(02-16-2015 10:21 PM)Owl 69/70/75 Wrote: First, the bunt in the 9th was with the score tied. One run wins it there.
As for the 10th down 1 run, there are three possible outcomes--you score 0 and lose, you score 1 and play on, or you score more than 1 and win outright. Not bunting increases the probability of scoring more than 1 and winning outright, but it also increases the probability of scoring 0 and losing. Bunting increases the probability of scoring exactly 1 and going to the 11th.
Going to the theoretical (I literally brought the audio up for the game to hear the last at bat, so I didn't hear the critical parts of the game).
If the odds of scoring one run bunting are X and odds of scoring one run hitting are Y and two runs Z, then you have to look at the likelihood a team wins once it goes to extra innings. If we were to say that odds of winning in extra innings is 55% for the home team, then you'd have to figure that bunting is better only if:
Likelihood of loss in current inning batting: (1-y)
Likelihood of loss in extra innings batting: (y-z)*.45
Likelihood of loss in current inning bunting: (1-x)
Likelihood of loss in extra innings bunting: x*.45
Scoring one run is greater with a bunt: X>Y>Z
and total number of runs is maximized with a hit: (y+z*2 > x), noting that 2 is a stand-in for any number of runs greater than one, but assume it's not going to be much more than 2... z>(x-y)/2
So, bunt if (1-x)+.45*x < (1-y)+(y-z)*.45, or (1-.55*x) < (1-.55*y-.45*z), or if (.55*x) > (.55*y + .45*z) or (x-y)*(5/4)>z or (x-y)*(5/4)>(x-y)/2 ... since 5/4 > 1/2, it seems like you're more likely to win by hitting.
Sorry that I can't organize it better or doublecheck my logic.