Friday, April 27, 2012

Why 10 runs equals 1 win

It's a rule of thumb that in baseball, every additional 10 runs you score turns one loss into a win.  When I first heard that, it seemed like 10 was a lot ... but I managed to convince myself that it made sense.  Here's how I explained it to myself.

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Imagine a reasonably large number of baseball games -- a team-season, or decade, or whatever.  Pick 10 games at random, and then pick one of the teams randomly in each of those 10 games.  Add 1 run to those ten teams' score.  

You've now added 10 runs.  How does that change things?

Well, for many of those games, it won't change things at all.  If the game  didn't go into extra innings, and was won by 2 runs or more, than adding one extra run can't change the outcome.

In the 1990s, 68.4 percent of games were decided by more than one run.  That means that 6.84 of those extra 10 runs are "wasted", and don't do anything.

Now, consider the 9-inning games decided by exactly one run.  That was 22.5 percent of all games.  Half of the time, the extra run will go to the winning team -- so that run doesn't do anything. 

That leaves 11.3 percent of games where the run goes to the team who lost by one run.  That 11.3 percent of the time, the game will now go into extra innings.  The team that gets the run will win half of those.  That means that 5.6 percent of those extra 10 runs turn a loss into a win.  That's 0.56 wins.

That leaves only games that went into extra innings.  In the 1990s, that was 9 percent of all games.

If we add a run to one of those teams, that team now wins the game outright.  It would have won half of them anyway, so half of those runs don't do anything.  But, the other half, the run turns a loss into a win.  That's 4.5 percent of all games, or 0.45 wins.

Add 0.56 wins to 0.45 wins, and you get ... 1.01 wins.

That's how every 10 runs leads to one win.

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Another way of getting the same answer (with all numbers rounded):

If you assign 10 extra runs randomly, 5 will be assigned to the team that won the game anyway, so those are wasted.  Another 3 will be assigned to teams that lost by two or more runs, so those are wasted too.  That leaves 2 runs. 

One of those runs will turn a nine-inning tie -- half a win -- into a full win.  So that's 0.5 wins.

The other one will turn a one-run deficit into an extra inning game -- which turns a loss into a half win.  So that's the other 0.5 wins.


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I guess you can generalize this to other sports: the number of wins per "point" is half the percentage of games that are tied in regulation, plus half the percentage of games that are lost in regulation by exactly one point.  

For basketball, you'd have to also adjust for deliberate fouls at the end of the game.  For hockey, you have to adjust for empty-net goals.  And for football ... well, I don't know how you'd do it for football, since points usually come in large bunches of 3 or 7.  Maybe you could add a field goal to get wins per extra 3 points.  


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Disclaimer: this analysis makes a few simplifications.  It ignores bottom-of-the-ninth issues.  It assumes all teams are .500.  It assumes that none of the extra runs were allocated to an extra inning.  And it assumes you never choose the same game twice, randomly (which is related to assuming all teams are .500).  

But, if you fix all those things, you'll still get a number close to 10 runs.


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3 Comments:

At Saturday, April 28, 2012 11:43:00 AM, Anonymous Anonymous said...

Not sure why who you are looking at this at such a micro level. Look at it at a macro level instead.

Per Pythagorus:
E(W)=162*((RF^2)/((RF^2)+(RA^2)))

So, if RF=720, RA=720, E(W)=81.

Adjust RF and/or RA in blocks of 10 and E(W) will change by 1.

 
At Saturday, April 28, 2012 8:02:00 PM, Anonymous Rodeo Jones said...

I like this analysis a lot, Phil. However, why are you assuming that runs are randomly distributed between games? Aren't a disproportionate number of runs scored in higher scoring games?

Say a team replaces player A with player B, who is 10 runs better on offense. I would think that:

1) additional runs from player B would be more likely to occur in high scoring games, and

2) high scoring games are less likely to be close.

 
At Tuesday, May 08, 2012 3:28:00 PM, Blogger Phil Birnbaum said...

Rodeo Jones: sorry I'm so late noticing your comment!

I don't think runs are really much more likely to be scored in higher-scoring games -- that is, a game with 11 runs after seven innings is not (much) more likely to see a 12th run than a game with 3 runs in seven innings is likely to see a fourth.

It's possible to check, though ...

 

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