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## Saturday, April 28, 2012

#### Sabermetric Research: Why 10 runs equals 1 win

Phil Birnbaum explains…I think.

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 a 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 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.

Repoz Posted: April 28, 2012 at 09:57 AM | 26 comment(s) Login to Bookmark
Tags: sabermetrics

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1.  Posted: April 28, 2012 at 10:43 AM (#4118386)
That means that 5.6 percent of those extra 10 runs turn a loss into a win.  That’s 0.56 wins.

I don't get the math here. 5.6% is 0.056 as a decimal. How does that end up being 0.56 wins?
2. Swedish Chef Posted: April 28, 2012 at 10:46 AM (#4118388)
. How does that end up being 0.56 wins?

5.6% of 10 runs.
3. Pasta-diving Jeter (jmac66) Posted: April 28, 2012 at 11:28 AM (#4118416)
I agree with 7% of the linked article
4. bjhanke Posted: April 28, 2012 at 11:49 AM (#4118436)
An excellent write-up for those who are new to sabermetrics and don't understand why ten runs. Good Job, Phil! And BTW, yes, what post #2 is trying to say is that 5.6% of ten runs is .056 x 10 = .56. - Brock Hanke
5. Delino DeShields & Yarnell Posted: April 28, 2012 at 12:02 PM (#4118444)
That leaves 11.3 percent of games where the run goes to the team who lost by a 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.

So to win these now-extra-inning-games don't they have to get another run?
6. FancyPantsHandle glistening with foreign substance Posted: April 28, 2012 at 12:07 PM (#4118450)
Hmm, the way I always figured it was: ~5 runs per team per game = 10 total runs.
1 win for each game; 10 runs per game => 10 runs per win

Am I missing something, or isn't it really just that simple?
7. Matt Clement of Alexandria Posted: April 28, 2012 at 12:09 PM (#4118454)
That leaves 11.3 percent of games where the run goes to the team who lost by a 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.

So to win these now-extra-inning-games don't they have to get another run?
Of course they do.

I guess you've found a small infelicity in the example - if you're keeping team runs scored constant other than the added 10 runs, then this team can't, by the terms of the example, score another run to win the extra-inning game. I think the example is still useful, though.
8. Matt Clement of Alexandria Posted: April 28, 2012 at 12:10 PM (#4118457)
Hmm, the way I always figured it was: ~5 runs per team per game = 10 total runs.
1 win for each game; 10 runs per game => 10 runs per win

Am I missing something, or isn't it really just that simple?
I think it is. I've always found that concept a bit hard to explain to people, though. It's a sort of "mathy" way of approaching the issue. Phil's method - which gets to the exact same place by a different route - feels more "basebally".
9. FancyPantsHandle glistening with foreign substance Posted: April 28, 2012 at 12:23 PM (#4118461)
I think it is. I've always found that concept a bit hard to explain to people, though. It's a sort of "mathy" way of approaching the issue. Phil's method - which gets to the exact same place by a different route - feels more "basebally".

Wait. 'Teams score 10 runs a game, and you get 1 win per game, therefore 10 runs equates to 1 win', is more mathy than this:

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 a 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 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.

Dicking around with percentages of games, and fractions of wins? I swear, I will never understand people...
10.  Posted: April 28, 2012 at 12:40 PM (#4118474)
Delino/5: Good call! I had to think for a minute to figure out the answer.

The rule is that adding 10 runs gives you 1 win assuming you don't also change the total of opposition runs.

For the extra inning case, you have to score a run in the extra inning to win it, but the opposition has to score a run in the extra inning for you to lose it. Those runs cancel out, and the net (you minus opposition) is still 10 runs.
11.  Posted: April 28, 2012 at 12:42 PM (#4118477)
6/Fancy Pants: That's just coincidence, that you need 10 runs for a win, and teams score 10 runs per game.

And even then, the coincidence is only very approximate. In the 1990s, there were only 9.35 runs scored per game, but you still needed very close to 10 runs to get an extra win.

For instance: in the NBA, you need an extra 30 points (or is it 33? I forget) for a win, but teams score a lot more than 30 points a game.
12. Delino DeShields & Yarnell Posted: April 28, 2012 at 12:49 PM (#4118480)
I guess you've found a small infelicity in the example - if you're keeping team runs scored constant other than the added 10 runs, then this team can't, by the terms of the example, score another run to win the extra-inning game. I think the example is still useful, though.

My tongue was in cheek. Infelicity is a good word, though.

10 games where on average 1 run extra per game was scored is a compound Poisson process and wouldn't be that hard to simulate I guess. 1/3 of the time 0 extra runs are added, 1/3 of the time it's the 1 added run universe detailed here, and with decreasing probability, 2,3,4 more runs. Those 0's probably outweigh the 2,3,4.. so, as intimated in other comments, actually takes more than 10 runs collectively to get the 1 extra win.
13. Delino DeShields & Yarnell Posted: April 28, 2012 at 12:56 PM (#4118484)
For the extra inning case, you have to score a run in the extra inning to win it, but the opposition has to score a run in the extra inning for you to lose it. Those runs cancel out, and the net (you minus opposition) is still 10 runs.

Or similarly, just imaging 11.3% of the games never ending.
14. Matt Clement of Alexandria Posted: April 28, 2012 at 01:19 PM (#4118497)
'Teams score 10 runs a game, and you get 1 win per game, therefore 10 runs equates to 1 win', is more mathy than this:
Yeah. The first sentence doesn't explain in actual baseball terms how the translation from runs to wins occurs. The second does.

To Phil, I thought that the relationship between runs and wins was linked, that in a lower-scoring environment where it takes fewer runs to win a game, as such runs translate into wins at a correspondingly higher rate.

At the limit case, where only one run is scored in every game, 1 run = 1 win. So at that limit case, the number of runs per game equals the number of runs it takes to translate to a win. It seems to work in the current game at about 10 runs per win and 10 runs per game. Is it just coincidence that both of those work?
15. Boxkutter Posted: April 28, 2012 at 01:23 PM (#4118502)

Now, I get the feeling that they're just making #### up.
16. bjhanke Posted: April 28, 2012 at 01:44 PM (#4118510)
#14 - More or less, it's a coincidence. Historically, teams have scored ABOUT (not exactly) ten runs a game. But Phil says above #11 that even when team runs scored roam a bit away from 10, that ten still works. This, I would imagine, comes from so few leagues with teams whose total scored are more than one run away from ten, so ten is the closest integer to what you need. There may also be something coming out of the math that moves the extra runs needed by game to ten. If we ever get a real league that performs anywhere near the limit rate of one run per game, then, of course, it won't take ten. - Brock Hanke
17. bjhanke Posted: April 28, 2012 at 01:48 PM (#4118515)
It's worth noting that the problems that have been brought up here are not really math problems. EVeryone can do Phil's math. They are responses to the question itself and the process within which the math fits. That's why Phil's explanation is so good, You can get someone through this relatively simple algebra. It's getting them to see how the word problem that the math comes from is set up. That's what Phil is doing. He's not presenting a theoretical math paper (although I've seen him do that - Phil ain't amateur hour) here. He's proposing an applied math word problem, using the techniques of technical writing. Understand the words, and the math is easy. - Brock
18. Petunia inquires about ponies Posted: April 28, 2012 at 02:05 PM (#4118523)
I'm with [15]. This whole discussion - you guys are having me on. Nobody thinks this way, do they?

The idea that the mental contortions described in TFA are in any way 'basebally' strikes me as patently ridiculous.

Adding runs to games is not a 'Poisson process,' and while the answer to the question posed in [5] couldn't be more obvious, larger infelicities abound. Has anybody given any thought to the fact that in order to 'add a run' to a game, somebody has to give it up, and somebody else has to score it? That this has a consequent impact on the game state and all the decisions that follow? I know I must be coming off as a Luddite but the facile back-of-the-envelope acrobatics leading up to a conclusion that's written in stone seems to me to be representative of the worst kind of basement behavior.

EDIT: No offense. You're all smarter than me. This just seems so... clinical, in a way that most other stat talk doesn't seem to me to be.
19. Matt Clement of Alexandria Posted: April 28, 2012 at 02:22 PM (#4118533)
Another reason I thought the relation was pretty direct - the pythagorean relation of runs to wins suggests that the number of runs needed to add a win in expectation is directly related to the run environment.

Take an 81-win team that scores and allows 5 runs per game. It takes just about exactly 10 runs to get them to be an 82 win team by Pyth. If that team scores and allows four runs per game, it takes eight added runs to add a win of expectation. This works at 3, 4, 5, 6, and 7 R/G.
20. FancyPantsHandle glistening with foreign substance Posted: April 28, 2012 at 02:23 PM (#4118534)
Brock, I hate to break up your love fest, but going over it, unless I am missing something, I am actually pretty sure Phil's approach is fundamentally flawed. His approaches presupposes that all "9-inning games" that if the losing team had scored an extra run, the game would have gone to extras 100% of the time. That's plainly false, since in many cases the game ends prematurely once the home team leads by 1 after 8.5 innings or later. In many cases, the home team would have won by more than 1 run, had they needed to. Especially considering many ninth inning walk-offs leave men on the bases, meaning there is a heightened probability of scoring.

Home team's are vastly more likely to win by 1 run than away teams, and the advantage extends way beyond simple HFA.
21. Boxkutter Posted: April 28, 2012 at 02:29 PM (#4118536)
I understand the idea behind this. That adding a run to one of the two teams in each game, will lead the losing team to eventually win the game a certain number of times. But what about the opposite? Wouldn't that also cause that same team to lose a game that they originally would have won? Thus making it a zero sum gain?
22. FancyPantsHandle glistening with foreign substance Posted: April 28, 2012 at 02:41 PM (#4118540)
Also, I have a problem with selection bias in this method. Since you are just focusing on 1 run games, it allows you to catch all extra inning games, and all 1 run games, essentially giving you 2 tiers. If you add 2 runs to a team instead of 1, you should expect to win twice as many extra games. But that would mean that there have to be as many 2-run games, as there are EI games and 1-run games combined. That seems wrong to me.
23. KJOK Posted: April 28, 2012 at 03:15 PM (#4118552)
Take an 81-win team that scores and allows 5 runs per game. It takes just about exactly 10 runs to get them to be an 82 win team by Pyth. If that team scores and allows four runs per game, it takes eight added runs to add a win of expectation. This works at 3, 4, 5, 6, and 7 R/G.

It is pretty close for RPG environment to equal "Runs needed to add a win" but it's not exact. In a 7 RPG environment you need around an 8 run improvement....in a 15 RPG environment you need around 14 more runs for an additional win... so in-between those two extremes at 10 RPG it happens to hit right around 10 runs needed..

24.  Posted: April 28, 2012 at 05:23 PM (#4118625)
20/Fancy: Yes, the analysis is a bit oversimplified by ignoring bottom-of-the-ninth issues. I mentioned that in the original post.
25.  Posted: April 28, 2012 at 05:26 PM (#4118628)
14/Matt: when only (exactly) one run is scored in every game, you need four more runs to turn a loss into a win, not one.

Suppose you add four random runs. You turn two 1-0 games into 2-0 games; no change. Then, you turn two 0-1 games into 1-1 games.

You've turned 2 wins and 2 losses into 2 wins and 2 ties. After extra innings on the ties, you'll have 3 wins and 1 loss. That's turning one loss into a win.
26. Sunday silence Posted: April 29, 2012 at 05:47 PM (#4119291)
But wouldnt this only be true if the runs were added at random? If you were to produce more runs in say the 4th slot in the batting order, wouldnt this be adding 10 runs non randomly? Because for example you would be planning around that, using pinch hitters, bunting, taking pitches etc. if you know the 4th place hitter is creating (or helping to create) the extra runs.

I can sort of see the basic thinking here but I think once you have intelligent beings controlling things, those added runs are likely coming in less than random situations, so it's probably somewhere less than 10.

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