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Baseball probability

On opening day of baseball season, your favorite team loses its game. Its arch-rivals win their game. What are the odds that you will win more games than them this season, and thereby win the division title?

Criteria: Each team plays 162 games per season. If there's a tie at the end, there is a one-game playoff.

For simplicity's sake, assume all of the following:

Each team has a 50% chance of winning any given game

All the other teams in your division stink, so as long as your team beats the arch-rival, they're sure to win the division

Don't worry about the fact that in real life, your team would play its rival now and then, and in those games, exactly one of you will win and one of you will lose. Pretend all games are independant of all others, and so you never actually play your rival.

Extra credit: Come up with a formula that takes as input the number of games your team has won and lost and the number of games your rival has and outputs the odds.

I've simulated this with Monte-Carlo, and i found the results rather surprising.

Mike Schiraldi
Wednesday, April 24, 2002

Sounds like statistics homework to me, so here are some hints rather than a complete solution.
Write the ditribution of wins for both teams, and approximate as a normal distribution.  The difference between two normals is normal.  Find its mean and variance.  Given its mean and variance, what is the probability that it is greater than zero?

Unless I'm overlooking something, there is no "clever" approach to this problem.

Thursday, April 25, 2002

Also, this problem really has very little to do with baseball, and trying to cram it into a baseball framework makes for some unrealistic constraints.
Why not just say that we are flipping coins?

Thursday, April 25, 2002

Well, if there's a clever solution, i haven't found it. But the reason i posted this is because i've found that falling behind even by a single game at the very beginning of the season can have a substantial effect on your chances of winning the division, even if there are hundreds of games left.

For example, right now the Yankees are just two games behind Boston. Between them, there are 274 games left to play. But assuming that each game has a 50% chance of going one way and a 50% chance of going the other way, the Yankees only have a 40% chance of winning the division. I find that pretty shocking.

This is based on a Monte Carlo simulation, but i'm working on a formula to calculate it. What i have so far is:

Let G be the number of games you have left to play plus the number your rival does.

Each of these games has an outcome that is favorable (you win or they lose) and another that is not. Let M be the number of positive outcomes you need in order to clinch
the division (M is often called your magic number).

G choose M / 2^G  = 
the probability you will get exactly  the minimum number of favorable decisions you need

G choose M+1 / 2^G  = 
the probability you will get one more than that

this goes up to:

G choose G /  2^G  = 
the probability you will win every remaining game and your rival will lose every one

So the probability you will clinch is the sum of all those.

p =  Sum from k=M to k=G of: G! / ( (G - k)! k! )

Actually, that doesn't account for a tiebreaker playoff game. To put it another way, if  M-1 of the games have a favorable outcome, you still have a 50% chance of winning the division anyway.

Mike Schiraldi
Thursday, May 2, 2002

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