# Will The Combined Score Of Tonight's Game Be Odd Or Even?

April 19, 2006 9:09 AM Subscribe

Baseball Score Probabilities: Is it any more likely that the combined score of a Major League Baseball game will be even than odd (or vice versa)?

At first glance it would seem that the liklihood of the combined number of runs scored by both teams being even is equal to that of it being odd. But the game can't end in a tie, which means that some even scores cannot occur. Does this mean, as a practical matter, that an odd combined score is more likely? If the game goes into extra innings is it, as a practical matter, likely that one team will win by a single run, making an odd combined score that much more likely? Baseball and probability fans, please hope me!

At first glance it would seem that the liklihood of the combined number of runs scored by both teams being even is equal to that of it being odd. But the game can't end in a tie, which means that some even scores cannot occur. Does this mean, as a practical matter, that an odd combined score is more likely? If the game goes into extra innings is it, as a practical matter, likely that one team will win by a single run, making an odd combined score that much more likely? Baseball and probability fans, please hope me!

From 2000-2004 (~12,000 games), about 58.8% of games ended in an odd number.

For extra inning games in that period, I have 1,012 games. Games that ended with a difference of:

1 run: 732

2 runs: 146

3 runs: 74

4 runs: 33

5 runs: 15

6 runs: 9

7 runs: 3

posted by milkrate at 9:29 AM on April 19, 2006 [1 favorite]

For extra inning games in that period, I have 1,012 games. Games that ended with a difference of:

1 run: 732

2 runs: 146

3 runs: 74

4 runs: 33

5 runs: 15

6 runs: 9

7 runs: 3

posted by milkrate at 9:29 AM on April 19, 2006 [1 favorite]

While I have not seen evidence to back it up, I can say with relative confidence the answer to your question is yes.

Here's why:

Assume that without tie games, the probability for the score to be even or odd is equal. This is a safe assumption - there's no reason to believe otherwise.

So the only games that can throw off probability are those that go to extra innings, right?

If a game goes into extra innings, the following are the possible outcomes:

- 1 run homer ends game [odd combined score]

- 2 run homer ends game [even combined score]

- 3 run homer ends game [odd combined score]

- Grand slam ends game [even combined score]

- Run driven in by base hit/double/triple -- here's the one that throws it off -- I could be wrong on this, but I'm pretty sure that once the first guy crosses the plate, the game is over. [odd combined score]

For example: Bases loaded, and there's a base hit -- the first guy who crosses the plate ends the game and even if the other guys could score before being thrown out, I'm pretty sure the runs are not counted.

Assuming that's the case - there's 3 ways to make an odd score, and 2 ways to make an even score. It could easily be argued that the run being driven in (i.e. not a home run) is also probably more common than the game ending on a homer - meaning the probability leans more heavily toward an odd finish.

How much statistical difference does this make in the grand scheme of things? Probably not much -- but some.

posted by twiggy at 9:29 AM on April 19, 2006

Here's why:

Assume that without tie games, the probability for the score to be even or odd is equal. This is a safe assumption - there's no reason to believe otherwise.

So the only games that can throw off probability are those that go to extra innings, right?

If a game goes into extra innings, the following are the possible outcomes:

- 1 run homer ends game [odd combined score]

- 2 run homer ends game [even combined score]

- 3 run homer ends game [odd combined score]

- Grand slam ends game [even combined score]

- Run driven in by base hit/double/triple -- here's the one that throws it off -- I could be wrong on this, but I'm pretty sure that once the first guy crosses the plate, the game is over. [odd combined score]

For example: Bases loaded, and there's a base hit -- the first guy who crosses the plate ends the game and even if the other guys could score before being thrown out, I'm pretty sure the runs are not counted.

Assuming that's the case - there's 3 ways to make an odd score, and 2 ways to make an even score. It could easily be argued that the run being driven in (i.e. not a home run) is also probably more common than the game ending on a homer - meaning the probability leans more heavily toward an odd finish.

How much statistical difference does this make in the grand scheme of things? Probably not much -- but some.

posted by twiggy at 9:29 AM on April 19, 2006

(58.8% ended with an odd number of combined runs scored, I mean)

posted by milkrate at 9:31 AM on April 19, 2006

posted by milkrate at 9:31 AM on April 19, 2006

Thanks Milkrate, that's exactly what I was looking for.

posted by The Bellman at 9:39 AM on April 19, 2006

posted by The Bellman at 9:39 AM on April 19, 2006

If you have Access/ some database familiarity, the Lahman Database is a great resource.

posted by yerfatma at 10:01 AM on April 19, 2006

posted by yerfatma at 10:01 AM on April 19, 2006

It's not just a matter of the number of different ways a game can end. It's the likelihood of each of those ways. As MrMoonPie says, 1-0 is a likely score.

posted by weapons-grade pandemonium at 10:17 AM on April 19, 2006

posted by weapons-grade pandemonium at 10:17 AM on April 19, 2006

The Lahman Database

posted by TimeFactor at 10:17 AM on April 19, 2006

*is*great but it won't help with this question as it only has player data, not game data. Retrosheet is the place for game scores and play-by-play data.posted by TimeFactor at 10:17 AM on April 19, 2006

twiggy: each of your final cases are not equally likely, right? The distribution of runs from a home run is probably heavily weighted towards the low side.

posted by RustyBrooks at 10:29 AM on April 19, 2006

posted by RustyBrooks at 10:29 AM on April 19, 2006

Thanks, TimeFactor, Retrosheet is totally cool. Because the various outcomes are

posted by The Bellman at 11:01 AM on April 19, 2006

*not*equally likely, Milkrate's large-sample empirical data seems to me the easiest and best way to answer the question. I love AskMe.posted by The Bellman at 11:01 AM on April 19, 2006

If you're going on a purely mathematical basis I think an even score is more likely based on how addition of odds and evens works.

Odd + Odd = Even

Even + Even = Even

Odd + Even = Odd

So if baseball scores were statisically nice variables then even would win 2 to 1. In the real world its probably not so cut and dry, but I'd bet even still wins out on odd by a decent margin.

posted by hamhed at 11:17 AM on April 19, 2006

Odd + Odd = Even

Even + Even = Even

Odd + Even = Odd

So if baseball scores were statisically nice variables then even would win 2 to 1. In the real world its probably not so cut and dry, but I'd bet even still wins out on odd by a decent margin.

posted by hamhed at 11:17 AM on April 19, 2006

Sorry hamhed, that's not right at all. You forgot Even + Odd which is just as likely as the other three.

What you said is equivalent to saying that when two coins are flipped it's twice as likely for them to be the same as different because there are three equally likely outcomes HH, TT, and HT. But that's WRONG.

posted by Xalf at 11:27 AM on April 19, 2006

What you said is equivalent to saying that when two coins are flipped it's twice as likely for them to be the same as different because there are three equally likely outcomes HH, TT, and HT. But that's WRONG.

posted by Xalf at 11:27 AM on April 19, 2006

Oh no, logical fallacy! In that case odd should win (by 100/n percent where n is the max score) since the only non-allowable cases are ones that would end up with an even result.

posted by hamhed at 11:49 AM on April 19, 2006

posted by hamhed at 11:49 AM on April 19, 2006

Something like Benford's Law (in tables of statistics, numbers tend to start with number 1) also will bias the results.

Suppose there's an artificially smooth distribution of runs-per-game -- for example, there's a 10% chance of a 1-run game, then, a 9% chance of a 2-run game, 8.1% chance of a 3-run game, etc.

Then the Podd = .10 + .081 + ... = .1 (1 + .9^2 + .9^4).

And the Peven = .09 + .729 + .. = .1 ( .09 + .09^3 + ... ).

Obviously for this distribution Podd > Peven.

Now, it's doubtful it's a simple geometric distribution like this. But whatever the distribution actually is, it'll impart a slight bias toward odd or even. It'll happen wherever the distribution peaks.

The only way to avoid the bias is a (unnaturally) flat distribution at the peak, and a even distribution between the odds and evens elsewhere. This is unlikely, as the distribution probably tails off geometrically.

So, yeah, even without the effects of extra innings, the distribution would still be biased. My guess would be towards odds.

posted by maschnitz at 1:21 PM on April 19, 2006

Suppose there's an artificially smooth distribution of runs-per-game -- for example, there's a 10% chance of a 1-run game, then, a 9% chance of a 2-run game, 8.1% chance of a 3-run game, etc.

Then the Podd = .10 + .081 + ... = .1 (1 + .9^2 + .9^4).

And the Peven = .09 + .729 + .. = .1 ( .09 + .09^3 + ... ).

Obviously for this distribution Podd > Peven.

Now, it's doubtful it's a simple geometric distribution like this. But whatever the distribution actually is, it'll impart a slight bias toward odd or even. It'll happen wherever the distribution peaks.

The only way to avoid the bias is a (unnaturally) flat distribution at the peak, and a even distribution between the odds and evens elsewhere. This is unlikely, as the distribution probably tails off geometrically.

So, yeah, even without the effects of extra innings, the distribution would still be biased. My guess would be towards odds.

posted by maschnitz at 1:21 PM on April 19, 2006

oops, I meant:

Peven = .09 + .729 + .. = .1 ( .9 + .9^3 + ... ).

not .09.

But still, Podd > Peven.

posted by maschnitz at 1:22 PM on April 19, 2006

Peven = .09 + .729 + .. = .1 ( .9 + .9^3 + ... ).

not .09.

But still, Podd > Peven.

posted by maschnitz at 1:22 PM on April 19, 2006

**RustyBrooks**writes

*"twiggy: each of your final cases are not equally likely, right? The distribution of runs from a home run is probably heavily weighted towards the low side."*

Also your analysis only covers the home team winning. The visitors can win with a 15 run 10th inning.

posted by Mitheral at 2:01 PM on April 19, 2006

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posted by MrMoonPie at 9:25 AM on April 19, 2006