Trivia Questions

[Melottfan]

Allright Fans here you go,

How many ways can a baseball manager arrange the 9-man batting order?

What is the record for the most doubleheaders played by a club in one season?

What 20 game winner lost the most games in a season, the same season he won 20?


Good Luck:

[RuthMayBond]

What 20 game winner lost the most games in a season, the same season he won 20?


Good Luck:

Jim McCormick

[KCGHOST]

How many ways can a baseball manager arrange the 9-man batting order?

362,880 ways.

[Utter Chaos]

How many ways can a baseball manager arrange the 9-man batting order?

There could be 3 different answers here:

1. by postions would yield what KCGhost said, 362,880 (9 factorial)
2. using an optional DH would yield 3,628,800 (10 factorial)
3. using the full 40 man roster would yield approximately 815,915,283,247,897,734,345,611,269,596,120,000,00 0,000,000,000 (40 factorial)

[KCGHOST]

There could be 3 different answers here:

2. using an optional DH would yield 3,628,800 (10 factorial)
3. using the full 40 man roster would yield approximately 815,915,283,247,897,734,345,611,269,596,120,000,00 0,000,000,000 (40 factorial)

2. The DH doesn't matter. Only 9 players can bat. If you have DH the pitcher doesn't bat and vice versa.

3. For a full blown line up the manager can only select players on the 25 man active roster. that gives a calculation of 25! / 16! (25 x 24 x 23 x 22 x 21 x 20 x 19 x 18 x 17) because you can only select 9 batters. This gives him 7,413,547,680 permutations of the line up. I guess if it is September and you have called addition players up the calculation would N! / (N-9)! where N equals the number of players on the active roster.

[Brooklyn]

just to throw more numbers in, if you take 9 players and they can all bat in a different position in the lineup and play a different position on the field, there would be 362,880 squared combinations, or over 10 billion cominbations.

[Melottfan]

Trivia Questions

Gald to see the response to these questions. Makes one proud to be a baseball fan and a SABR member.

Answers:

1. Right. 362,880 ways

2. 1943 Chicago Cubs. They played 44 doubleheaders( That's 88 games)
They finished 5th, 74-79.

3. Right. Jim Mcormick, 1879 Cleveland Blues. His record is 20-40.

Thanks.

[west coast orange and black]

oh, what's this?
from a matter-of-fact 362,880 you present 815,915,283,247,897,734,345,611,269,596,120,000,00 0,000,000,000 only as an approximation?

the idea of considering all 25 players is right on, man.

[Utter Chaos]

2. The DH doesn't matter. Only 9 players can bat. If you have DH the pitcher doesn't bat and vice versa.

3. For a full blown line up the manager can only select players on the 25 man active roster. that gives a calculation of 25! / 16! (25 x 24 x 23 x 22 x 21 x 20 x 19 x 18 x 17) because you can only select 9 batters. This gives him 7,413,547,680 permutations of the line up. I guess if it is September and you have called addition players up the calculation would N! / (N-9)! where N equals the number of players on the active roster.

That's why I put "optional DH". Pitchers could bat in the AL if the manager chooses to do so. So I believe it would be 10!/(10-9)!, right?

It's been a while since I took statistics in college, I knew I forgot something. For my 40 man roster number I did 40! instead of 40!/(40-9)! The new number would be (I believe): 99,225,500,774,400 which is very close to our national debt!

[Chicoutimi CP]

Jim Mcormick, 1879 Cleveland Blues. His record is 20-40.

In the modern era, the most losses for a 20 game-winner were by Wilbur Wood, with 20 L (and 24 W) for the 1973 Chicago White Sox and by Phil Niekro who also had 20 (plus 21W) for the 1979 Braves. (Aaaah... those knuck-ballers : Wood had 48 starts and 359,1 innings pitched in '73 and Niekro, 44 starts and 342 in '79.)

I'll also add that before them, the most losses for a 20 W pitcher was Walter Johnson in 1916. He was 25-20.

Claude.

[KCGHOST]

That's why I put "optional DH". Pitchers could bat in the AL if the manager chooses to do so. So I believe it would be 10!/(10-9)!, right?

No, it wouldn't be right, particularly as the question is phrased. If you have a DH you don't have a hitting pitcher. That still leaves only 9 players to form the line-up from. If you have a hitting pitcher you don't have a DH. Still just nine guys, still 9!.

Also the question was phrased "arrange the 9-man batting order".

[Buzzaldrin]

In the modern era, the most losses for a 20 game-winner were by Wilbur Wood, with 20 L (and 24 W) for the 1973 Chicago White Sox and by Phil Niekro who also had 20 (plus 21W) for the 1979 Braves. (Aaaah... those knuck-ballers : Wood had 48 starts and 359,1 innings pitched in '73 and Niekro, 44 starts and 342 in '79.)

I'll also add that before them, the most losses for a 20 W pitcher was Walter Johnson in 1916. He was 25-20.

Claude.

Do you count the modern era from 1920? Or 1901? because, for example, Bill Dineen was 21-21 in 1902, and I don't know, but I think that it was the most losses for a 20 game winner since 1901. There were several other 20 game winners who lost an even 20 between 1901-1920 other than Johnson, though Johnson was the last one before Wood