The Art of Stealing Bases

[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.

[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%.

[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.

[Antitrust32]

Originally Posted by Cannon Shell

Your math is not valid

Why not?

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

[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.