Massive ERA's

[OldFatGuy]

I don't have a screen, mainly cause the point is just easy to make.

I just saw the highest ERA I've ever seen and was wondering what you guys have seen.

I could laugh at it since this guy wasn't on my team, but he used to be, and as such I had him shortlisted, so when he got hurt I got an email, and looked into his profile and....

It's only April 11, and Jim Gordon is a MR. He has gotten in three games so far.

Game 1) Couldn't get a single batter out. Final tally 5 runs allowed, 3 earned in 0 IP.

Game 2) Managed to get one batter out. Final tally, 4 runs allowed, 4 earned in 1/3 IP.

Game 3) He's getting better!!! He got TWO guys out in this game. Final tally, 5 runs allowed, 5 earned in 2/3 IP.

All told he's allowed 12 earned runs in a grand total of 1 inning pitched for an ERA of 108.00, and at one point it was over 189.

I honestly don't remember ever seeing one over a hundred before, but certainly not 189.

I didn't see the cause of the injury, but I have to wonder if it was the manager or a fan breaking a bat over his head after those games....

EDIT: Added screen shot from the game log screen showing his 3 games so far...

ADDED: Ok, never mind about not seeing higher ERA's before as I guess it's another case of my memory being bad again. I just looked in the league and there is actually one who gave up 8 earned runs in 1/3 of an inning with an ERA over 200. So I guess it does happen.

Still, an ERA at 108.00 after being in three games, ugh, that's bad.

[NYY #23]

Looks like 54.00 is the highest I've got, although not as bad as Joe Cleary.

[OldFatGuy]

Originally Posted by NYY #23

Looks like 54.00 is the highest I've got, although not as bad as Joe Cleary.

LOL, didn't know there was a real life one with a 189.

I wonder though, simple math says that 7 runs in a third of an inning means 21 runs in a full inning times 9 equals 189 even.

And in the Joe Cleary case, he clearly () has an ERA of 189.00. Wonder why OOTP assigned an ERA of 189.19 to 7 earned runs in 1 third of an inning???

[NYY #23]

Originally Posted by OldFatGuy

LOL, didn't know there was a real life one with a 189.

I wonder though, simple math says that 7 runs in a third of an inning means 21 runs in a full inning times 9 equals 189 even.

And in the Joe Cleary case, he clearly () has an ERA of 189.00. Wonder why OOTP assigned an ERA of 189.19 to 7 earned runs in 1 third of an inning???

Yeah, you are right on that one. It should be 189.00, not 189.19.

[VanillaGorilla]

Originally Posted by NYY #23

Yeah, you are right on that one. It should be 189.00, not 189.19.

When the program performs the calculation it is using .333 instead of 1/3 for the 1/3 IP.

[RchW]

Originally Posted by VanillaGorilla

When the program performs the calculation it is using .333 instead of 1/3 for the 1/3 IP.

Picking nits but 1/3 is 0.333333.....infinitely. You would never use 0.3, the error would be large, 23.3333....

Typically one would go 7/0.333 = 21.021021 but round off to 3 sig figures so you get 21.021 X 9 = 189.189

[JohnHoward]

Originally Posted by NYY #23

Looks like 54.00 is the highest I've got, although not as bad as Joe Cleary.

Thanks NYY #23 for posting Joe Cleary so I didn't have to. I had him waiting in the bullpen for a "have you seen" question like this.

[BIG17EASY]

Originally Posted by NYY #23

Looks like 54.00 is the highest I've got, although not as bad as Joe Cleary.

But his K/9 ratio is epic!

[VanillaGorilla]

Originally Posted by RchW

Picking nits but 1/3 is 0.333333.....infinitely. You would never use 0.3, the error would be large, 23.3333....

Typically one would go 7/0.333 = 21.021021 but round off to 3 sig figures so you get 21.021 X 9 = 189.189

Since we are picking nits...using .333 gives you the correct number to 3 sig figs, which is 189. If you want 5 sig figs, which would get you to the hundredths place for 189.00 use .33333.

For ERAs below 10, the use of .333 for the calculation is fine because the common display is for three sig figs. This can cause a discrepancy if there is a close race for the ERA title and we need to go past the hundredths place (assuming an ERA not less than 1).

[The Game]

Highest one I had with a guy with more then 5 IP was 81.00 highest with a guy with more then 10IP was 54.00

[RchW]

Originally Posted by vanillagorilla

since we are picking nits...using .333 gives you the correct number to 3 sig figs, which is 189. If you want 5 sig figs, which would get you to the hundredths place for 189.00 use .33333.

for eras below 10, the use of .333 for the calculation is fine because the common display is for three sig figs. This can cause a discrepancy if there is a close race for the era title and we need to go past the hundredths place (assuming an era not less than 1).

nm

[OldFatGuy]

oops, nevermind, my math was off.

lol, getting old is a bummer.

[RchW]

Originally Posted by OldFatGuy

Well MLB got the correct ERA down to the hundreds place. I was thinking though instead of using .3333 that it would be even more accurate to just convert all IP into 1/3 of inning, then divide by three.

In other words, ERA =27XER/number of 1/3 IP /3

so in this case it would be 27X12X1 divided by 3. If someone had pitched 7 1/3 innings and gave up 5 earned runs his ERA would be

27 times 5 divided by 22 (7 IP =21 plus the 1/3 =22).

Then it's just a matter of rounding to whatever place you want and it will be accurate. I figured that's the way it was done everywhere to get around the 1/3 problem, but I guess Markus didn't do it that way.

Whatever, but 189.19 is the right ERA.

[OldFatGuy]

Originally Posted by RchW

Whatever, but 189.19 is the right ERA.

Sorry RchW, I had the formula wrong and edited out but not before you quoted me I guess.

The correct formula to get rid of the .33333 or .666666 problem is to convert all IP to thirds and then it's

27 x ER / # of 1/3 IP

27 X 7 / 1 in the case of Cleary equals 189.00

That is the correct ERA.

If someone gave up 4 runs in 7 2/3 IP it would be

27 x 4 / 23 (23 is the number of 1/3 IP for 7 2/3) and the answer is 4.70.

Converting it fractions to make all the calculations does away with the round errors of the 1/3 and 2/3 IP calculations and it's just a matter of where one wants to round. I thought that's how everyone did it, but obviously not Markus.

I'm not understanding why you say the right ERA is 189.19 when it seems obvious to me the "right" ERA is 189.00 .

If a pitcher gave up 7 runs every third of an inning he pitched when he had pitched 9 innings he would have allowed 189 runs exactly, not 189.19 runs. I might be missing/forgetting something though.

[VanillaGorilla]

Originally Posted by OldFatGuy

If a pitcher gave up 7 runs every third of an inning he pitched when he had pitched 9 innings he would have allowed 189 runs exactly, not 189.19 runs. I might be missing/forgetting something though.

That is correct. You are not missing or forgetting anything.

[pennywise]

time to demote that guy