The Art of Stealing Bases

[King Glorious]

Originally Posted by Cannon Shell

The intent is the same, to move the runner over. While it is surely not the same the 2 are part of a theory of small ball that go hand in hand.

The theory is the same but the way they are carried out is far different. In one, you are going to be successful 75% of the time without giving up an out and in the other, you are almost 100% of the time going to give up an out and it's probably less than 75% of the time that the runner advances without being the sacrificee.

[Cannon Shell]

http://www.baseballprospectus.com/ar...articleid=2607


"A runner on first with no one out is worth .9116 runs. A successful steal of second base with no one out would bump that to 1.1811 runs, a gain of .2695 expected runs. If that runner is caught, however, the expectation--now with one out and no one on base--drops to .2783, a loss of .6333 expected runs. That loss is about 2.3 times the gain.

Not all steals come with a runner on first and no one out, of course, and there's a lot of math that goes into the 75% conclusion. Michael Wolverton covers the concept in this excellent piece. The main point is that in considering stealing bases, you have to consider both the benefit and the cost. In all but the most specific situations, outs are more valuable than bases, which is why the break-even point for successful base-stealing is so high. "

[Antitrust32]

Originally Posted by Cannon Shell

http://www.baseballprospectus.com/ar...articleid=2607


"A runner on first with no one out is worth .9116 runs. A successful steal of second base with no one out would bump that to 1.1811 runs, a gain of .2695 expected runs. If that runner is caught, however, the expectation--now with one out and no one on base--drops to .2783, a loss of .6333 expected runs. That loss is about 2.3 times the gain.

Not all steals come with a runner on first and no one out, of course, and there's a lot of math that goes into the 75% conclusion. Michael Wolverton covers the concept in this excellent piece. The main point is that in considering stealing bases, you have to consider both the benefit and the cost. In all but the most specific situations, outs are more valuable than bases, which is why the break-even point for successful base-stealing is so high. "

that number seems appropriate for people who only convert 50% of the time.. @ 75% it seems like it should be half of that.

[King Glorious]

Originally Posted by Cannon Shell

http://www.baseballprospectus.com/ar...articleid=2607


"A runner on first with no one out is worth .9116 runs. A successful steal of second base with no one out would bump that to 1.1811 runs, a gain of .2695 expected runs. If that runner is caught, however, the expectation--now with one out and no one on base--drops to .2783, a loss of .6333 expected runs. That loss is about 2.3 times the gain.

Not all steals come with a runner on first and no one out, of course, and there's a lot of math that goes into the 75% conclusion. Michael Wolverton covers the concept in this excellent piece. The main point is that in considering stealing bases, you have to consider both the benefit and the cost. In all but the most specific situations, outs are more valuable than bases, which is why the break-even point for successful base-stealing is so high. "

All of that would be fine if the chances of successfully stealing a base were 50/50. Sure, if the guy is only 50/50, you don't want him running wild on the bases. But when your success rate is 75%, and usually up around 85% for the best base stealers, that throws those expected run numbers out of the window. If the chances of getting an out and a base are even (as with a 50% base stealer or in the case of sacrifice bunts), then it's not worth it. When the chances of getting that same base without giving up an out is 80%, I think it's worth it. The numbers giving expected runs are based on all base runners being equal. They aren't.

[Cannon Shell]

Originally Posted by Antitrust32

that number seems appropriate for people who only convert 50% of the time.. @ 75% it seems like it should be half of that.

It has nothing to do with the runner, just the avg run expectation between no outs and a runner on versus one out and no runner on

[Cannon Shell]

Originally Posted by King Glorious

All of that would be fine if the chances of successfully stealing a base were 50/50. Sure, if the guy is only 50/50, you don't want him running wild on the bases. But when your success rate is 75%, and usually up around 85% for the best base stealers, that throws those expected run numbers out of the window. If the chances of getting an out and a base are even (as with a 50% base stealer or in the case of sacrifice bunts), then it's not worth it. When the chances of getting that same base without giving up an out is 80%, I think it's worth it. The numbers giving expected runs are based on all base runners being equal. They aren't.

That isnt truye. It is simply run expectation of a man on and no outs versus no one on and one out. You may believe the average player is in the 50% range but that isnt true. In 2009 there have been 1900 SB and just 694 CS. In 2008 2739 sb and 1035 cs. In 2007 2918 sb and 1002 CS. In 1980 during the SB era the numbers are 3294 and 1610.

[dalakhani]

Interesting stuff guys.

I really dont have anything to add although I do have a question. Does anyone have the stats on double steals of second and third?

[SniperSB23]

If you are an 85% base stealer your expected runs from attempting a steal is (0.85*1.1811) + (0.15*0.2783) which equals 1.04568 runs which is considerable higher than 0.9116 runs if you don't attempt a steal. Even at 75% it would be .9554 expected runs. At 70% they are equal. So basically if you have a 70% chance of stealing you won't help or hurt your team in the long run. Having more than a 70% chance will help your team, less than 70% will hurt your team. Yet there are a considerable amount of players well over 75% that should be running more.

[Cannon Shell]

Originally Posted by SniperSB23

If you are an 85% base stealer your expected runs from attempting a steal is (0.85*1.1811) + (0.15*0.2783) which equals 1.04568 runs which is considerable higher than 0.9116 runs if you don't attempt a steal. Even at 75% it would be .9554 expected runs. At 70% they are equal. So basically if you have a 70% chance of stealing you won't help or hurt your team in the long run. Having more than a 70% chance will help your team, less than 70% will hurt your team. Yet there are a considerable amount of players well over 75% that should be running more.

Your math is not valid

[SniperSB23]

Originally Posted by Cannon Shell

Your math is not valid

How do you possibly come to that conclusion? It's 100% valid. It's a simple calculation of expected value.

[Antitrust32]

Originally Posted by Cannon Shell

Your math is not valid

Why not?

the paragraph you posted before is implying a 50% success rate.

[Antitrust32]

Originally Posted by Cannon Shell

http://www.baseballprospectus.com/ar...articleid=2607


"A runner on first with no one out is worth .9116 runs. A successful steal of second base with no one out would bump that to 1.1811 runs, a gain of .2695 expected runs. If that runner is caught, however, the expectation--now with one out and no one on base--drops to .2783, a loss of .6333 expected runs. That loss is about 2.3 times the gain.

Not all steals come with a runner on first and no one out, of course, and there's a lot of math that goes into the 75% conclusion. Michael Wolverton covers the concept in this excellent piece. The main point is that in considering stealing bases, you have to consider both the benefit and the cost. In all but the most specific situations, outs are more valuable than bases, which is why the break-even point for successful base-stealing is so high. "

this even proves snipers math is correct

"which is why the break-even point for successful base-stealing is so high"

aka 70-75%. a person who steals at an 85% clip would be above the break even point.

[Cannon Shell]

Originally Posted by SniperSB23

How do you possibly come to that conclusion? It's 100% valid. It's a simple calculation of expected value.

You are only using the no out scenario and there are three scenarios that are in play in your sb%.

[Cannon Shell]

Originally Posted by Antitrust32

Why not?

the paragraph you posted before is implying a 50% success rate.

No there is no implication of anything. There is no indication of HOW the runners got to the bases, just their chance of scoring under different scenarios.

[Cannon Shell]

Originally Posted by Antitrust32

this even proves snipers math is correct

"which is why the break-even point for successful base-stealing is so high"

aka 70-75%. a person who steals at an 85% clip would be above the break even point.

However in running more their overall % will drop, quite possibly to below the break even point.

[SniperSB23]

Originally Posted by Cannon Shell

You are only using the no out scenario and there are three scenarios that are in play in your sb%.

(0.85*1.1811) + (0.15*0.2783)

That is the 85% chance of the steal plus the 15% chance of getting out. In the 85% chance the steal is successful your expected values of runs is 1.1811. In the 15% chance the steal is successful your expected value of runs is 0.2783. It factors in both possibilities if you attempt a steal.

For a 75% base stealer:

(0.75*1.1811) + (0.25*0.2783)

The break even point is the 70% base stealer;

(0.70*1.1811) + (0.30*0.2783)

That equals .91026 which is almost identical to the .9116 expected runs if you don't attempt the steal.

[Cannon Shell]

Originally Posted by SniperSB23

(0.85*1.1811) + (0.15*0.2783)

That is the 85% chance of the steal plus the 15% chance of getting out. In the 85% chance the steal is successful your expected values of runs is 1.1811. In the 15% chance the steal is successful your expected value of runs is 0.2783. It factors in both possibilities if you attempt a steal.

For a 75% base stealer:

(0.75*1.1811) + (0.25*0.2783)

The break even point is the 70% base stealer;

(0.70*1.1811) + (0.30*0.2783)

That equals .91026 which is almost identical to the .9116 expected runs if you don't attempt the steal.

yes with no outs and a man on first which isnt the only scenario that an 85 or 75% stealer will face.

[SniperSB23]

Originally Posted by Cannon Shell

yes with no outs and a man on first which isnt the only scenario that an 85 or 75% stealer will face.

No one is saying they should be stealing home with one out.

[Cannon Shell]

Originally Posted by SniperSB23

No one is saying they should be stealing home with one out.

Or second with two.

[SCUDSBROTHER]

How can you not mention the pitcher n' catcher involved? Some of these teams really struggle at throwing runners out at 2nd.