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Friday, June 06, 2014

Odds that one of Cueto, Wainwright, Felix, Tanaka, Buehrle will win Cy Young in 2014?

By Tangotiger 01:34 PM

That's my question.  Is it at 50/50?  More?  Less?

We're about sixty games in so far.  If an aspiring saberist wants to tackle this, go to Pinto's site, take the top 2 Cy contenders in each league for the last five years (2009-13), which is 10 Cy Youngs, plus the best 5th guy? (either after around 60 games, or through games of June 5 of that year), and tell us how many Cy Youngs were in that group.  Was it 5?  3?  8?  What exactly?


#1    Tangotiger 2014/06/06 (Fri) @ 14:12

Checked real quick…. looks like the odds are 30% that one of those 5 will win the Cy Young.  However….

***

Nine of the 10 winners were also within 12 Cy Young points of the leader. 

That’s 7 NL pitchers that fit the bill there:
Cueto, Wainwright, Teheran, Hudson, Bumgarner, Greinke, Lohse

I’d give Cueto and Wainwright 20% each, and 10% for each of the other 5.

In the AL:
Tanaka, Felix, Buehrle, Darvish, Gray, Kazmir

So, I’ll give 20% for each of the top 3, and 10% for each of the other three.

Therefore, I’m going with a 50/50 shot that one of the five guys I listed will win a Cy Young.

***

10% chance of a long-shot winner from among this group: Wacha, Strasburg, Ross in the NL, and Kluber, Keuchel, Lester, Scherzer in the AL.


#2    Tangotiger 2014/06/06 (Fri) @ 14:16

Hmmm… I should check my math.

If there’s a 40% chance that Cueto+Wainwright will win in the NL and a 60% chance that Tanaka+Felix+Buehrle win in the AL, then:

24% chance that we get both winners in that group
52% chance that we get exactly one from that group
24% chance that we don’t get any winners of that group

So, the odds of AT LEAST one winner is 76%.

The expected number of winners is:
.24 x 2
+ .52 x 1
= 1.00 winners (out of 2)

That’s why I had said 50/50.  But I wasn’t answering my own question!


#3    rosen380 2014/06/06 (Fri) @ 14:17

For simplicity, I just used 6/5 of each year.

2013: Kershaw, Buchholz, Iwakuma, Wainwright, Zimmerman (1)
2012: Beachy, GGonzalez, Verlander, James McDonald, Sale (0)
2011: Jered Weaver, Halladay, Haren, Jurrjens, Lohse (0)
2010: UJimenez, Halladay, Wainright, Price, Lester (1)
2009: Greinke, Santana, Halladay, Billingsley, Jered Weaver (1)

So, for this sample, three of 10 awards…


#4    Tangotiger 2014/06/06 (Fri) @ 14:18

As for the more brute force way, Greinke (2009), Halladay (2010), Kershaw (2013) all were in the top 5 after June 5 games.  That’s three pitchers in 5 years, or 60% chance.

***

Anyway… I’m going with around 75% chance that one of those 5 will win the Cy Young in 2014.


#5    rosen380 2014/06/06 (Fri) @ 14:22

2008 (2)
2007 (1)
2006 (1)
2005 (0)
2004 (1)
2003 (0)
2002 (1)
2001 (1)
2000 (2)

Back to 2000, 12 of 26, 46%

 

 


#6    rosen380 2014/06/06 (Fri) @ 14:25

And that is 10 in 13 seasons, so a 77% chance that at least one of the five wins a Cy Young in any given year…


#7    Tangotiger 2014/06/06 (Fri) @ 14:27

2000-2013 is 14 years, or 28 awards.

So, we have 12 winners in 14 years, or 86%.

***

We have 12 in 28 awards, or 43%.  If that 43% were random, we’d have expected:
18% nailing both
50% getting exactly 1
32% getting neither

That’s a total of 68%

***

So, somewhere between 68% and 86%.  That’s an average of 77%.

Therefore, let’s go with 75% chance.


#8    rosen380 2014/06/06 (Fri) @ 14:27

Make that 14 seasons and 28 Cy Youngs… 😊


#9    Tangotiger 2014/06/06 (Fri) @ 14:29

“So, we have 12 winners in 14 years, or 86%.”

Oops… that’s 10 years with at least 1 out of 14 years, or 71%.

Couple that with the 68%, and that puts the odds at 70% that at least one of those 5 guys will win the Cy Young in 2014.

Thanks for rolling up your sleeves!


#10    Tangotiger 2014/06/06 (Fri) @ 14:31

If you want to see something cool, compare Scherzer’s position to the leader in 2013 and his position in 2014.

Go ahead!


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