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September 5, 2015 / Nat Anacostia

Update (with head-to-head) – How many games do the Nats need to win?

For the last two weeks, I’ve posted estimates of how many games the Nats will need to win the rest of the way in order to beat the Mets.

This week, the teams are playing first of two head-to-head series which will have a significant impact on the outcome. So this time, I’ve decided to tweak my calculations to show how the team’s performance in these head-to-head match ups will affect the performance needed in their non-head-to-head games.

My starting point, as it was in my earlier two posts, is to make my calculations conditional on the Mets doing as well as expected in their remaining games. I’m looking at two statistics sites, Fangraphs and Baseball Prospectus. Fortunately, they agree almost exactly on the Mets expected performance (though Fangraphs is more optimistic than BP about the Nats). Both sites expect the Mets to win on average (based on repeated simulations) about 89.5 games.

This time I’m going to split out the non-head-to-head games. Both teams (as of Saturday September 5th) have 28 games left to play. With 6 head-to-head games, that leaves 22 non-head-to-head games. The reasonable assumption is that these simulations are assuming that the Nats and Mets split their 6 head-to-head games, which means that they expect the Mets to win 12.5 of their remaining 22 non-head-to-head games.

So, rounding up and assuming that the Mets go 13–9 in their remaining non-head-to-head games, I calculate the following based on how the Nats do in the head-to-head match ups:

  • If they split the head-to-head 3 games each, then the Nats would need to go 19–3 in their remaining non-head-to-head games. That’s pretty unlikely, so the Nats need to do better than split the head-to-head. (The other possible way to win would be if the Mets collapsed and did much more poorly than expected, but given their easy schedule, that’s a slim hope.)
  • If the Nats win 4 of 6 in the head-to-head games, then they would need to go 17–5 in non-head-to-head games to beat the Mets. That’s still tough, but certainly more plausible than going 19–3. Winning 4 of the head-to-head match ups seems like the minimum acceptable goal for the Nats.
  • If the Nats win 5 of 6, then they would need to go 15–7 in their non-head-to-head games. That’s roughly equivalent to winning two-thirds of their remaining games, and with a relatively easy schedule, that seems feasible.
  • If the Nats sweep all 6 head-to-head games, then they’d need to go 13–9 in their non-head-to-head games. Sweeping the head-to-head games would clearly be the best route to try to guarantee a division championship.

So we see how critical this week’s series against the Mets will be. If the Nats fail to win at least two, they will find themselves on the periphery of the race, with little chance of contending. If they go 2–1, they’ll stay in the race, but still at fairly long odds. If they manage to sweet the series, they’ll vault directly into a tight race for the division championship. They need to go for the sweep!

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