How many times has this happened to you? You’re driving down the street and your spouse or significant other drops a bombshell:

“Honey, how many runs are expected to score in the rest of the inning if there are runners on first and third with two outs?”

Or you’re at a cocktail party, and someone who’s had a little too much punch starts arguing in a loud voice”

“I’d rather have the bases loaded with one out than runners on second and third with nobody out!”

And silly, careless you! You left your copy of “The Hidden Game Of Baseball” at home. You don’t even have your Blackberry handy to find Palmer’s base-out expected runs matrix on the Internet. And unlike Clay Davenport and James Click, you haven’t committed the matrix to memory. What will you do? WHAT… WILL… YOU… DO?

Baseball Prospectus to the rescue. Here’s a quick and dirty way you can approximate the expected number of runs given the bases that are occupied and the number of outs. We’ll use an example to demonstrate–runners on first and third with one out:

**Method A)**

Step 1) Count up the “total bases” represented by the runners on base–that is, a runner on third counts for 3 bases, a runner on second counts for 2 bases, and a runner on first counts for 1 base.

Runners on first and third are 1 + 3 = 4 total bases.

Step 2) Multiply by the number of outs left in the inning.

In our example, there’s one out, so there are two outs left in the inning.

4 total bases * 2 remaining outs = 8

Step 3) Divide by 10.

8 / 10 = 0.8

Step 4) Add 0.45 if 3 outs remain, 0.25 if 2 outs remain, or 0.10 if 1 out remains. This is approximately the expected number of runs.

0.8 + 0.25 = 1.05 runs

This approximation yielded 1.05 runs for 1st & 3rd, 1 out. The actual Hidden Game matrix has 1.088. In fact, this method has only one situation that yields an error of more than 0.1 runs (that situation is 2nd and 3rd, 1 out, which computes as 1.25 runs instead of 1.371 runs given in Hidden Game). The average absolute error across all situations is 0.037 runs.

So, to summarize in mathematical notation:

Exp_Runs ~= TB * Outs_left/ 10 + {0.10 0.25 0.45}[Outs_left]

where {0.10 0.25 0.45}[Outs_left] is intended to be the Outs_left element of the array {0.10 0.25 0.45}

Hidden Game table Method A approximation outs 0 1 2 0 1 2 --- 0.454 0.249 0.095 0.450 0.250 0.100 1-- 0.783 0.478 0.209 0.750 0.450 0.200 -2- 1.068 0.699 0.348 1.050 0.650 0.300 12- 1.38 0.888 0.457 1.350 0.850 0.400 --3 1.277 0.897 0.382 1.350 0.850 0.400 1-3 1.639 1.088 0.494 1.650 1.050 0.500 -23 1.946 1.371 0.661 1.950 1.250 0.600 123 2.254 1.546 0.798 2.250 1.450 0.700

**Method B)**

If you’re willing to do a little more arithmetic, we can improve the accuracy some more:

Step 1) Start again with the “total bases” of the runners on base.

Runners on first and third are 1 + 3 = 4

Step 2) Subtract 0.15 for each runner on base from the total.

Two runners are on base, so: 4 – 0.15 – 0.15 = 3.7

Step 3) Multiply by the number of outs left in the inning,

3.7 * 2 outs = 7.4

Step 4) and divide by 9

7.4 / 9 = 0.822

5) As before, add 0.45 if 3 outs remain, 0.25 if 2 outs remain, or 0.10 if 1 out remains to get the answer:

0.8222 + 0.25 = 1.072 runs

This method gives 1.072 runs vs. 1.088 from the Hidden Game Matrix. We’ve lowered the absolute error versus the previous method by 58%. On average, this shaves the absolute error by about 10%, and reduced the maximum error for any base-out state from 0.121 to 0.081. The state with the greatest error in this method is bases-loaded, 2 outs, which computes as 0.717 instead of the correct 0.798.

To summarize:

Exp_Runs ~= (TB – 0.15 * #Runners) * Outs_left / 9 + {0.10 0.25 0.45}[Outs_left]

Hidden Game table Method B approximation outs 0 1 2 0 1 2 --- 0.454 0.249 0.095 0.450 0.250 0.100 1-- 0.783 0.478 0.209 0.733 0.439 0.194 -2- 1.068 0.699 0.348 1.067 0.661 0.306 12- 1.38 0.888 0.457 1.400 0.883 0.417 --3 1.277 0.897 0.382 1.350 0.850 0.400 1-3 1.639 1.088 0.494 1.683 1.072 0.511 -23 1.946 1.371 0.661 2.017 1.294 0.622 123 2.254 1.546 0.798 2.300 1.483 0.717

Now, armed with either of the two formulae above, you can be the life of the party and a hero(ine) to your loved ones.

And, please, there’s no need to send thank-you gifts to me. That look in your eye is reward enough.