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