Would Pete Alonso improve Red Sox first base situation all that much?

Pete Alonso is one of the best bats in baseball, no question about that. But to get his bat at first base, you have to take with it his awful fielding and awful baserunning. That waters down his value to some extent. But by how much? And how does he compare to players currently on the Red Sox, and other available options?

I’ll present the data here and some other observations, so that you can compare. I’ll finish by talking about whether it makes sense for the Red Sox to add a player or stick with who they have. All the data shown here is from BaseballSavant.

The players to be compared

I picked the 3 first base free agents who were considered the best on the market when this offseason began, and put their stats on the top row of the comparisons below. (I started putting these together weeks ago when Josh Naylor was still a free agent.) On the bottom row I put a top first base trade possibility in Yandy Diaz, and the two top major league first base options on the Red Sox currently, Romy Gonzalez and Triston Casas.

For 5 of the players, I show their 2025 numbers. For Triston Casas, who didn’t play enough in 2024 and 2025 to give us a good idea of what he is, I show his 2023 numbers.

Expected stats

Let’s start with the expected stats. This is where they look at the velocity, launch angle and trajectory of every ball a player put in play, and tally up the probable results based on those numbers.

Focusing primarily on xwOBA, we see that all six players did well, although when you look at the actual values instead of the percentiles, Alonso is clearly separated from the pack, with only Triston Casas giving him a challenge there.

Quality of contact

Now we’ll look at quality of contact.

Alonso had the best overall contact, however Romy Gonzalez had more hard-hit balls. In fact he had the 5th highest Hard Hit% in baseball for players with over 300 PA. (Who was ahead of him? 1. Roman Anthony 2. Kyle Schwarber 3. Shohei Ohtani 4. Aaron Judge.) Yandy Diaz also hit it hard frequently.

But neither Gonzalez nor Diaz get an ideal launch angle (“LA Sweet Spot %”) as much as Alonso does. For both of them it turns out it’s because they hit too many ground balls – Diaz especially. This is likely the reason Diaz’s results aren’t as good as Alonso’s, and for Gonzalez, one of two reasons (we’ll see the other in the next section).

While O’Hearn and Naylor are limited by lower bat speed, O’Hearn improves his results by often having a good launch angle, and Naylor gets a better exit velocity by hitting on the sweet spot of the bat a lot.

Triston Casas’ 2023 comes the closest to Alonso’s 2025 among those pictured here. The differences may only be due to looking at a rookie season versus a veteran in his prime having his best season yet.

Non-contact stats

So that’s what happens when they swing and make contact. What about the numbers when they don’t make contact? Who chases pitches out of the zone too much (Chase %)? Who misses a lot when he swings (Whiff %)? Who walks too little or strikes out too much?

Alonso and O’Hearn are average in these categories. Yandy Diaz is above average, and Romy Gonzalez is well below average. Triston Casas has a great eye, but still manages to swing and miss at an above average pace. Josh Naylor doesn’t chase and doesn’t strike out, but walks an average amount.

Here we have what looks like the other reason Romy Gonzalez doesn’t get better results despite hitting the ball so hard. He chases too much. And while fixing that doesn’t necessarily fix his higher strikeout rate and low walk rate, it ought to at least help.

Fielding and Baserunning

What’s left? Fielding and baserunning.

Here again we see strong similarity between Triston Casas and Pete Alonso. They’re both terribly slow, and awful at both fielding and baserunning. But being slow isn’t the excuse for the rest, because look at Josh Naylor, who is even slower, but manages to be an average baserunner and a decent fielder.

When it comes to baserunning, Romy Gonzalez is the opposite of Josh Naylor. He’s the only one in this group that could be called “fast”, yet he’s still a poor baserunner. Maybe he should get a pointer or two from his teammate Trevor Story, who runs just as fast as Gonzalez but was one of the top baserunners in the game last year. Or maybe we should give him a little credit for being an average or above average baserunner in the past.

As for good fielders, it looks like Ryan O’Hearn is the only one, with Naylor and Gonzalez a little below average. But Gonzalez split his time between first and second base (and some other spots), and when you break his fielding down by position, both in his career and in 2025, he’s been an above-average fielding first baseman, and a below-average fielder everywhere else.

Categorizing these players

So to sum up, I see two basic types of player here.

Pete Alonso and Triston Casas are the power hitters who can get on base, too, but are awful at fielding and baserunning. Yandy Diaz is, too, but with a little less power and a little better baserunning.

In the other category are Josh Naylor and Ryan O’Hearn, who have some power, but not a lot, but still manage to have above-average impact as hitters. And at everything else, they’re average, on the whole.

The 2025 version of Romy Gonzalez belongs in the O’Hearn/Naylor camp, as a well-rounded player with an above-average bat. But he has the raw tools to become much better. He’s got enough speed to become a great baserunner. He’s one of the best in the game at hitting the ball hard, but he hits it on the ground too much, and he misses it too much. And here’s the thing: the parts of his game that are lacking and that are holding him back, are all things he can learn to be better at. He can learn to be a smarter baserunner. He can learn plate discipline. He can learn to hit the ball just a little lower than he does now, to get it into the air more.

The question is, will he?

If he does, he creates a new category, combining the best of O’Hearn/Naylor with the best of Alonso/Casas, and he’d be better than all of them.

Who’s on first?

So what should the Red Sox do? If they can get Pete Alonso in to play first base for them, should they?

Alonso would certainly help the lineup. But if Triston Casas has a healthy year, he’s basically a Pete Alonso clone for much less money.

What if Casas is injured again, though? He sure seems injury prone. Then your backup plan is Romy Gonzalez, who is as good as your second-or-third best first base free agents that were on the market at the start of this offseason. And with the right coaching and effort, could end up being better than all of them in the short term.

So regardless of whether Casas can or can’t play, the Red Sox will have a plus option at first base. They don’t need Alonso to play first base for them.

But Alonso would improve them at DH. But to make that room, they’d need to trade/drop Masataka Yoshida, to whom they owe $36M over the next two years, and probably one of their 4 top-notch outfielders. Not to say they won’t; they may. But they may not.

In the end, Alonso may not add as much value as people think he will, when compared to what the Red Sox would get from the current players who he would replace. All that may not be worth the expected $150M price tag.

My Red Sox posts on OverTheMonster

It’s become very hard to find a particular author’s FanPosts on OverTheMonster. So I’ve made this page so that I can find mine.

But before we get to mine, have a look at this post by another FanPoster that aged really well:

Mar 28, 2018 How the Red Sox will Achieve World Domination in 2018

posted over 7 years ago by Ricochet!

Here are links to every FanPost I made from 2017 through 2019. I hope to add the earlier ones in soon.

Date Article
Jul 28, 2019 This Fangraphs writer clearly doesn’t get how deep the Red Sox lineup is
Jun 22, 2019 Is JBJ about to turn into a pumpkin?
May 31, 2019 OPS contest thirdway checkin
Apr 6, 2019 We can throw away the first 10 Red Sox games
Mar 30, 2019 It’s not just that the Red Sox starters are still in spring training
Mar 15, 2019 Contest: Who will be the most improved Red Sox hitters this year?
Nov 2, 2018 Poll: which free agents should the Red Sox bring back?
Sep 14, 2018 Last night the Red Sox clinched home field advantage in the World Series
Aug 20, 2018 Are the Red Sox becoming victims of their own success?
Jul 24, 2018 The Red Sox do not need another reliever
Jul 6, 2018 Red Sox first half surprises and busts (OPS contest results)
Jul 5, 2018 A 10-day opportunity to build a lead over the Yankees
Apr 20, 2018 Absurd projections and feasible dreams after one ninth of a season
Apr 3, 2018 A quick look at how the first trip through the rotation went
Mar 24, 2018 Contest – predict the Red Sox’ first half OPS’s
Dec 27, 2017 Which Mitch? What an injury-free Moreland might produce in 2018
Oct 23, 2017 Get Big Papi into the Red Sox dugout as a coach
Oct 3, 2017 Fanpost Friday playoff predictions of tomisphere
Sep 15, 2017 The Red Sox could play 9 games in a row against the Houston Astros
Jul 28, 2017 Final-stretch thoughts on Rusney, Dustin, Mookie, Brock, Eduardo, Deven, Rafael, and everybody
Jul 14, 2017 Poll: How do we feel now about the Red Sox catchers?
Jul 4, 2017 Deven Marrero just had a fantastic week at the plate
Jul 3, 2017 Contest results: predict Red Sox first half OPS’s
Jun 15, 2017 Poll: Is it time to advance Devers & Chavis?
Jun 8, 2017 How I became a fan of the Boston Red Sox
May 24, 2017 Poll: Will the next Red Sox call up be … Ryan Court?
May 13, 2017 Cap extra innings at three, then declare a tie
May 8, 2017 Judging Surprising Aarons
May 1, 2017 Fix the Red Sox offense – mostly, by waiting
Apr 18, 2017 The coming year will determine whether to trade Leon or Swihart
Apr 4, 2017 2017 Red Sox Season in Review: What a Year It Was
Mar 26, 2017 Contest: predict Red Sox first half OPS’s
Mar 20, 2017 FF Favorite: Javier Baez, For Love of the Game
Mar 15, 2017 Poll: Who will be the Red Sox main catcher after the All-Star break?
Mar 4, 2017 What you think of the World Baseball Classic depends a lot on who you are
Feb 25, 2017 My rule change would hopefully reduce the chance of injury
   

The Texas Rangers have about a 1 in 4000 chance of making the playoffs

If you look at the playoff odds on FanGraphs.com right now, you’ll see the Texas Rangers listed as having a 0.0% chance of making the playoffs this year. But that doesn’t mean they have no chance. It just means their chance is so small that it doesn’t round up to 0.1%; instead it rounds down to 0.0%, as any chance less than 1 in 2000 will do. As it turns out, their chance of making the playoffs is about 1 in 4000 right now.

How we get to that number involves a lot of logical reasoning, complicated by the fact that the Rangers will play a series against one of the four teams they’re chasing, and there will be two series played this week between some of those same four teams.

Let’s set the stage properly. Here are the 8 remaining playoff contenders in the American League:

Only 6 teams in the American League may go to the playoffs. To be one of those 6, the Rangers must pass 2 of the 7 teams ahead of them in the standings (so long as one of them is not a division winner). Fortunately for the Rangers, there are 4 teams they still have a chance to reach. Unfortunately, they’ll be very difficult to reach.

Notice that if the Rangers win all 6 of their remaining games, and the Red Sox lose all 6 of theirs, that the Rangers would only manage to be tied with the Red Sox. But because they hold the tiebreaker over the Red Sox (having won 4 of the 7 games played between them this year), the Rangers would beat out the Red Sox in that case.

The same goes for Detroit. The Tigers must lose all 6 of theirs, and the Rangers must win all 6 of theirs, for the Rangers to tie; because they win the tiebreaker (having won 4 of 6 against the Tigers), the Rangers would beat out the Tigers.

The Rangers did not win their season series against the Astros, however, so must beat them by a game in the final standings, to pass them for a playoff spot. Because they are currently 5 games behind them, that could only happen if the Rangers win all 6 of their remaining games, and the Astros lose all 6 of theirs.

For the Rangers to catch the Guardians, they’ll have to win some of their remaining 3 games against them; those wins would give the tiebreaker to the Rangers. So the Rangers could stand to lose 1 game, or could stand the Guardians winning 1 game, and still beat them for a playoff spot.

Given that there’s only 1 team that isn’t forcing the Rangers to win all their remaining games, but that they need to beat at least 2 of these teams, the only option for the Rangers is to win all their remaining games.

Let’s start a list of requirements like this one:

We’re assuming here that every game a team plays the rest of the way has a 1/2 chance of being a win, and a 1/2 chance of being a loss. Because the Rangers have 6 games remaining, and there’s only 1 way to achieve the stated outcome (Rangers win all 6), that’s 1 outcome out of 26 possible outcomes, or a 1/64 chance of it happening.

What other outcomes must we consider?

Well if none of these teams were playing each other in these final 6 games, it would be a little less complicated. All the outcomes would be independent, so we could calculate the odds of each team’s win totals independently, as a starting point. Our list of requirements would look like this:

Because the Rangers would have to beat at least 2 of these teams, we’d take pairs of outcomes and calculate their odds:

[ (Red Sox lose all) AND (Tigers lose all) ]
OR [ (Red Sox lose all) AND (Astros lose all) ]
OR [ (Red Sox lose all) AND (Guardians lose 5 or 6) ]
OR [ (Tigers lose all) AND (Astros lose all) ]
OR [ (Astros lose all) AND (Guardians lose 5 or 6) ]

Notice that we didn’t include (Tigers lose all) AND (Guardians lose 5 or 6). That’s because one of those teams will win the central division; beating a division winner doesn’t help you win a wild card spot. They have to beat at least one of the Red Sox or Astros to get into the playoffs.

So we would multiply odds everywhere there’s an AND above, and then add them everywhere there is an OR above.

This would double-count or triple-count some cases though. For example, it would triple count the case where all three of these occur: (Red Sox lose all) AND (Tigers lose all) AND (Astros lose all). We’d have to subtract out double the odds of that happening.

After making a few more adjustments where 3 of those occur, we’d probably have one final adjustment to make for the case where all 4 occur:

(Red Sox lose all) AND (Tigers lose all) AND (Astros lose all) AND (Guardians lose 5 or 6).

Then we’d multiply our result by the odds of the Rangers winning all their games, because that has to happen in every case of the Rangers making the playoffs.

But we don’t live in that world. We live in a world where, in the final games of the season:

The Tigers play 3 games against the Red Sox
The Tigers play 3 games against the Guardians
The Rangers play 3 games against the Guardians

Oh my. This reduces the number of games that determine the Rangers’ fate from 30 down to 21. That’s good for the Rangers, because it means a lot fewer games would have to go a certain way for them to make the playoffs, and that gives them better odds.

It also changes how we do this. Now the outcomes we need to consider look like this:

I’ve used highlighting to show outcomes that are related to each other in that they cannot both happen. For example, looking at the two lines in gold, we see that the Red Sox cannot simultaneously lose all their remaining games while also winning all 3 against the Tigers.

Let’s consider those two middle lines right now. If the Tigers lose all their remaining games, that means both the Red Sox and Guardians win at least 3 games. And that means the Rangers can’t beat either of those teams. The only team left that they could beat is the Astros. So if the Rangers beat the Tigers, they must also beat the Astros (and only the Astros) to get into the playoffs. That gives us this:

(Tigers lose all) AND (Astros lose all)

Which is actually this:

(Red Sox win all 3 against the Tigers) AND (Guardians win all 3 against the Tigers) AND (Astros lose all)

And there is no chance of double-counting with other outcomes. This will very much simplify our work to compensate for double countings.

To this we add the following:

[ (Red Sox lose all) AND (Astros lose all) ]
OR [ (Red Sox lose all) AND (Guardians lose 5 or 6) ]
OR [ (Astros lose all) AND (Guardians lose 5 or 6) ]

But consider that in the end we’ll be multiplying everything by the odds of (Rangers win all), which must happen in every scenario. Because the Rangers play 3 of those games against the Guardians, that means three of the Guardian’s losses have already been accounted for by the (Rangers win all) outcome. So we only need to consider the additional chance that the Guardians lose 2 or 3 against the Tigers. So the above becomes:

[ (Red Sox lose all) AND (Astros lose all) ]
OR [ (Red Sox lose all) AND (Guardians lose 2 or 3 to Tigers) ]
OR [ (Astros lose all) AND (Guardians lose 2 or 3 to Tigers) ]

Notice that in all 3 of these scenarios, the Tigers become unreachable to the Rangers, because they win at least 2 games. The only double or triple counting in this trio of scenarios is where the Rangers beat everyone but the Tigers:

(Red Sox lose all) AND (Astros lose all) AND (Guardians lose 2 or 3 to Tigers)

That’s a triple-count, so we have to subtract double the odds of that happening.

We can put all this together, with odds, in a new chart:

We add the first four lines then subtract 2 times the last line to compensate for double counting:

2-12 + 2-12 + 2-7 + 2-7 – 2 x 2-13 = 2-6 + 2-12 = 65/4096.

Which we multiply by the odds of the Rangers winning all 6 of their remaining games, to give 65/262144. That’s about 1 in 4033, or 0.0248%.

Had it not been for teams playing each other, the odds would have been 1 in about 16,186. So the Ranger’s chances of making the playoffs are about 4 times better because of these teams playing against each other.

The Toronto Blue Jays have about a 99.99987% chance of making the playoffs

There is a very very narrow range of circumstances under which the Toronto Blue Jays do not make the playoffs. So narrow, in fact, that if we assume every game remaining in the MLB this year has a 50% chance of being won by either team, the odds of the Blue Jays failing to make the playoffs are 1 about 793,072. That equates to a 99.999874% chance that they make the playoffs.

So how do we work out such numbers? Buckle up for a logic roller coaster ride.

To fail to get into the playoffs, every one of the Yankees, Red Sox, Mariners, Astros, Tigers, and Gaurdians would have to beat the Blue Jays, and these are the only 6 teams capable of surpassing the Blue Jays at this point.

At this point, the Blue Jays can end up with at most 73 losses, if they lose all 7 of their remaining games. So surpassing them would especially require a lot of winning by the Gaurdians and Astros (with 71 losses each currently) and the Red Sox and Tigers (with 70 losses each).

But these teams are limited in how much winning they can do the rest of the way, because there will be 6 games played between them. The Tigers and the Gaurdians will play 3 games against each other, and the Tigers and the Red Sox will play 3 against each other. That means there will be at least 6 losses spread around among those 3 teams.

So let’s consider the possible outcomes of the Tigers/Gaurdians series. For each outcome, let’s assume the Blue Jays lose all 7 of their remaining games, ending with a record of 89-73. Let’s also assume both the Tigers and the Gaurdians win all 4 of their other remaining games.

Except that we can’t assume that. Because if the Tigers win all their other 4 games, that means they deliver 3 losses to the Red Sox, who end up at best 89-73, the same as the Blue Jays. Because the Blue Jays end up with 7 wins and 6 losses against the Red Sox, they win the tiebreaker with the Red Sox and are in the playoffs. So the Tigers must lose a game to the Red Sox, and the Red Sox must win their other 4 games not against the Tigers, for the Blue Jays to have a chance at elimination here.

So we’ll assume the Tigers lose 1 more game (versus the Red Sox) outside of the Tigers/Gaurdians series, and the Guardians don’ t lose any others.

If the Gaurdians sweep the Tigers, the Tigers end up 88-74, a game behind the Blue Jays, and the Blue Jays are in the playoffs.

If the Tigers sweep the Guardians, the Gaurdians end up 88-74, a game behind the Blue Jays, and the Blue Jays are in the playoffs.

If the Gaurdians win 2 of 3, the Tigers end up 89-73, tied with the Blue Jays. Because the Blue Jays had 4 wins and 3 losses in their games against the Tigers this year, they win the tiebreaker between the teams, and are in the playoffs.

That leaves the scenario where the Tigers win 2 of 3. Then the Tigers end up 90-72, ahead of the Blue Jays, while the Guardians tie the Blue Jays at 89-73. So as a tiebreaker we look and see that the Blue Jays and Gaurdians each won 3 games against each other this year. We have to use the second tiebreaker, which is records within their own divisions. The Gaurdians end up 36 and 16 against their weaker division; the Blue Jays 25 and 27 against their stronger division. The Gaurdians therefore win this tiebreaker, and the Blue Jays are out of the playoffs – if the other 3 teams surpass them too, that is.

That’s the only scenario in which the Blue Jays are eliminated.

What if The Tigers lose one more game against another opponent? Then they end up with the same record as the Blue Jays, and the Blue Jays are in because they win the tiebreaker with the Tigers. So the Tigers must only lose the one game against the Red Sox.

That covers what must happen with the Tigers, Gaurdians, and Red Sox. What of the Yankees, Mariners, and Astros?

The Blue Jays hold the tiebreaker over the Yankees, so the Yankees must get to at least 90 wins, and therefore must win at least 3 of their last 7 games.

The Blue Jays hold the tiebreaker over the Mariners, so the Mariners must win at least 4 of their last 7 games.

The Astros hold the tiebreaker over the Blue Jays, so they must win at least 5 of their last 7 games.

So now we must get the odds of all these things happening and multiply them together to get the odds that the Blue Jays miss the playoffs. We assume in every game that the teams have an equal chance of winning. The following table contains all the odds:

The reason the Mariners and Astros are lumped together in the last line is that they play one more game against each other, so their odds of reaching their respective win totals are linked because of that game.

When you multiply all these odds together you get 693,198 divided by 2 to the 39th power, which is about 1 in 793,072, or 0.000126%. That’s the odds that they don’t make the playoffs; so the odds that they make the playoffs are about 99.999874%.

Garrett Crochet pitched the best start of the year but didn’t get the win. We can fix that.

The merit method of awarding wins fixes all the injustices that the current method of awarding wins punishes starting pitchers with. The latest of these injustices happened to Garrett Crochet, ace of the Boston Red Sox, last night. He threw 8 scoreless, 3-hit innings before giving up a game-tying solo home run to baseball’s best hitter, Aaron Judge, with one out in the top of the ninth. With only one run of support from his teammates, that tie score made Crochet ineligible to earn a win, as he was removed from the game at that point, and the current rules say the win goes to the pitcher who was the active pitcher when the winning team took its final lead. So the win went to Garrett Whitlock, who retired the two batters he faced in the top of the 10th inning. Whitlock performed well. But who did more to earn the win? The guy who faced 2 batters over 1 inning of work, giving up no runs, or the guy who faced 30 batters over 8 and 1/3 innings, giving up 1 run to a top offense and the Red Sox’ arch rivals?

If you agree with me that Crochet did more to earn that win, then you’ll like the merit method for awarding wins.. I explain the method in my post The how and the why of awarding wins to pitchers by the merit method.

Using that method, we take the 2 runs that the Red Sox scored in Friday’s game and divide by 9 and 2/3, which is the number of innings the Red Sox were at bat. This gives us the average number of runs the Red Sox scored in each inning, the fraction 6/29. Then we simply credit each Red Sox pitcher with this number of runs for every inning they pitched. And then we subtract from this the number of runs they gave up. This gives each pitcher a number of “Runs Ahead”. Then we give the win to the pitcher with the greatest number of Runs Ahead.

The cool thing about this method is that adding the Runs Ahead of all the winning team’s pitchers always gives you a positive number, and adding all the Runs Ahead of the losing team’s pitchers always gives you a negative number. This method also assigns the losing pitcher as the one with the most negative Runs Ahead value.

Here’s a table showing the numbers discussed above for the Red Sox pitchers in last night’s game. In the first three columns in the table below we see IP, RCr/IP, and RCr, which stand for Innings Pitched, Runs Credited per inning pitched, and Runs Credited, respectively. You get the third column (Runs Credited) by multiplying together the first two.

PitcherIPRCr/IPRCrRRA
Garrett Crochet8 ⅓6/2950/29121/29
Aroldis Chapman6/294/2904/29
Garrett Whitlock16/296/2906/29
Runs Ahead (RA) calculations for Red Sox pitchers in victory over New York Yankees, June 13, 2025

Then you subtract runs allowed (R) from this to get each pitcher’s number of Runs Ahead (RA) for that game. Because Garrett Crochet had the highest number of Runs Ahead for the winning team, he would be awarded the win by the merit method. But by current rules, the win went to the other Garrett (Whitlock).

I hope someday to convince MLB league officials to change to the merit method for awarding wins. It fixes so many things that are just not right about the current method.

Who was the “most .500” team of 2024?

Of course, one could say that the “most .500” Major League Baseball team in any given year is the one whose record is closest to .500, or 81-81 in a full season. But even a hypothetical team that always had exactly a 50% chance of winning would sometimes end up, by luck of the “coin flip”, a few games away from .500.

And what about a team that’s great for the first half of the season, then awful for the second half, ending up with a .500 record? They weren’t really a .500 team at any point in the season, in that their chance of winning games wasn’t actually close to 50% at any point, nor were they winning about half their games in any given week.

So here are a few different ways to measure how .500 a team was, along with the top teams by each method.

Final Record

We can just look at a team’s final record and see how many games away from .500 it was, above or below.

The Boston Red Sox had the only .500 record, but several other teams were close.

Run Differential

This was mentioned in Sarah Langs’ 2020 article What does a true .500 team look like?

The run differential of a team is the runs it scores over the entire season minus the runs it allowed in that same time. A small run differential is a good predictor of a team that will have a record near .500. (There is even a stat called Pythagorean expectation which estimates what record a team should have based on it totals of runs scored and runs allowed.)

Whose run differential was closest to 0 in 2024?

Four teams had a run differential close to 0. Of these, again, the Boston Red Sox were the closest to 0, just barely. It seems we have a frontrunner.

Number of times at .500

A team that plays “a .500 brand of baseball” throughout the season is likely to have a winning percentage of exactly .500 at several times during the course of the season. The most times this could possibly happen is 81, though even for a hypothetical team that always has a 50% chance of winning, the odds of that happening 81 times are over 2,000,000,000,000,000,000,000,000 to 1 against. The most times it’s ever been done, at least before 2020, is 35 by the 1959 Chicago Cubs.

The Tampa Bay Rays came close to that this year, tying for 3rd most. The Padres, Red Sox, and Cardinals also had a lot.

For fun: consecutive times at .500

This last one is more about the luck of streaks than anything else. But there was an interesting streak this year in this regard, so I thought I’d throw it in.

When a team is at .500 in the middle of the season, the next game they play takes them off of .500; it’s only 2 games later that they can be back at .500 again. So a streak of consecutive times at .500 means that at the end of every 2 games played after being at .500, they’re back at .500 again.

The Red Sox were at 26-26 on May 25, 2024 – 26 wins and 26 losses. Two games after that they were 27-27, then 28-28, 29-29, and so on up to 35-35. That’s ten times in a row at .500. The likelihood of that happening, once a team has reached a .500 record, is more than 500-to-1 against.

Here are the longest such streaks in the majors in 2024:

The “Winner”

It’s gotta be the Boston Red Sox as The Most .500 Team of 2024. They top every list except number of times at .500, and they did pretty well there, too. Runner up goes to the Tampa Bay Rays.

Interestingly, these two teams played each other in their last 3 games of the season, with the Rays winning the first two but losing the final game. Had they won it, they would have replaced the Red Sox atop the Final Record list, probably solidified the Red Sox hold on the Run Differential list, but strengthened their own position atop the Times At .500 list. That game was something of a battle for Most .500 Team of 2024. Congratulations, Red Sox, on your “victory”!

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

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.

Predictions for all the division series

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

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