Vin Scully has had a hell of a run. Who else can say they worked the same job for sixty-seven years? Who else has given this kind of joy to millions of people? So many better writers have told the tale of Vin, I don’t know that I can do a better job. I’ll just summarize some of Vin’s attributes that we have come to enjoy for many decades: (1) his smooth baritone voice, (2) his story telling, (3) his impartiality, (4) his teaching fans the finer points of baseball, (5) his researching players from other teams and telling us about them, (6) his consistent excellence in play-by-play announcing [in contract to Charlie Steiner among others], (7) his reaching into our family rooms to be part of our families while we watch the game, and (8) his humility. Vin Scully never ever irritated nor annoyed me. I’ve enjoyed every minute that I listened to him.
Meanwhile, we’ve got some baseball to play this weekend (accompanied by Vin Scully). Scott Kazmir (10-6, 4.59 ERA) will pitch against Rockie righty Jon Gray (10-8, 4.42 ERA). Gray last pitched against the Dodgers on August 29th in Denver, and he pitched six shutout innings giving up four hits and three walks while striking out eight Dodgers. The Rockies won that game by a score of 8-1.
What are the chances that the Dodgers clinch the NL West this weekend? About 1 in 3. What are the chances that the Dodgers clinch before playing the Giants again? About 93%. For those interested, the math follows.
To make the math a little easier, let’s assume that the Dodgers have a 50% chance of winning each game, the Giants have a 50% chance of winning each game, and each game is independent of the others. So each game is equivalent to a coin flip. To clinch this weekend, we need a total of four Dodger wins or Giants losses. There are six games. Essentially, we have six coin flips and need heads at least four times (that is four times or five times or six times). The chances of clinching this weekend is [6!/(6!x0!)+6!/(5!x1!)+6!/(4!x2!)]/(2^6) = 34.38%.
Similarly, there are six Dodger games and six Giant games before the Dodgers play the Giants again, for a total of twelve games. We need a total of four Dodger wins and Giants losses to clinch. So the probability of clinching before next Friday is [12!/(12!x0!)+12!/(11!x1!)+12!/(10!x2!)+12!(9!x3!)+12!/(8!x4!)+12!/(7!x5!)+12!/(6!x6!)