Wednesday, February 03, 2010
Chase Utley: SB wizard?
In his career, he has 83 SB and 11 CS. Last year, he was 23-0, as Tommy points out. I remember many many many years ago, when I was a teenager, Raines was still in Montreal, and I just got my first computer. One of the first things I did was try to calculate the breakeven point for steals, by quality of stealer. For example, let’s say that Tim Raines tried to steal only one base all season. Obviously, he would take the most favorable one, the one where the pros outweighed the cons the most. I would give him .999 SB and .001 CS. Now, what if he stole in the two most obvious steal situations. Maybe in this second one, he would be successful .995 times and CS .005 times. See where I’m going here?
So, I fired up my Lotus 1-2-3 spreadsheet, and put in some numbers and tried stuff out. I tried to get to 70 SB and 9 CS, which is a Raines staple. In order to get there, I started him at .999, and dropped by .003 for each steal attempt (0.00289548847776695 if you want to be Excel-exact). By the time I got to the 79th steal attempt, I was at .773 SB. The sum of these 79 was 70 SB. So, I reasoned, Raines was still leaving steals on the table, based on this analysis. What if I kept going? On the 80th steal attempt, he’d be at .770 SB. By the time I get to his 104th attempt, he’s at .701. Basically, we might say that he reaches his personal breakeven point at the 104th attempt. And, adding up all of these steals gives us 88 SB and 16 CS. In some sense, I’m saying that a guy who steals 70 and gets caught 9 times would steal 88 and get caught 16 times.
Interestingly, Raines once had a 90-14 season. Sweet, right?
And, what if Raines decided that his body couldn’t take it, and instead, attempted only 55 steals? In that case, I simply keep the same pattern, but stop him on the 55th attempt. In that case, the sum of his 50 best steal situations is 51 SB and 4 CS (actually 50.6, 4.4). And, interestingly, Raines once had a 50-5 season. I know, I know. I love this stuff, especially if I can apply it to Raines.
Back to Utley. He averages about 15 steals and gets caught 2 times. In order to get that, I can start him as a .999 base stealer, like Raines, but drop him by .0146 after each attempt (about 5 times the degradation rate of Raines), so that he ends up at 15 SB, 2 CS. His last attempt has a success rate of .766. This is what his chart looks like:
Attempt Success
1st 99.9%
2nd 98.4%
3rd 97.0%
4th 95.5%
5th 94.1%
6th 92.6%
7th 91.1%
8th 89.7%
9th 88.2%
10th 86.8%
11th 85.3%
12th 83.8%
13th 82.4%
14th 80.9%
15th 79.5%
16th 78.0%
17th 76.5%
TOTAL 15.00 SB
If I extend it until his last attempt is just over 70%, I get 18 SB, 3 CS.
18th 75.1%
19th 73.6%
20th 72.2%
21st 70.7%
That basically becomes the maximum attempts he should make. UNDER THIS ASSUMPTION. There’s no reason I needed to start him at .999. I could have started him at .9055, and dropped him at .003 like Raines. In that case, his 17th attempt is at 85.9%. In order to get to 70% as the marginal rate, I’d have to extend him to 72 attempts (and he’d end up with 58 SB and 14 CS). As you can see, it depends.
I have to believe that runners are mostly running at their optimal frequency levels, if not pretty close to it. Anyway, I spent (or wasted) lots of hours trying different runners like this. It was loads of fun.


For those who want to get more technical, you should limit it to steals of 2B, and make it steals attempts per opportunity to steal.
So, a baserunner might find himself on 1B with 2B open say for 150 PA, and if he attempts 75 steals, that’s an attempt rate of 50%. So, you might have a chart like this:
Attempt Success
0.01 99.9%
0.02 99.4%
0.03 98.8%
...
0.48 74.5%
0.49 74.0%
0.50 73.4%
The average of the success in this case is .8667. So, 150 opps times 50% attempts = 75 SB attempts. At .8667 success rate, that’s 65 SB, 10 CS.
This mythical runner would get extended like this:
0.51 72.9%
0.52 72.4%
0.53 71.8%
0.54 71.3%
0.55 70.7%
0.56 70.2%
That’s an average success rate of .850, and gives us 71.4 SB, and 12.6 CS.