Why is stealing bases negative unless you are successful at least 75% of the time?

Question

I heard this from Billy Beane and Moneyball. Please explain their reasoning to me, thanks.

Answer 1

The "net stolen bases" rate fluctuates year to year, a little bit, because it is empirically derived and not all stolen bases are perfectly equal (stealing home is better than merely stealing second, though far less common), but the baseline value tends to be at 67%, for a rather simple reason. Getting caught stealing costs your team one out AND one baserunner -- and since getting men on base is hard, they are very much worth preserving. That's two penalties (of approximately equal value) weighted against one potential base gained.

Therefore, it takes essentially two successful steals to counterbalance the detrimental effects of one caught steal.

The "net stolen bases" formula is

netSB == SB - 2*CS

If a runner cannot swipe the base two times out of three, he is overall hurting his team's offense with his poor baserunning.

Of course, 67% is the minimum; I don't remember that exact topic in Moneyball, but Beane may take a higher value into consideration as his desired acceptable minimum success rate when evaluating players.

Answer 2

It's negative when you make an out in any play. The steal is just as it says. Theft. If you get caught, you are out and your team can suffer for it. It's the only form of cheating that is allowed in professional sports and it is legal by the rules. It only looks bad when you fail.

Answer 3

Moving a base (usually first to second) isn't nearly as helpful as losing a baserunner and taking an out is harmful. Thus, you need to succeed most of the time for it to be worth it.

Answer 4

you dont want to ever get caught when you steal.