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Friday, October 23, 2009
Deconstructing the ARod walk
Here is the scenario: the Angels are up by 1, in the top of the 9th, at home, with 2 outs, bases empty, and the opposing team’s best hitter (ARod) batting.
What are the implications of letting ARod bat, or walking him?
Part 1:
Let’s start with walking ARod. You have a runner on 1B with 2 outs. How often should you expect to end the inning with 1 run to tie, and 2 runs, and 3 runs, etc? Well, I presume all of you have this chart in your back pocket, so pull it out:
http://www.tangotiger.net/RE9902score.html
We expect a scoreless inning 86.4% of the time. 1 run to score 6.2% of the time. 2 runs to score 4.9% of the time. 3 runs to score 1.6% of the time. And 4 or more to score 0.9% of the time. That’s if you have average batters facing average pitchers. Which, in this case, it might be fair to say that the talent level of Matsui, et al, is on par with the pitching level of Fuentes.
Part 2:
What if instead, we pitch to ARod? Let’s give ARod some ridiculous numbers: 10% chance of HR, 15% chance of walk, 15% chance of single, 5% chance of double. That’s a .450 OBP, and .765 SLG, or a .505 wOBA. It’s fairly superlative.
Let’s take them one-at-a-time:
- the 30% times that he walks or singles, we end up with the numbers we noted with him being IBBed.
- the 55% times that he gets an out, the inning is scoreless 100% of the time (and game is over)
- the 5% of the time that he doubles, then we take out our back-pocket chart and get this: 77.7% chance of scoreless inning, 14.7% chance of 1 run, 4.9% chance of two runs, 1.7% chance of 3 runs, and 1% chance of 4 or more runs
- the 10% of the time that he HR, then we get: 92.3% chance of exactly 1 run scoring, 5.1% of 2 runs, 1.7% of 3 runs, 0.9% of 4+ runs
Part 2a
Now we just have to add everything up. The chance of a scoreless inning in pitching to ARod is:
= 30% x 86.4
+ 55% x 100.0
+ 5% x 77.7
+ 10% x 0
= 84.8%
Compare this to Part 1 where we said that walking ARod means you get a scoreless inning 86.4% of the time. So, on this point, Sciosica is right. If you are trying to make sure that no runs score, then the best thing to do is to walk ARod.
Part 2b
The chance of exactly 1 run scoring pitching to ARod is:
= 30% x 6.2
+ 55% x 0
+ 5% x 14.7
+ 10% x 92.3
= 11.8%
Once again, if you let them tie it up, this will happen 12% of the time pitching to ARod, and 6% if you walk him.
Part 2c
The chance of exactly 2 runs scoring:
= 30% x 4.9
+ 55% x 0
+ 5% x 4.9
+ 10% x 5.1
= 2.2%
This is the bad part. The Yanks take the one-run lead 4.9% of the time with ARod walking, and 2.2% of the time with Arod hitting.
Part 2d
The chance of exactly 3 runs scoring:
= 30% x 1.6
+ 55% x 0
+ 5% x 1.7
+ 10% x 1.7
= 0.7%
Another bad part. With ARod walking, they take the 3-run lead 1.6% of the time. With Arod batting, they take the 3-run lead 0.7% of the time.
Part 2e
The chance of 4+ runs scoring: 0.4% with ARod batting and 0.9% with Arod walking.
Part 3
As you can see, it’s similar to the small ball strategies of giving up the chance for big innings, in return for a greater chance at small innings. Is this tradeoff good?
Let’s compare:
When the Yanks score 0 runs, the Angels win. This happens 86.4% of the time with the IBB and 84.8% of the time facing ARod.
When the Yanks score 1 run, the game goes to the bottom of the 9th tied. Normally, this means that the home team has a two-thirds chance of winning (30% of the time they score, and 70% of the time it goes to extra innings, and they win half of those, for a total of 0.65 wins). But, Mariano is pitching. Maybe only 15% of the time they will score, and the other 85% of the time they will win it 45% of the time, for a total of a bit over 50%. Let’s presume a tied game with Mo pitching means that the Angels win 55% of the time entering the bottom fo of the 9th. They will enter such an inning 6.2% of the time walking ARod and 11.8% of the time facing Arod. That gives the Angels .034 wins walking ARod and .065 wins facing Arod.
So far our tally is that Angels win:
.864+.034 = .898 walking ARod
.848+.065 = .913 facing ARod
If the Yanks score exactly 2 runs, and take a 1-run lead into the bottom of the 9th, lets give the Angels a 10% chance of winning the game. So, the 4.9% of the time that this happens means .005 wins for the Angels if walking ARod, and for the 2.2% of the time that this happens facing Arod means .002 wins for the Angels. Our current tally is that Angels win:
.864+.034+.005 = .903 walking ARod
.848+.065+.002 = .915 facing ARod
If the Yanks score 3 runs to take a 2-run lead, let’s give the Angels a 5% chance of winning the game. Doing similar calculations, and the tally:
.864+.034+.005+.001 = .904 walking ARod
.848+.065+.002+.000 = .915 facing ARod
And let’s give the Angels almost no chance of winning if the Yanks have a 3-run lead with Mo.
So, there are your odds of the Angels winning: 90.4% walking ARod and 91.5% facing Arod. And we made ARod a monster.
The breakeven point is if you give ARod 12.5% HR per PA, and make him a roughly league-average hitter if you take away all his HR (making him a .550 wOBA hitter, .475 OBP, .882 SLG). That’s even better than Bonds.
(Hopefully I didn’t make any calculation mistakes.)