Off to the Races

by
April 6, 2010

I find these early season posts harder to write than the off-season posts. Part of it is the desire to write about the games occurring, while at the same time trying to provide something worth reading. There are only so many ways to write about the improbability of Yuniesky Betancourt’s at-bat against Justin Verlander yesterday*. With that in mind, I’m going to shamelessly steal a topic from Dave Cameron’s 2009 early season posts: Wins in the bank.

Dave’s original post was followed by Sky Kalkman expanding on the topic by applying it to the CHONE standings through that point. The concept is explained by both, but I’ll rephrase it here for originality’s sake. Say the Orioles start off 6-4. CHONE projected the Orioles to win 75 games, or 46% of their games. That 60% win rate seems to be overachieving, but don’t trip into the Gambler’s Fallacy line of thinking that the Orioles will go 3-7 at some point to ‘even things out’.

No, instead you give the Orioles credit for those earned wins while respecting the projections heading forward. 46% of 152 equates to 70 wins. Add those six they already racked up, and the Orioles solid start improved their expected record by a whole game. Of course, this can be applied at just about any time throughout the season. Say the Rays go 35-20 and we still have reason to expect them to be a 90 win team, then they would improve their expected record by four games. In that division, in that race, that’s a huge swing.

It’s early and easy to get swept away in some paranoia and hyperbole. But yeah, the results matter, and they can make a difference.

*I posed the question: How improbable was that?

Betancourt has 2,473 career plate appearances. He has 32 home runs, which means 1.3% of his total plate appearances have ended in jogs … okay, that’s not true, let’s say trots. Per Baseball-Reference, Betancourt has 141 plate appearances that went to a full count. That’s, oh, 5.7%.

On to Justin Verlander. He’s faced 3,580 batters throughout his career and has allowed 81 homers; or a shade over 2%. He’s went to a full count 500 times, or 14%.

Multiply that out and the probability that all of it happens during one plate appearance is roughly: .0002%.





 

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Rusty
14 years ago

That math is atrocious.

Betancourt going to a full count and Verlander going to a full count are NOT independent events when they’re facing each other, so you can’t just multiply the probabilities!

Even going to a full count and hitting a home run are not completely independent.

A reasonable estimate of the probability of Betancourt hitting a full count home run in that particular at-bat would be ~10% chance of count going full, times ~1% chance of a homer ~= .1%. About 500 times more likely than your calculation.

4
jscape2000
14 years ago
Reply to  Rusty

Verlander has allowed 11 full count homers in his 3578 PA career (.003).
Betancourt has 2 full count homers in his 2478 PA career (.0008).
Can I just multiply those numbers together?

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Rusty
14 years ago
Reply to  jscape2000

No.

But if you assume that .003 and .0008 are their true probabilities, and you know the league rate of full count homers, you can come up with an estimate.

leagueRate * (.003/leagueRate) * (.0008/leagueRate)

which reduces to

(.003 * .0008) / leagueRate

0
DavidCEisen
14 years ago
Reply to  jscape2000

@Rusty: I don’t think thats correct either. You can’t multiply the probability of hitting a home run by the probability of giving up a home run.

To simplify: Say a player has a 10% chance of hitting a HR and a pitcher has a 10% of giving up a HR. There isn’t a 1% of a HR occurring, because these aren’t independent events.

I’ll put forward my own guess as to the best way to figuring this out. First we need rate Verlander gets into a full count compared to the league average. Then since Betancourt gets into a full count 5.7% of the time (assuming this means against the league average pitcher), we can find a way to normalize Betancourt’s rate against Verlanders–don’t feel like thinking this through so I’m not going to write a formula. This would give us the probability of getting into a full count in this situation.

After this we would need to figure out the probability of hitting a home run in a full count situation. This is more difficult. It might be that we could find the rate Verlander gives up home runs in full count situations compared to the league, but this might be a small sample size. So maybe we should just consider home run rates (like R.J. did). Either way at the end we would have two probabilities: one for a full count the other for a home run.

These may or may not be independent–are home runs more or less likely to occur in full count situations or does it not matter at all? If they are independent, multiply away. If not…

1
Rusty
14 years ago
Reply to  jscape2000

@DavidCEisen

Yes, it is correct if you are dividing by the league rate.

To use your own example, say a player has a 10% chance of hitting a HR and a pitcher has a 10% of giving up a HR. If the league rate is also 10%, the probability is ( 10% * 10% ) / 10%, = 10%.

If the league rate is lower, say 8%, then the batter is better than average and the pitcher is worse than average, so the probability is higher: (10% * 10%) / 8% = 12.5%.

If the league rate is higher, say 12%, then the batter is worse than average and the pitcher is better than average, so the probability is lower: (10% * 10%) / 12% = 8.3%

So if you wanted to assume that a batter’s chance of hitting a home run on a given pitch is unaffected by the count, so that reaching a full count and hitting a home run on the full count pitch are independent, you can come up with a formula:

VF = Verlander full count rate
YF = Yuni full count rate
LF = League full count rate
VH = Verlander home run rate
YH = Yuni home run rate
LH = League home run rate

Chance of a full count home run =

((VF * YF) / LF) * ((VH * YH) / LH)

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vivaelpujols
14 years ago
Reply to  jscape2000

The odds of Verlander going to a full count and the odds of Betancourt going to a full count are inclusive, so instead of multiplying them together, you should do a weighted average. So the odds of a Verlander/Betancourt at bat going to a full count is 10.5%. You then figure out the odds of a Verlander/Betancourt at bat ending in a home run, which is 1.9%. Multiply the two together to get .2%. That assumes that the odds of a 3-2 count ending in a home run are the same as in all counts, but whatever.

When I use Rusty’s method,

((.14 * .06) / .13) * ((.02 * .01) / .03)

I get .04%

1
Jon
14 years ago
Reply to  jscape2000

..and the math jocks come out of the woodwork to flex their skills. haha

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Stovokor
14 years ago
Reply to  Rusty

more or less agree. the math in the article is completely wrong.

to illustrate it by example, lets say the rays have a 60% chance of winning a given game and the white sox have a 60% chance of losing a given game. you would NOT multiply 60% by 40% to get a 24% chance of the rays winning a game against the chisox.

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MikeS
14 years ago
Reply to  Stovokor

Hey! I don’t expect a world championship out of the White Sox bot they better win more than 40% of their games this year.

0