Royals won game 1:

Down to 35 possible win/loss combinations:

The Mets can still win, but not in 4:
3% chance of winning in 5
11% chance of winning in 6
29% chance of winning in 7 (same as before)
Down to 43% overall

Conversely, the Royals are still alive for the sweep:
3% chance of a sweep
9% chance of winning in 5
17% chance of winning in 6
29% chance of winning in 7 (same as before)
Up to 57% overall
After the Royals won game 2:

Only 15 possible win/loss combinations remain:

The Mets now only have a 33% chance of winning overall:
7% chance of winning in 6
27% chance of winning in 7 (same as before)

The picture for the Royals is much rosier (67% chance of winning overall):
7% chance of a sweep
13% chance of winning in 5
20% chance of winning in 6
27% chance of winning in 7

After the Mets won game 3:

Now it gets interesting - only 10 possible win/loss combinations remain:

The Mets now have a 40% chance of winning overall, a decent shot:
10% chance of winning in 6
30% chance of winning in 7 (same as before)

The Royals are a little disheartened after losing their sweep possibility, but at least now they could win in at home (60% chance of winning overall):
10% chance of a sweep
20% chance of winning in 5
30% chance of winning in 6
40% chance of winning in 7

After the Royals won game 4:

Uh oh, the bottom is falling out for the Mets - only 4 possible win/loss combinations left, each with an equal chance of happening:
The Mets have a 25% chance of winning (in 7 games) - this is their only option.

But the Royals can follow any of 3 paths (although they obviously like one more than the others): 
25% chance of winning in 5
25% chance of winning in 6
25% chance of winning in 7

All of this is/was assuming that the 2 teams were evenly matched.  Which as we saw was a horribly incorrect assumption.



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