Question

In the bottom of the 9th inning in a tie game, the leadoff batter for the home team hits a double. According to researcher Tom Tango, teams in this situation have a win probability of 0.807. Because the home team only needs one run to win the game, some managers will choose to have the next player attempt a sacrifice bunt, which allows the runner on second base to reach third base while the hitter is thrown out. If the sacrifice bunt is successful, then the home team has a runner on third base with 1 out and a win probability of 0.830. However, if the sacrifice is unsuccessful and the runner is thrown out at third, there is a runner at first with one out (win probability of 0.637). What is the minimum probability of a successful bunt that would warrant using the bunt?

Answer 1

• 0.881

Step-by-step explanation:

The minimum probability of a successful bunt that would warrant using the bunt is that probability that, at least, does not decrease the probability of winning after the batter hit the double: 0.807.

Call p the probability of a succesful sacrifice bunt.

Using a probability tree diagram:

• successful sacrifice bunt: p

- win: 0.830

- loose: 0.17

• unsucessful sacfifice bunt: ( 1 - p)

- win: 0.637

- loose: 0.363

From that, the probability of winning is 0.830(p) + 0.637(1 - p)

You want to determine p, such that 0.830(p) + 0.637(1 - p) ≥ 0.807

Solve for p:

• 0.830p + 0.637 - 0.637p ≥ 0.807

• 0.193p ≥ 0.170

• p ≥ 0.8808

Rounding to thousandths, the minimum probability of a succesful bunt that would warrant using the bunt is 0.881.