4
   

World Series Statistics

 
 
LionTamerX
 
  2  
Reply Tue 2 Mar, 2010 06:45 am
@Setanta,
Red Schoendist ?
0 Replies
 
Setanta
 
  1  
Reply Tue 2 Mar, 2010 06:53 am
Bingo ! ! !
0 Replies
 
Region Philbis
 
  1  
Reply Tue 2 Mar, 2010 11:19 am
@Setanta,
Quote:
Name the only player ever to hit into an all Cuban double play combination.
which cuban players were involved?
i googled, came up empty...
Setanta
 
  1  
Reply Tue 2 Mar, 2010 12:40 pm
@Region Philbis,
It was the Washington Senators, in, i think, 1961 or -62. One and possibly two of them were on the field having pinch hit in the previous half inning. I read about it many moons ago, and didn't make a point of trying to remember the names of the infielders.
0 Replies
 
Setanta
 
  1  
Reply Tue 2 Mar, 2010 01:17 pm
Well, i'm not having any luck finding it, either. But it was not likely to have been against the Senators, since Schoendienst only every played for National League teams.
0 Replies
 
wandeljw
 
  0  
Reply Tue 2 Mar, 2010 01:50 pm
I searched the internet and found an essay on post World War II baseball. Here is a relevant excerpt:
Quote:
Before Castro, there were a lot of Cuban players in Major League Baseball, and "Red" Schoendienst, better known to most baseball fans as the manager of the St. Louis Cardinals, once made baseball history while he was a player, when, due to substitutions during a game against the Washington Senators, he hit into the only all-Cuban double play combination in major league history (that means that the entire infield, four of the nine players, were Cuban).

Source
realjohnboy
 
  2  
Reply Tue 2 Mar, 2010 02:18 pm
I suspect that someone on A2K is mad at us. We were asked to do homework for him/her re odds. And look where we are now!
Fred "Red" Schoendienst did indeed play his entire career in the National League. His team was in the World Series 3 times: against Boston in 1946 and against the Yankees in 1957 and 1958.
I checked his stats for those 3 Series and only in 1958 did he hit into a double play.
How will we ever resolve this and shouldn't we have better things to do with our time?
I think the "Red" Schoendienst connection is a red herring.
0 Replies
 
Setanta
 
  1  
Reply Tue 2 Mar, 2010 02:20 pm
@wandeljw,
Now that's interesting because it raises the question of why his team were playing the Senators.
0 Replies
 
Setanta
 
  1  
Reply Tue 2 Mar, 2010 02:21 pm
Oops . . . that doesn't help. That's a post by me from a few years ago.
wandeljw
 
  1  
Reply Tue 2 Mar, 2010 02:25 pm
@Setanta,
Sorry, it was the only reference I could find, using Google. Smile
Setanta
 
  1  
Reply Tue 2 Mar, 2010 02:29 pm
@wandeljw,
I did manage to find an article which said that the Senators had an unusually high proportion of Cuban players . . . but didn't quote it here, because it would not explain how Schoendienst would be playing against the Senators. I read this maybe thiry years ago or more.

Of course, in those days, you could read any old tripe, and didn't have Wikipedia to set you straight . . .
0 Replies
 
markr
 
  5  
Reply Wed 3 Mar, 2010 11:41 pm
@engineer,
I'd agree with engineer if the question were, "What is the probability that a team (either A or B) wins the series in exactly 6 games, and one team (either A or B) won both the first and second game?"

However, we are given that one team has already won the first two games. This eliminates all six-game series where the first two games are split.

A wins the first two games with probability 0.68^2 = 0.4624
B wins the first two games with probability 0.32^2 = 0.1024

Possible series and final four-game probabilities when A wins first two games:
AA ABBA 0.68^2 * 0.32^2
AA BABA 0.68^2 * 0.32^2
AA BBAA 0.68^2 * 0.32^2
AA BBBB 0.32^4

AA total is 3 * 0.68^2 * 0.32^2 + 0.32^4 = 0.15253504

Possible series and final four-game probabilities when B wins first two games:
BB BAAB 0.68^2 * 0.32^2
BB ABAB 0.68^2 * 0.32^2
BB AABB 0.68^2 * 0.32^2
BB AAAA 0.68^4

BB total is 3 * 0.68^2 * 0.32^2 + 0.68^4 = 0.35586304

The totals need to be weighted based on the relative probabilities of AA and BB in the first two games. Therefore the grand total is:

0.15253504 * [0.4624 / (0.4624 + 0.1024)] + 0.35586304 * [0.1024 / (0.4624 + 0.1024)] = 0.18939904
engineer
 
  1  
Reply Thu 4 Mar, 2010 09:19 am
@markr,
I agree.
0 Replies
 
fbaezer
 
  2  
Reply Wed 17 Mar, 2010 01:32 pm
Cuban players, contemporaries of Schoendist, who played for the Senators:
Ossie Alvarez IF
Julio Becquer 1B
Zoilo Versalles SS
José Valdivieso IF

Of course, there's a probability some older Cuban player was still in Washington (Minnie Miñoso, for example).
0 Replies
 
 

Related Topics

The Baseball Thread - Discussion by Rockhead
Are You Ready For Fantasy Baseball - 2009? - Discussion by realjohnboy
All Time "Ethnic" Baseball Teams - Discussion by fbaezer
10 Best Baseball Books - Discussion by tsarstepan
ARE YOU READY FOR SOME BASEBALL? - Discussion by realjohnboy
The Diceman Cometh - Discussion by Region Philbis
 
Copyright © 2014 MadLab, LLC :: Terms of Service :: Privacy Policy :: Page generated in 0.07 seconds on 12/21/2014 at 05:13:18