Percent Chance of Select Royals Hitters Hitting a Certain Number of Homers given 650 PAs in 2009,...

Percent Chance of Select Royals Hitters Hitting a Certain Number of Homers given 650 PAs in 2009, drawing on CHONE projections. If you need a closer look, click here.

I already posted this at Royals Review, but given the recent fun with Excel's BINOMDIST function and the accompanying charts (not to mention how long it took it to get through my thick head), I thought I'd post it here.

Those more in the know: is something like a binomial distribution roughly how stuff in projections systems like CHONE's percentiles and ZiPS' "Oddsinator" are generated?

about 3 years ago Matt Klaassen 15 comments 0 recs |

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So, care to post the formula you posted in order to project this?

by R.J. Anderson on Jan 14, 2009 7:09 PM EST reply actions

It just hit me, you probably used the percentiles.

by R.J. Anderson on Jan 14, 2009 7:19 PM EST up reply actions

no, actually, I didn't

Would that have given similar results?

What I did was simply go to the main projection page for the team, and took the number of number runs/number at bats as the %, then 650 as the number of trials, so, for Alex Gordon, CHONE projects 15 HR in 548 ABs, so the equastion (excel) is:

=(1-(BINOMDIST))

Where B3 is the number of home runs. So for the chart, I did 1, 5, 10, 15, 25… you get the picture.

Make sense?

by the way, using the “FALSE” Cumulative leads to some cool results when running this with win%

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by Matt Klaassen on Jan 14, 2009 7:54 PM EST up reply actions

I have no idea, I don't think so, given Bloomquist's only goes to 0%.

Thanks.

BINOMDIST has so many functions.

by R.J. Anderson on Jan 14, 2009 8:45 PM EST up reply actions

WFB, baby

it looks like the curve dips below 0% for a bit – that’s true grit

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by Jack Moore on Jan 14, 2009 7:12 PM EST reply actions 1 recs

yeah, that's just ... something

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by Matt Klaassen on Jan 14, 2009 7:54 PM EST up reply actions

Not sure on whether they use binomial distributions

They might binomial, they might be normal, not sure. But, the key to any “odds” projection is the standard error of estimatation. It works much like a standard deviation does. So, if my projection for (name here) is .10 HR/FB, with an SEE of .01, then my model says there is a 68% chance (1 standard deviation either way) of the result being .09 or .11, and a 95% chance of it being between .08 and .12

http://mvn.com/mlb-stats

by pizzacutter on Jan 14, 2009 9:05 PM EST reply actions

I like the non-cum graphs for comparing two teams or players.

The area of the graphs overlapping is related to the probability that the worse team will outperform the better team.

Beyond the Boxscore // Calling BJ Upton lazy is lazy.

by Sky Kalkman on Jan 14, 2009 9:06 PM EST reply actions

can you show us an example of said graph?

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by Matt Klaassen on Jan 14, 2009 9:23 PM EST up reply actions

Not right now.

Just imagine two bell curves side by side with the right tail overlapping the left tail of the other one. The more they overlap, the better chance the worse team has of outperforming the better team (or worse HR hitter has of out-performing the better HR hitter).

Beyond the Boxscore // Calling BJ Upton lazy is lazy.

by Sky Kalkman on Jan 14, 2009 9:39 PM EST up reply actions

yeah, just did one

I did one earlier today for win % for the Royals, where I overlapped the cum and non-cum win curve, thanks to one of the brainiac economists who frequent RR…

Here’s the rough version, too small, of course:

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by Matt Klaassen on Jan 14, 2009 9:45 PM EST up reply actions

Binomial distributions

I do something to this effect, but the results are going to be slightly too narrow and frequently too symmetrical if you don’t take the error of the probability into account.

As an example let’s say that a player is projected to hit 20 homers in 650 PA. The problem we run into is that the distribution is based off our best guess of the probability, not the actual probability.

The player’s “true” homer distribution is going to look different based on the probability of the projection itself. With a coin, you’re reasonably sure that the it’s a 50/50 bet to get heads or tails. With a 20-homer hitter, you’re not entirely sure if he’s not really an 18-homer or a 22-homer hitter and there’s a small chance that he’s actually a 10-homer or a 30-homer hitter.

Luckily, my projection for Guillen seems to also be 20 homers per 650 PA, which makes it easier for me. He’s more likely to “truly” be a 15 homer hitter than a 25 home run hitter (per 650 PA), so you get a more skewed distribution.

Once everything’s combined (I use statistica to simulate), i get

(per 650 PA)

<5 HR – 0.2%
<10 HR – 3.4%
<15 HR – 18.4%
<20 HR – 47.2%
<25 HR – 76.3%
<30 HR – 91.5%
<35 HR – 97.7%
<40 HR – 99.5%

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by D.Szymborski on Jan 14, 2009 10:35 PM EST reply actions

thanks, Dan

even though my head now hurts

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by Matt Klaassen on Jan 14, 2009 10:55 PM EST up reply actions

Maybe I can put it simpler

Think of it this way.

We’re going to do 100 coin flips. Coins are typical 50/50 propositions. Let’s say we want to know the odds of flipping 45 heads or more. Assuming that 50/50 split, we’d guess that it would happen 86% of the time.

But what if someone tells us that the coin is weighted and will never come up heads? In that case, we would never flip 45 heads or more out of 100.

Then again, we don’t know this guy too well and think he might be lying. We’re evenly split between whether he’s lying or telling the truth. So, depending on what’s true, we have a good chance of getting 0 coin flips. But if we just took the average of the two probabilities of the coin, 0.5 and 0.0, our coin would simply “project” to being a 1-in-4 heads coin.

If we have a coin that comes up heads 1 time in 4, the odds of flipping tails 100 times in a row is 3.2 trillion-to-1. But in this scenario, our chances of flipping tails 100 times in a row is actually over 50%!

Does this make more sense?

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Dan Szymborski
dan@baseballprimer.com

by D.Szymborski on Jan 15, 2009 12:29 AM EST up reply actions