Boston Red Sox

Xander Bogaerts back on pace to reach 200 hits, win AL batting title

tomisphere1 Comment

Back on Wednesday morning, I showed that Xander Bogaerts and Miguel Cabrera were hitting at paces that would cause Bogaerts to (most likely) surpass Cabrera for the AL batting title. Though I didn’t mention it at the time, these projections also showed that he’d reach 200 hits even if he sat out a couple of games, and a few more than that if he played all the remaining games. After a pair of low-hit games knocked Bogaerts off that pace, his 3-for-4 performance last night has put him right back on it.

In trying to project future totals using “the pace at which a player is producing right now”, how many games do you use to determine what that pace is? The last 5? The last 10? 20?

I circumvent that question by using all of them … I calculate his pace of production over his last 5, 6, 7, 8, etc. games, then use that pace applied over the remaining number of games to be played to see what final numbers he’s headed for. This gives a big collection of possible final numbers; you then choose one in the middle.

On Wednesday I did that for Cabrera and Bogaerts using their paces of production as established by their last 8, 9, 10, etc. up to their last 20 games. That gave 13 paces of production for each player. I then applied these to their remaining games assuming they’d not sit out any games, and then again assuming they’d each sit out two games. I got these results:

If playing all remaining games
Bogaerts Cabrera
Low 0.327 0.324
Median 0.329 0.326
High 0.332 0.331
If sitting out two games
Bogaerts Cabrera
Low 0.327 0.326
Median 0.329 0.328
High 0.331 0.332

In all but one of these 26 projections, Bogaerts would end up with at least 200 hits.

I just updated these numbers, and now they look like this:

If playing all remaining games
Bogaerts Cabrera
Low 0.327 0.325
Median 0.329 0.326
High 0.330 0.332
If sitting out two games
Bogaerts Cabrera
Low 0.327 0.327
Median 0.328 0.328
High 0.329 0.332

Here are Bogaerts’ projected numbers of hits:

Bogaerts projected 2015 hits
# of recent games used If playing all games If sitting two games
20 204.0 200.8
19 203.3 200.2
18 203.0 200.0
17 203.3 200.2
16 203.6 200.5
15 204.0 200.8
14 205.1 201.7
13 204.9 201.5
12 203.8 200.7
11 204.4 201.1
10 204.0 200.8
9 204.7 201.3
8 204.3 201.0

Longer term projections (based on his last 40 or more games) almost all have him finishing with 200 hits exactly if he sits out 2 games, 203 hits if he plays all remaining games, and a .327 average.

If they play it out, and stay on pace, Bogaerts probably will win the batting title and will get to 200 hits.

Thanks to Baseball-Reference.com for the gamelog data I used for this article.

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Predictions for all the division series

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So far of the 3 predictions I’ve made this October that have been tested, 2 ended up being correct:

  1. Rays beat Rangers on the strength of David Price’s performance: correct.
  2. Pirates beat Reds because it’s just the right ending: correct (the Pirates fans pretty much willed them to win).
  3. Indians defeat Rays: incorrect.

So, not bad so far.  I am emboldened to make some division series predictions now!

I’ve already called the Red Sox and A’s as winners.  Let’s add the Pirates and the Dodgers to the mix.  But let’s also get a little more specific.

Red Sox’s “rust” from not having played live baseball since Sunday could cost them game 1 against the Rays, despite their efforts to create some game-like intensity for Wednesday’s scrimmage, including letting fans come watch, a move I have publicly encouraged.  We’ve seen the effects of this many times before; perhaps none so clear as in the 2004 ALCS (also notable in my memory is the 2007 World Series).  So I won’t call game 1 either way, despite the Red Sox having home field and having their pitching lined up the way they like.  I’ll just say that neither team scores more than 5 runs in the first 9.  I will predict that the Red Sox will take every game starting with Game 2.

Rust won’t be a factor for A’s and Tigers who’ve had equal amounts of rest.  It’ll be a good matchup, so A’s in 5 games.  I won’t call specific games except as implied by the series going 5 games … so basically games 1-3 will be split, game 4 will be taken by whoever trails in the series, and game 5 will be taken by the A’s.

The Pirates will have a better chance against St. Louis than some may think, and I don’t think they can lose at home in this series with the best “10th Man” going in their very enthusiastic fans.  I think they can take 1 of 3 in St. Louis, so it’s just a question of which one.  I’ll play the rust card here again (hmm, but “rust” and “cardinal” are shades of red … interesting) and say Cardinals take game 2, and Pirates take games 1, 3, and 4.

The Dodgers and Braves: the Dodgers’ injuries make them vulnerable, but their 1-2 punch of Kershaw and Greinke makes them favorites.  Starting pitching is huge in the playoffs, and these two ought to be able to handle the Braves’ lineup.  In this series, the road team may be the victor each time.  I’ll go with that bold prediction: the road team wins each game.  Dodgers in 5.

So, if I count correctly, that’s 14 or 15 distinct predictions, depending on whether the Rays win game 1 against the Red Sox (15) or the Red Sox win (14).  We shall see how it goes!

Red Sox chances of having home-field advantage about 7 out of 11

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The Red Sox will lose any tiebreaker against the Oakland A’s this season, should their records end up tied.  As each team has only 4 games remaining, and the Red Sox have just a 1 game lead over the A’s, to end up ahead of the A’s, the Red Sox must at least match the A’s win for win.

There are 256 possible outcomes of the Red Sox and A’s remaining 8 games (16 outcomes for the Red Sox’s 4 games times 16 outcomes for the A’s 4 games).  Outcomes here means, for example, win-loss-win-win, or win-win-win-loss (order matters).

When you examine all possible combinations of these outcomes, and even factor in their likelihoods of occurring, and keep in mind the statements made above, it turns out that the Red Sox have about a 7 out of 11 chance of securing home-field advantage over the A’s.  When you factor in the remote possibility of the Red Sox beating out Oakland for home-field advantage, but not beating out Detroit, the odds are slightly lower, but still about 7 out of 11.

Now let’s break down some of the above statements to see what’s behind them.

First, why is it that the Red Sox will lose any tiebreaker against the Oakland A’s?

The first tiebreaker is record in head-to-head games between the teams.  With each team winning 3 of the 6 games they’ve played against each other, that tiebreaker has no effect.

The next tiebreaker is intradivision record (record against the other teams within their own division).  Each team will finish with 76 intradivision games, and currently each team has 30 intradivision losses.  So this tiebreaker will go to the team that gets fewer intradivision losses the rest of the way.

This may seem hard to predict, but we can use two facts to our advantage here: 1) To end up tied, the Red Sox must lose exactly one more game the rest of the way than do the A’s. 2) There is only one non-intradivision game left for either team, and that is tonight’s Red Sox-Rockies game.

If the Red Sox win tonight’s game with the Rockies but end up tied with the A’s, it will be because they lost one more of their other games, all intradivision games, than did the A’s.  Since both teams currently have the same number of intradivision losses, that will give the Red Sox one more intradivision loss than the A’s, and the worse intradivision record, and so they lose this tiebreaker to the A’s.

If on the other hand the Red Sox lose tonight’s game with the Rockies but end up tied with the A’s, it will be because they lost the same number of their other games, all intradivision games, than did the A’s.  Both teams end up with the same intradivision record in this case, and so we move on to the next tiebreaker.

The next tiebreaker is higher winning percentage in the last half of intraleague games – games against other teams in the American League.  Currently, the Red Sox are 40-28 in these games, and the A’s are 40-27.  The Rockies game is again the only one of the remaining games that doesn’t contribute to this tiebreaker, which means, since we are considering the case in which the Red Sox and A’s lose the same number of the other games, all of which are intraleague games, the Red Sox would end up with one more loss in the same number of games for this tiebreaker, thus having the worse record, and losing the tiebreaker.

All of the above considered, the Red Sox will lose any tie with the A’s for best record, and therefore must maintain or grow their current one-game lead over the A’s to get home-field advantage.

Next we consider the question of why this means the Red Sox have a 7 out of 11 chance of getting home field advantage.

Of the 256 possible outcomes of the 8 remaining Red Sox and A’s games, in 163 of them the Red Sox at least match the A’s win for win, if not surpass them.  This represents 63.7% of the 256 possible outcomes.  If we assume all outcomes have an equal possibility of occurring, that means the Red Sox have a 63.7% chance of ending up with home field advantage over the A’s, or about 7 out of 11 (which is 63.64%).  But saying all outcomes have an equal possibility of occurring is to assume that both teams have a 50-50 chance of winning each of their remaining games, and that’s probably not the case.  If instead we assume a 60% chance for each team winning each remaining game (which essentially matches their winning percentages on the season), we can redo the calculation, weighting less likely outcomes (like loss-loss-loss-loss) lower than more likely outcomes (like win-win-win-win).  When you do this, the Red Sox’s odds turn out to be just a little bit better, 64.0%, to end up with home field advantage over the A’s.  Still pretty much 7 out of 11.

To get their chances of ending up with home field advantage, period, we have to subtract the likelihood of the Tigers tying or surpassing the Red Sox, while the A’s do not.  These odds, right now, are very small.  Assuming 50-50 chances in the games, the Red Sox have a 63.5% chance of ending up with home field advantage, and assuming 60-40 chances in the remaining games for each division leader, the Red Sox have a 63.9% chance.

No matter how you slice it, it’s pretty much 7 out of 11.

Of course you could argue that due to schedules, the odds are better now for one team or the other.  But I think the schedules are not too slanted for one team or the other right now, so I’m sticking with 7 out of 11.

I hope somebody out there enjoys reading this even half as much as I enjoyed producing it.

My sources for the data and tiebreaker information in this post were:

http://www.overthemonster.com/2013/9/23/4761626/red-sox-athletics-playoff-tiebreaker-home-field-advantage http://en.wikipedia.org/wiki/Major_League_Baseball_tie-breaking_procedures