MLB Off-Topic

[WaitTilNextYear]

Re: MLB Off-Topic

A little binge on MLB-N programs today and they are showing Joe DiMaggio's "56" documentary. Still shocking about the probability math. Even accounting for the fact that DiMaggio batted .408 over that stretch and he was putting the ball in play an absurd 3.98 times per game, he had just a 6 in 10,000 chance at doing what he did. Comparing to some other hitters like Bryce Harper (6 in 100 million odds), Jose Altuve (3 in 10 million odds), Giancarlo Stanton (7 in 1 billion odds) is comical.

[CMH]

Re: MLB Off-Topic

A little binge on MLB-N programs today and they are showing Joe DiMaggio's "56" documentary. Still shocking about the probability math. Even accounting for the fact that DiMaggio batted .408 over that stretch and he was putting the ball in play an absurd 3.98 times per game, he had just a 6 in 10,000 chance at doing what he did. Comparing to some other hitters like Bryce Harper (6 in 100 million odds), Jose Altuve (3 in 10 million odds), Giancarlo Stanton (7 in 1 billion odds) is comical.

What are the odds based on?

Sent from my Pixel 2 using Tapatalk

[WaitTilNextYear]

Re: MLB Off-Topic

What are the odds based on?

Sent from my Pixel 2 using Tapatalk

Probability math. Essentially, you take batting average and number of chances per game into account, assume it to be a constant rate, and calculate the chance of a hit. Then compute the chance to do it 56 times in a row.

This is what I remember about the math from the documentary...

[1 - (1 - .408)^(3.98)]^56

DiMaggio batted .408 during the streak and averaged 3.98 balls in play per game. The quantity in the brackets is DiMaggio's expected hit probability in a single game (88%). The probability of 56 in a row is 0.000598, or about 6 in 10,000. I'm not a mathematician so I can't vouch for the accuracy of the calculation, but this is what I saw in the documentary.

[Blzer]

Re: MLB Off-Topic

That is valid math, assuming .408 is his BABIP and not his actual average during that time.

Thankfully they are able to use a binomial distribution function and do not have to resort to integral trial values because they only have to compute it once, but it would still be fine because that 3.98 is very close to 4, so it wouldn't really change the odds considerably regardless.

Here is what the formula is saying:


[1 - (1 - .408)^(3.98)]^56

if .408 is the probability of success for any given trial, then 1 - .408 = .592 is the probability of failure. This is the binomial condition: either you succeed, or you do not (which is failing, of course). Since we are only looking at "balls in play" for this, walks apparently are not a part of the figure anyway. The only thing that would upset this is if strikeouts are a part of the .408 number and we are excluding them in the trial results. We really should be looking at his BABIP here, or also make sure to exclude sacrifices on top of it. Regardless, let's just continue.


[1 - (.592)^(3.98)]^56

When we are generating the probability of failure, we are doing so because we want to consider the probability that he will fail in every at-bat of that game, or ball-in-play opportunity rather. If he gets a hit zero times in a game, he has failed each time. This is really easy math to do if you are looking at "succeeding every time" or "failing every time," because you just multiply by the number of trials you are doing... furthermore, because we aren't using an integer number of trials, we kind of have to do it this way. I'd try and explain if you really wanted to hear it out, but I would have to do some drawings to explain that. Let's just say that for now, based on the numbers, his probability of never getting a hit with 3.98 balls in play per game (provided that he has a 40.8% chance of success given any one at-bat) is roughly 12.4% (the exact value being 0.124119607271461, but I'll round it for display purposes only).


[1 - .124]^56

Remember, in the binomial condition, every time he doesn't fail means that he succeeds. So if he has a 12.4% chance of failure to do what he's doing in-game, then that means he has a 87.6% chance of success. In any given game, he has an 87.6% chance of getting at least one hit. The subtraction happening there is just doing it all in decimal form instead.


[0.876]^56

Just like before, any time that you are increasing the number of trials and hoping to strike luck again, you have to multiply that value. If you wanted the probability of getting a hit in Game #1 and in Game #2, you would multiply .876 * .876, which is roughly 76.7%. You have to multiply .876 (again, which is rounded for display purposes) by itself 56 times. The decimal that you get is the one that you typed out, which gives you that value.



None of this would be as simple (or maybe not possible) to do without assuming we are doing all-or-nothing, unless you'd be willing to round to the nearest integer when it comes to number of trials. So if we moved DiMaggio's number of balls in play per game to 4, you could compute anything you wanted including the odds of him getting at least three hits per game over a certain course of time. Maybe you could if you are okay with fractally splitting up your "number of trips" evenly, and that would probably provide the best prediction... but it's still shoddy because we now don't know what maximum number of at-bats we use to compute anymore are. For example, if we kept the number at 3.98 and wanted to assume he gets at least three hits a game, "at least three hits" means three or more, but where do we stop? At three or at four? 3.98 is in between them. Not so easy to compute by hand anymore.

Anyway, for hitting streak purposes you'll never have that issue so this is a cool theoretical number to spit out, and it just goes to show how awesome Joe D. was at doing what he did. There are so many external factors and pressures involved with it that make this thing pretty much an impossibility, but the simple math already shows you just how ridiculous it happens to be, plus ain't nobody getting 3.98 balls in play per game anymore.

[WaitTilNextYear]

Re: MLB Off-Topic

I think I got two things from the math on Joe D's streak (other than the obvious)...

1. The math here takes it as a constraint how good he was. The 6 in 10,000 isn't the probability for just any schmoe with a bat hitting in 56 straight games, but the very specific probability for DiMaggio that bakes into the formula how good he was at the time (.408 BABIP and so few strikeouts that he's putting damn near 4 balls in play per game). He struck out 13 times in 1941. In 621 plate appearances. And then still, at that otherworldly level, his odds were just 6 in 10,000 of doing what he did. And then as if he wasn't living a charmed enough life as it was, he went and married Marilyn Monroe.

2. The other thing is although you can basically compute the same probability for any player over any period of time using their stats/BABIP etc, the game has made it even harder for anyone to do what Joe did. The baseline K-rate is more than double what it was in 1941. Bullpens are a thing now, with a unique array of actual good pitchers throwing a zillion miles an hour. Shifts are a thing now, well beyond one team trying to stick it to Ted Williams as a gimmick in 1941. Hitters are taught to value BBs more than ever, so the balls in play are going down, even independent of massive amounts of strikeouts. I'd hesitate to estimate how much worse the odds of 56 would be for a hitter like Joe D in today's game, but it'd be easy to imagine how it'd be much more difficult today than in 1941.

[CMH]

Re: MLB Off-Topic

I didn't expect math today.

Sent from my Pixel 2 using Tapatalk

[BigOscar]

Re: MLB Off-Topic

I think I got two things from the math on Joe D's streak (other than the obvious)...

1. The math here takes it as a constraint how good he was. The 6 in 10,000 isn't the probability for just any schmoe with a bat hitting in 56 straight games, but the very specific probability for DiMaggio that bakes into the formula how good he was at the time (.408 BABIP and so few strikeouts that he's putting damn near 4 balls in play per game). He struck out 13 times in 1941. In 621 plate appearances. And then still, at that otherworldly level, his odds were just 6 in 10,000 of doing what he did. And then as if he wasn't living a charmed enough life as it was, he went and married Marilyn Monroe.

2. The other thing is although you can basically compute the same probability for any player over any period of time using their stats/BABIP etc, the game has made it even harder for anyone to do what Joe did. The baseline K-rate is more than double what it was in 1941. Bullpens are a thing now, with a unique array of actual good pitchers throwing a zillion miles an hour. Shifts are a thing now, well beyond one team trying to stick it to Ted Williams as a gimmick in 1941. Hitters are taught to value BBs more than ever, so the balls in play are going down, even independent of massive amounts of strikeouts. I'd hesitate to estimate how much worse the odds of 56 would be for a hitter like Joe D in today's game, but it'd be easy to imagine how it'd be much more difficult today than in 1941.

To be fair, I'd imagine Ichiro had a similar chance of getting that streak as DiMaggio, as he also put the ball in play constantly, while stealing singles. I'd have to look up the numbers, but I'd imagine Ichiro averaged more hits per game in the 262 year than DiMaggio did in the streak season, but despite that I don't think Ichiro ever had a particularly long hit streak. It just comes down to luck at a certain point, but I don't think it's made impossible by the modern game

[Majingir]

Re: MLB Off-Topic

What are the odds based on?

Sent from my Pixel 2 using Tapatalk

Probability math. Essentially, you take batting average and number of chances per game into account, assume it to be a constant rate, and calculate the chance of a hit. Then compute the chance to do it 56 times in a row.

This is what I remember about the math from the documentary...

[1 - (1 - .408)^(3.98)]^56

DiMaggio batted .408 during the streak and averaged 3.98 balls in play per game. The quantity in the brackets is DiMaggio's expected hit probability in a single game (88%). The probability of 56 in a row is 0.000598, or about 6 in 10,000. I'm not a mathematician so I can't vouch for the accuracy of the calculation, but this is what I saw in the documentary.

[WaitTilNextYear]

Re: MLB Off-Topic

To be fair, I'd imagine Ichiro had a similar chance of getting that streak as DiMaggio, as he also put the ball in play constantly, while stealing singles. I'd have to look up the numbers, but I'd imagine Ichiro averaged more hits per game in the 262 year than DiMaggio did in the streak season, but despite that I don't think Ichiro ever had a particularly long hit streak. It just comes down to luck at a certain point, but I don't think it's made impossible by the modern game

I know. That's the point, lol. Even doing what Joe D did, he was only going to succeed 6 chances in 10,000. Even if Ichiro had similar odds at some point during one of his many fantastic seasons, it's easy to picture him hitting on one of the 9,994 unsuccessful tries in 10,000.

[dubcity]

Re: MLB Off-Topic

Is there a probability equation for the effect being married to Marilyn Monroe had on his OBP?

[TripleCrown9]

Re: MLB Off-Topic

Finishing up 2018 by watching Moneyball and playing OOTP.

Have a great 2019, fellas. May your teams not suffer too greatly from the Red Sox repeating.

[SPTO]

Re: MLB Off-Topic

Just a reminder for all of us baseball fanatics that MLB-N will be celebrating 10 years with a special at 6 PM EST then re-airing their very first program which was a sitdown interview with Bob Costas, Yogi Berra and Don Larsen then they'll be showing the actual Larsen perfect game.

[ShowTyme15]

Re: MLB Off-Topic

[SPTO]

Re: MLB Off-Topic



Getting closer and closer to a modern fauxback version of the '80s Phillies look.

[ShowTyme15]

Re: MLB Off-Topic

The Rays are closing the upper decks at the Trop.

https://mlb.nbcsports.com/2019/01/04/rays-to-close-upper-deck-reduce-capacity-to-26000/