An Alternative Universe in Which the Mets Made the Playoffs

Early last week, I wrote about whether the Nationals should rest Max Scherzer or maximize their chances of retaining home field advantage in the NL Wild Card game. Well, joke is on me, because they easily did both.

With that question centering around the Nationals' top tier pitching options, my esteemed colleague (a suffering Mets fan) posed this: would the Mets be the rare underdog that actually would have a better chance against the Nationals in a 5-game series, as opposed to a one game playoff?

Generally, an underdog wants a shorter time frame to try to pull off the upset - less time equals more randomness, and less of a chance for talent to win out. But the Mets are one of the few teams in baseball that could match the Nationals' pitching staff, with three aces in the top 10 of NL pitcher WAR (plus the NL Rookie of the Year, and still didn't make the playoffs).

So consider an alternative reality in which the Mets make the playoffs, and face the following proposition: they have to face the Nationals in a one game playoff, or a full five game NLDS (in both cases Washington has home field advantage).

Do you want to roll the dice on one game, with your best pitcher (Jacob DeGrom) up against their's (Scherzer)? Or would you rather have your lineup of DeGrom (#1 in NL pitcher WAR), Wheeler (#7), and Syndergaard (#10) go up against Scherzer (#2), Strasburg (#3), and Corbin (#6)?

Over 10,000 simulations, the Nationals win 61.8% of the time in the one game playoff (this also serves as Game 1 of the NLDS). Here are the game-by-game probabilities over a five game series:

Game	Home	Away	Home Pitcher	Away Pitcher	WSN Win Prob	NYM Win Prob
Game 1	WSN	NYM	Scherzer	DeGrom	61.80%	38.20%
Game 2	WSN	NYM	Strasburg	Wheeler	69.99%	30.01%
Game 3	NYM	WSN	Syndergaard	Corbin	41.45%	58.55%
Game 4	NYM	WSN	DeGrom	Scherzer	39.43%	60.57%
Game 5	WSN	NYM	Strasburg	Wheeler	69.12%	30.88%
Series	WSN	NYM			62.30%	37.70%

It's actually incredibly close - Washington would only have a 0.5% edge (62.3% favorites) in the NLDS scenario compared with the wild card game.

Outcome	In...	Prob
WSN	3	17.93%
WSN	4	17.39%
WSN	5	26.98%
NYM	3	6.7%
NYM	4	18.9%
NYM	5	12.1%

But, as expected, the Mets would still (barely) prefer the randomness inherent in one game over a five game series.

Should the Washington Nationals Rest Scherzer, Risking Home Field Advantage?

The Washington Nationals have clinched a wild card spot, but are only one game up on the Milwaukee Brewers for home field advantage in the NL Wild Card game. The Brewers have been red hot, with a 19-4 record in the month of September. 

Meanwhile, the Nationals have one of the best rotations in baseball, with Max Scherzer and Stephen Strasburg ranking #2 and #3 in the NL, respectively, in pitcher WAR this season.

Scherzer is their best pitcher. And he's currently in line to start the last game of the regular season on Sunday. However, that would burn him for the one-game playoff next Tuesday. So which is more valuable? Scherzer pitching in the NL Wild Card game, or the Nationals retaining home field advantage?

My colleague answered this question via FiveThirtyEight's pitcher ratings:

Using 538's Elo rating, they have Scherzer (30) as worth 11 more points than Strasburg (19). However, they have HFA listed at 24 points, which is actually doubled to 48 points if you take the HFA away from WAS and give it to MIL. So Strasburg at home is worth 43 points and Scherzer on the road is worth 6 points, a difference of 37 points! So they theoretically should keep Scherzer on normal rest this week as long as HFA is still on the line.

Per the Elo formula and FiveThirtyEight's ratings, the Nationals would be 54.4% favorites on the road with Scherzer, and 59.7% favorites at home with Strasburg. So they should maximize their chances at retaining home field.

I checked this using my play-by-play MLB simulator, assuming that the Brewers' top arm, Brandon Woodruff, would start in either wild card scenario. I found similar results: the Nationals win 57.5% of the time on the road with Scherzer, and 59.9% of the time at home with Strasburg. Both pitchers are really good! So home field is the deciding factor.

And for the Brewers to maximize their chances of winning, they have to chase either pitcher early: Scherzer's average start in Milwaukee is 6.08 innings in a Brewers win, but 6.49 innings in a Nats win. Strasburg's average start in Washington is 5.79 innings in a Brewers win, but 6.25 innings in a Nats win.

"What are the odds?" Of Rolling 7 Doubles in a Row (Dice)

In backgammon, each game begins with each player throwing a single die to determine who goes first. In the event that each player's roll is the same, "then both players roll again until they roll different numbers". The following question was posed to me: what are the odds that each player rolls the same number... 7 times in a row?

There are 6 potential numbers on each die, so over two dice there are 6 * 6 = 36 possible combinations. Additionally, there are 6 possible pairs: (1,1), (2,2), (3,3), (4,4), (5,5), (6,6). So 6 / 36 = 1 / 6 ~= 0.167% chance of tying on one roll. But when this is replicated 7 times in a row: (1 / 6) ^ 7 = 0.00036%, or 1 in 279,936.

N	Chances	1 in ...
1	16.67%	6
2	2.78%	36
3	0.46%	216
4	0.08%	1,296
5	0.01%	7,776
6	0.002%	46,656
7	0.00036%	279,936
8	0.0001%	1,679,616
9	0.00001%	10,077,696
10	0.000002%	60,466,176

That's far less likely than having a doctor on your flight or getting the same taxi driver twice in a large city, but way more likely than two teams sharing identical box scores on the same day or three people in the same family all being born at the exact same time (on different days).

What if you play backgammon a lot though? How long would you have to play in order to expect to see a streak of 7 dice ties in a row?

The math is explained on this nice "Probability of Runs" calculator, but you would need to play 232,845 games of backgammon until there would be a better than 50/50 chance of seeing a run of ties like this. The average number of games needed until you would see this happen is even larger: 335,922 games of backgammon.