Is The All-Star Game The Biggest Remaining Game for Dodgers?
By Sky Andrecheck

A week from today, the Major League All-Star Game will be played between the American and National leagues. Traditionally the greatest exhibition in all of sports, the game changed in 2003 when Bud Selig decreed that the winner of the All-Star game would have home field advantage during the World Series. While the move (which I love) has largely accomplished its purpose of rejuvenating the game and inspiring competitive instincts in the players, the game is still somewhat treated as an exhibition. Players still occasionally beg out of the game and the managers still try to get everybody in the game, even if it means taking some of the best players out. Terry Francona actually said he was hoping for a game ending NL homer last year as the game went to extra-innings. For him, the All-Star game was secondary to regular season games.

The All-Star game's slogan is "This Time It Counts", but how much does it really count? Obviously Francona and company don't think it counts for very much. Is home field advantage in the World Series really worth playing for? Or are players better off focusing on the regular season?

Earlier this season, I debuted Championship Leverage Index, an attempt to measure the impact and importance of a game on a team's chances of winning the World Series. We can apply this same methodology to the All-Star game. Of course, during the regular season, an additional win adds to the probability of winning the World Series by increasing the team's chances of making the playoffs and thus winning a championship. In the case of the All-Star game, winning the game helps only if the team makes it to the World Series.

How Much Is An All-Star Game Win Worth?

Assuming that a team does make it to the Fall Classic, how much does having the home field help? Historically, having home field in an individual game adds about 4% to a team's probability of victory. This number has been larger during the playoffs, but this likely has something to do with the best teams playing more home games, so it's a misleading guide. Taking this 4% mark and assuming the teams are even, a World Series team would have a 54% chance of victory in its home games and a 46% chance of victory in its road games. How does this translate during an entire 7-game series? Turns out that the mathematics show that a team which has the home field advantage in the series as a whole will win 51.26% of series.

Overall, the extra 1.26% is pretty small - there's probably a reason that MLB doesn't tout this number in its "This Time It Counts" promos for the All-Star game - but in something as big as the World Series, it helps to have every advantage possible. Of course, the game only adds 1.26% for a team actually playing in the World Series - to other teams, the game is worthless. Apportioning this advantage blindly among an average of 15 teams per league, an All-Star win adds 0.084% to each team's chances of winning the World Series. Thus, if you were oblivious to the standings, and your league won the All-Star game, you could rejoice that your team's chances had just gone up by 8 one hundredths of one percent. This time it counts, eh?

Indeed, 0.08% sounds impossibly small, until you consider that the average regular season win doesn't help you much more, clocking in at a mere 0.28% according to my prior work linked to above. Dividing these figures, we find that the All-Star game has a Championship Leverage Index of .30, meaning that the game is about 30% as meaningful as the average regular season game.

You can be the judge of whether 30% of an average regular season game is more or less than you might have thought. I suspect it's more. 30% of an average game is not a lot, but it's more meaningful than any Nationals game has been since May, and it's more meaningful than many teams' regular season games will be in about a month's time. What's more, I doubt that Francona would be willing to wish away even 30% of a regular season Sox game.

All-Star Championship Leverage Index by Individual Team

Of course, while on average the All-Star game is worth about 30% of an average regular season game, we can calculate this separately for individual teams, with dramatically differing results. Teams which are far back in the race need every win they can get their hands on, and home field advantage in the World Series means little. The chance that one regular season win will prove decisive is low, but the chance that home field advantage in the World Series will make a whit of difference is even lower. For teams in the thick of a pennant race, the World Series advantage is useful, however each regular season win has a fairly high chance of being crucial, rendering a regular season win far more valuable than an All-Star game win.

However, the All-Star game takes on the most importance to teams which are far ahead in the standings and have a high likelihood of making it to the Fall Classic. For these teams, a regular season game also means relatively little, since a playoff berth is all but locked up. In this scenario, is the All-Star game actually more important than a regular season game?

Looking at this year's Dodgers, let's aim to find out. As of Sunday night, the Dodgers were sitting at 52-30 with a 7.5 game lead in the NL West and a 9 game lead in the NL Wild Card. When I previously calculated Leverage Index, I assumed each team had a 50% chance to win each game and that each team had a 50% chance to win each playoff series. Here I used a more refined method, using Baseball Prospectus' PECOTA-adjusted winning percentages as the probability of victory and playing out the play-off series according to baseball's home field advantage rules.

Using this methodology and simulating 10 million seasons, I calculated that the Dodgers have a 39.0923% chance of advancing to the World Series. After a regular season win, this chance increased to 39.2781%, while after a regular season loss, this chance decreased to 38.8368%. On average, this resulted in a change of 0.215% in probability to advance to the World Series and hence, a 0.108% change in probability of winning the World Series (0.215% times 50%). Thus, a regular season game is worth about 0.108% in championship probability to the Dodgers (only about 39% as important as an average game). An All-Star win, however, will increase their chances of winning the World Series from 19.5462% (39.0923*.5) to 20.0387% (39.0923*.5126), a difference of 0.4925%. Comparing this mark to 0.108% for a regular season game and we find that the All-Star game is not only worth more than a regular season game to Los Angeles, it is worth vastly more than than a regular season game. In fact, the All-Star game is worth somewhere on the order of 4 to 5 times more than a current regular season Dodgers game.

A fuss was made over Chad Billingsley bowling over the catcher during a Dodgers-Padres game this past weekend. Would Billingsley have done so at the All-Star game? Likely not. But this analysis shows that if there is one game the rest of the season in which LA players should sacrifice life and limb, it is Tuesday night's "exhibition" contest.

The following chart shows a few other teams, and the All-Star game vs. regular season game impact on the team's chances of winning the World Series. The Championship Leverage Index for each was computed relative to the baseline of the average game (0.28%).


As you can see, only the Dodger players have more incentive in the All-Star game than in their next regular season game. With an All-Star game Championship Leverage Index of 1.76, the All-Star game is not only more important than their next regular season game, but is likely the most important game of the entire season. Regular season Champ LI's don't usually get that high until at around mid-season, and considering the way that the Dodgers have run away with the NL West, they likely have not had a game this important during the entire year.

However, the game also means a great deal to Boston and other AL East contenders who expect to be playing October baseball. For these teams the game is not merely an exhibition, but a game nearly as important as any other on the schedule. For the Red Sox, the All-Star game is about 66% more important than the average game, although not as important as the key games they are playing now. Still, for Francona's Sox, the All-Star game is about 85% as important as their next regular season game. For the other teams listed, the All-Star game is not nearly as important as their next regular season game, drifting to nearly meaningless for fringe teams like the Astros.

For baseball fans, the All-Star game is must-see TV because it's the one chance to see baseball's stars compete against each other. For fans of the Dodgers, it's must-see TV because it's by far the most important game LA will play until October.


Excellent work, Sky. Insightful and timely. Thanks.

Very cool article. I passed this link on to some friends.

Really cool question, Sky, and great work getting the answer.

awesome work

Interesting article. I came up with almost the same advantage for the home team but just slightly different, at 51.52%

Thanks Cyril,
I went back and calculated HFA for a 7-game series exactly (the first time I used Monte Carlo method). I get 51.2548%. I can't find a flaw in your tables, but I do notice one thing - you have a 6 and 7 game series at equal probability. This would be true if each team had a 50% chance to win each game, but with a HFA, the series should be slightly more likely to go 7. I suspect the discrepancy may lie there somewhere.

I recently did some analysis on this which I haven't posted yet (I was unaware of Cy's piece from 2006 and Sky's post hadn't been published yet, so in the end I guess I wasted my time). Anyway, I get the exact same result (51.2548%) as Sky.

Cy, I believe that the discrepancy arises from the second last line of your 6-game series probabilities. Your list of sequences lists:

W, W, W, W, L, W

Of course this would actually be a 4-game sweep. I believe that what you meant was:

W, L, W, W, L, W

which in probabilities would be:

.54, .46, .46, .46, .54, .54

for a probability of .015 versus the .018 you list, which would reduce the probability of a six-game series to 16.6%, and the probability of the team w/ HFA winning the series to .512, which is line with Sky's result except for rounding error.

Nice catch Patriot. Great minds think alike!