World Series Statistics

Is one of these teams the Milwaukee Brewers? Because that would affect my answer.

Name the only player ever to hit into an all Cuban double play combination.

What are the odds of that?

Gazillions to one . . . hint, it was the infield of the Washington Senators . . .

Joe made me laugh.



(I'm taking other than the cubbies in this one.)

OK . . . as a hint, that sucked . . . here's a more helpful hint . . . this unhappy player later became the manager of the Cardinals . . .

Joe Torre?

No.

I'm surprised, i'd have thought that hint would be a dead give away . . .

You are assuming we know something about baseball. How about Whitey Herzog?

No.

Knowing something about politics would help . . . Castro came to power in 1960 . . .

Mike Jorgensen?

No.

I'm flabbergasted . . . pretty soon, someone will get it by process of elimination . . .

Johnny Keane? What are the odds of my going 0-3?

No.


The odds, apparently, were 100%.

Jose Valdivieso and Reno Bertoia...I think I was at that game...

Let's look at from the point of view of one of the teams whose probability of winning is P and the probablity of losing is P'. In order to meet your requirements, then the team must either a) win the first two games, win only one of the next two, then win the sixth game or b) lose the first two, then win the next four.

a) Prob of winning the first two = P^2
Prob of winning only one of the next three = 3 P P'^2
Prob of winning the last game = P
-- total probability of this scenario = 3 P^4 P'^2

b) Prob of losing two then winning four = P'^2 P^4

Compute for team A = 3 (.68)^4 (.32)^2 + (.32)^2 (.68)^4 = 8.76%
Compute for team B = 3 (.32)^4 (.68)^2 + (.68)^2 (.32)^4 = 1.94%

Total = 10.7%

umless it's the cubbies...

(it is a world series question)

tony larussa?

Nope

Jeeze . . . i really thought that clue would be a dead give away . . .