Wednesday, October 28, 2015

World Series Math

In case there are any other nerds out there who like sports out there, this article is for you.  In a recent post I mentioned that my brother-in-law got tickets to game 3, so even if his team (the Mets) win he won't be able to see it.

That got me thinking about probabilities, so I crunched some number for you (you can thank me by reading this):

  • There are 70 win/loss combinations possible in the World Series (and incidentally, in any other 7-game series, such as the NHL or MLB playoffs) - this 70 includes winning (or losing) in 4 games, 5, 6, or 7 games.
  • Assuming 50/50 win probabilities, there is a:
    • 2% chance of a 4 game series
    • 12% chance of 5 games
    • 28% chance of 6 games
    • 58% chance of 7 games
  • This absolutely makes sense from a statistical perspective, b/c if the teams are perfectly evenly matched, you would not except a sweep - you would expect each team to win at least a couple.
That means a 1% chance of a Mets (or Royals) win in 4, a 6% chance of them winning in 5, a 14% chance of them winning in 6, and a 29% chance of them winning in 7.  So all things being equal, if you have limited resources but want to see your team win the Series (not just play in it), your best chance statistically is to buy a ticket for game 7 and cross your fingers.

Of course if you like a sure thing, you should buy tickets for games 4-7; just be prepared to spend some dough.
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