For simplicity (at least relative simplicity), we'll assume the visiting team is shutout (which is the single-most likely outcome, and the math is straightforward on this):
| Team | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | Total |
| Visitor | 69.78% | 76.71% | 73.60% | 70.41% | 71.69% | 73.90% | 73.70% | 75.44% | 76.06% | 6.2162% |
So for now we'll focus on the other team (in this case the home team, the Giants). Overall, the total probability two teams match is 0.6022%. Over all runs, the results are as follows:
| Total Runs | Total Prob | Frequency |
| 1 | 0.0687% | 11.41% |
| 2 | 0.0813% | 13.50% |
| 3 | 0.0896% | 14.88% |
| 4 | 0.0785% | 13.04% |
| 5 | 0.0724% | 12.02% |
| 6 | 0.0531% | 8.82% |
| 7 | 0.0397% | 6.59% |
| 8 | 0.0276% | 4.58% |
| 9 | 0.0208% | 3.45% |
| 10+ | 0.0326% | 5.41% |
As seen in the chart below, the likelihood increases up to 3 runs, and decreases as runs increase from there (with the final bar showing the frequency of 10 or more runs):
There is in fact a pattern to this distribution, and the (still very small) chances of the scores matching do change with respect to the total number of runs scored.
Back to the instance in which one team is shutout: 6.2057% * 6.2057% is 0.3851%, but it seems two teams in two different games get shutout more often than once in every 260 tries. However, this 1 in 260 refers to the number of pairs, so consider a typical day of baseball: there are 15 games, and 1 team has to win (and thus score 1 or more runs) in each game, which leaves us with 15 possible teams to match. 15 choose 2 is represented as follows:
which is equal to 105. So there are actually 105 possible pairs in any given (typical) night of MLB games, meaning we should see 0.003851 * 105 = 0.404 matches of shutouts on average. That's about 1 every 2.5 nights, which seems to fit reality.
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