I tweeted the results but I thought I’d show my work in case anyone was interested. The possible outcomes if he sends him are run scores and you have 1st and second one out or you have 1st and second and 2 outs. Those outcomes have Run Expectancies (RE) of 1.971 and o.466 according to the table found here. The alternative is that you’d have the bases loaded with one out, which has an RE of 1.65. Now all you have to do to find the equilibrium point is solve the equation 1.971*x+(1-x)*0.466=1.65 where x is the probability of success. That comes out to ~0.79. If you assume that the runners behind the play will advance then you get something closer to ~0.76.

Clearly the RE tables are generic, so you’d want to factor in Holliday coming up, but this gives a general feel for the problem.