It's the Red Thread !!

[Frayed Knot]

Might as well have one focusing on this one team for the week seeing as how, y'know, we have to be at least one game better between now and Sunday.

Pirates 4 - Reds 2, end of 2 in Cincy

Pitt put up a 4-spot in top-2. The Reds' Elly De La Cruz (EDLC to his friends) went yard with/a man on in the bottom half.

Oddly that's only EDLC's 21st HR on the season but just his second since the end of July.

[Frayed Knot]

Still 4-2, end of 7

[Frayed Knot]

Bottom 9, still 4-2

Spencer Steer facing Dennis Santana -- K

Elly De La Cruz -- Single (4th hit of the night for EDLC)

Matt McLain --

[Frayed Knot]

Bottom 9, still 4-2

Spencer Steer facing Dennis Santana -- K

Elly De La Cruz -- Single (4th hit of the night for EDLC)

Matt McLain -- Game Ending Double Play

[Edgy MD]

Buccos!!

Buccarooonies!!!

[Frayed Knot]

So we are, however temporarily, back in the WC #3 lead (they lost tonight and we haven't).

[Frayed Knot]

Game 2 Wednesday 6:40 -- Paul Skenes vs Hunter Greene

[Edgy MD]

The Bucsters!

Buccarya!!!

[Frayed Knot]

So we started the day having to out-win the Reds by at least a margin of one over a six game span.

We start tomorrow having to out-win the Reds or just play them even over the next five.

And we prevent the DBacks from getting any closer no matter what happens out west tonight.

Small victories, sure ... but I'll take them.

[Centerfield]

Small victories? That was huge!

[Frayed Knot]

Sure it was. But it was HUGE mainly because the opposite result would have been a disaster: gain ground on no one while losing ground to Arizona and being behind TWO teams in the fight for one spot with another day erased off the schedule. So even though the odds of finishing ahead of either pursuer is now in the Mets favor, the fact that they still have to hold off both still makes the odds very tenuous.

I may have time later to do a deep dive into the numbers as it gets complicated when three clubs are involved.

But to short-cut it: there are 32 possible permutations* when playing five games and right now the Mets stay ahead of EITHER Cincy or Arizona in 21 of them or about two out of three. Good, right? Well, yeah, except that the Mets wind up on the outside looking in if just one or the other plays their next five even one game better than them so that likely keeps their odds under the 50/50 mark.

So while our position is better than it was at this time yesterday ... that still doesn't make it good. So, yeah, small victories all around.

* in one of those 32 you go 0-5 and one where you go 5-0

1-4 & 4-1 each have 5 possibilities (win/lose only the first game, only the 2nd ... etc.)

and there are ten different ways each of finishing 3-2 or 2-3 [so 10 + 10 + 5 + 5 + 1 + 1 = 32]

So if you treat each game as a 50/50 proposition, the odds of going

5-0 = 3.12% (1/32)

4-1 = 15.6% (5/32)

3-2 = 31.25% (10/32)

2-3 = 31.25% (10/32)

1-4 = 15.6% (5/32)

0-5 = 3.12% (1/32)

[Frayed Knot]

To ball park it quickly: if there's a 65.6% chance* at staying ahead of Cincy and a 65.6% chance of staying ahead of Arizona, then there's roughly a 43% shot (.656 x .656) at pulling off both. Or about 3-in-7

* Figuring 21 'Good' outcomes out of 32 possibilities while treating each game as a generic 50/50 shot without regards to opponent, starting pitcher, home field, etc.

[Edgy MD]

You are, I think, a little bit overthinking it.

The problem with that calculation is that you are looking at them as outcomes produced independently of one another, which they are not.

[Frayed Knot]

The Mets', Reds' and Dbacks' games aren't independent of each other?

[Centerfield]

Paul Skenes vs. Hunter Greene. A matchup of two aces.

Skenes was roughed up in his last outing against the Cubs. 3 ER in 3.2 IP. Had been pretty solid before that. Pitched 6 scoreless in his last start against the Reds in August.

Hunter Greene was roughed up in Sacramento two starts ago (2.1 IP, 5ER). But has otherwise been lights out in September. 2 ER in 22.1 IP.

Coin flip as far as pitching, but the Reds are at home, and the Pirates are last in runs scored in all of baseball. (that one weekend against the Mets notwithstanding)

[Edgy MD]

The Mets', Reds' and Dbacks' games aren't independent of each other?

 

The Mets half of the equation affects both games.

If the Mets win, it affects both outcomes. If the Mets lose, it affects both outcomes.

It is not an independent variable. The Mets effort to stay ahead of one directly impacts their effort to stay ahead of the other.

I might have a 50% chance of marrying in my lifetime and a 40% chance of having one or more children, but that does not compound to a mere 20% chance of marrying and having children, because each act directly affects the other.

[Benjamin Grimm]

You're so old-fashioned! You don't need to be married to have kids!

I think what FK is doing is he's seeing all of the possible combinations of outcomes of the 15 games remaining among the Mets, Reds, and Diamondbacks. And since none of these teams play each other, each outcome is independent of the others.

[Frayed Knot]

The Mets half of the equation affects both games.

If the Mets win, it affects both outcomes. If the Mets lose, it affects both outcomes.

It is not an independent variable. The Mets effort to stay ahead of one directly impacts their effort to stay ahead of the other.

 

None of this makes a lick of sense to me.

The Mets half of the equation affects both games. -- not the outcome of those games, it doesn't.

If the Mets win, it affects both outcomes. If the Mets lose, it affects both outcomes. -- affects them how?

It is not an independent variable. The Mets effort to stay ahead of one directly impacts their effort to stay ahead of the other. -- ?????????????

 

I think what FK is doing is he's seeing all of the possible combinations of outcomes of the 15 games remaining among the Mets, Reds, and Diamondbacks. And since none of these teams play each other, each outcome is independent of the others.

 

YES!!!!!

[Edgy MD]

A Reds loss is a good outcome. A Diamondbacks loss is a good outcome, but a Mets win is two good outcomes. It affects both the Diamondbacks chances, and the Reds chances. Therefore, these two possibilities have some interdependence.

 

You're so old-fashioned! You don't need to be married to have kids!

 

But having kids does increase the likelihood that a person will marry at some point.

[Benjamin Grimm]

For the 15 games, there are 32,768 possible outcomes. Some of these will have the outcome you mentioned, where there are days when the Mets win and the Diamondbacks and Reds both lose. And there are others where that won't happen. But you can look at each of the 32,768 combinations and determine in how many the Mets win the wild card, and in how many it's the Reds or the Diamondbacks.

[Edgy MD]

I understand.

[MFS62]

I have rooted for both the Reds and the Bucs in past years.

After my team moved from Brooklyn, I was looking for a team to replace them. The MFYS were never even considered.

My high school friend (and later the best man at my wedding)was a Pirates fan, so I started rooting for them, through that wonderful year of 1960.

But then there were stories that the Reds would move to New York to fill the National League void. So I rooted for them in 1961.

I'll only be rooting for the Pirates tonight.

Later

[Edgy MD]

And, oddly enough, losing two players to paternity leave last week has presumably suggests that someone marrying and/or having children has (if only at a miniscule level) affected the likelihood of the Mets staying ahead of the Reds and/or Diamondbacks.

[Frayed Knot]

A Reds loss is a good outcome. A Diamondbacks loss is a good outcome, but a Mets win is two good outcomes. It affects both the Diamondbacks chances, and the Reds chances. Therefore, these two possibilities have some interdependence.

 

Which "two possibilities" are you talking about? On each of the next five days there will be 8 different possible outcomes* which, if you assume each is a 50/50 proposition, are all equally likely to occur. But the outcome of each game has no effect on the outcome of each of the other two games.

You seem to be trying to 'weight' one as being more important than others which it is to the team playing it but that applies also to the other two games as well, each of their wins will be 'two good outcomes' to them too.

So I still haven't the slightest idea what you're trying to say here.

* Eight daily possibilities:

1) NYM Win, Reds Win, DBacks Win

2) Mets Win, Reds Win, DBacks Lose

3) Mets Win, Reds Lose, DBacks Win

4) Mets Lose, Reds Win, DBacks Win

5) Mets Win, Reds Reds Lose, DBacks Lose

6) Mets Lose, Reds Win, DBacks Lose

7) Mets Lose, Reds Lose, DBacks Win

8) Mets Lose, Reds Lose, DBacks Lose

[Edgy MD]

I am sorry you do not have the slightest idea what I am asserting.

I do not know why you have printed the eight daily possibilities, so there!

OE: My "two possibilities," which I thought was clear, refers to the two season outcomes of (1) the Mets finishing ahead of the Diamondbacks, and (2) the Mets finishing ahead of the Reds. These are related, as a single Mets performance affects both outcomes. Even if you disagree, and disagree strongly, it is easy enough to understand that thesis, having been stated several times.

[Benjamin Grimm]

This is what Microsoft Copilot has to say:

🧮 Total Scenarios

Each of the 15 remaining games (5 per team × 3 teams) has 2 outcomes →

2^15 = 32,768  total scenarios

🎯 Goal

Count how many of those scenarios satisfy:

Mets win ≥ Reds wins

Mets win ≥ Diamondbacks wins

🧠 Strategy

We’ll iterate over all possible win totals for each team (0 to 5 wins), and count how many combinations satisfy the condition.

Let’s define:

𝑀

: Mets wins (0–5)

𝑅

: Reds wins (0–5)

𝐷

: Diamondbacks wins (0–5)

Each team’s 5 games have:

(5/𝑘) ways to win 𝑘 games

So total combinations for a given (M, R, D) triple:

(5/𝑀)⋅(5/𝑅)⋅(5/𝐷)

We sum over all triples where

𝑀≥𝑅 and 𝑀≥𝐷

✅ Valid Triples

Let’s compute the total:

plaintext

For M = 0:

R = 0, D = 0 → 1×1×1 = 1

For M = 1:

R = 0–1, D = 0–1 → 2×2 = 4

For M = 2:

R = 0–2, D = 0–2 → 3×3 = 9

For M = 3:

R = 0–3, D = 0–3 → 4×4 = 16

For M = 4:

R = 0–4, D = 0–4 → 5×5 = 25

For M = 5:

R = 0–5, D = 0–5 → 6×6 = 36

Now weight each by the number of ways to get that win count:

{Complicated formula that doesn't paste well}

Let’s compute that:

M Ways (M) Valid R Valid D Total Combos

0 1 1 1 1×1×1 = 1

1 5 1+5=6 1+5=6 5×6×6 = 180

2 10 1+5+10=16 1+5+10=16 10×16×16 = 2,560

3 10 1+5+10+10=26 same 10×26×26 = 6,760

4 5 1+5+10+10+5=31 same 5×31×31 = 4,805

5 1 1+5+10+10+5+1=32 same 1×32×32 = 1,024

🔢 Final Total

1+180+2,560+6,760+4,805+1,024=15,330

✅ Answer

Out of 32,768 possible scenarios, there are 15,330 where the Mets win as many or more games than both the Reds and Diamondbacks.

[MFS62]

That works out to 46.7834% that the Mets will equal or beat both.

I don't rely on the math.

I rely on the bookies.

What are the odds they are giving?

(I trust them more than Fangraphs.)

Later

[Edgy MD]

Let us calculate it as if it was a one-game race, as that is clearer.

If my team (the Mets) have a one-game lead on the Diamondbacks and a one-game lead on the Reds with a single game to go, I indeed have a 75% chance of finishing ahead of the Diamondbacks and a 75% chance of finishing in front of the Reds. But I do not have a mere 56.25% (75% x 75%) chance of beating them both. This is because a win by my team guarantees both of these outcomes. (These are not two separate races — in which case my chance of two victories would indeed be 56.52% — but one race including three opponents.)

So:

• Mets win, Reds win, Diamondbacks win = Mets as ultimate winners
• Mets win, Reds win, Diamondbacks lose = Mets as ultimate winners


• Mets win, Reds lose, Diamondbacks win = Mets as ultimate winners


• Mets win, Reds lose, Diamondbacks lose = Mets as ultimate winners


• Mets lose, Reds win, Diamondbacks win = Mets as ultimate losers


• Mets lose, Reds win, Diamondbacks lose = Mets as ultimate losers


• Mets lose, Reds lose, Diamondbacks win = Mets as ultimate losers


• Mets lose, Reds lose, Diamondbacks lose = Mets as ultimate winners

 

That gives the team a 62.5% chance (5/8) of beating both teams, which is greater than the 56.25% chance that they would have had if they had to beat them in two separate one-game races.

Over multiple games, the math is compounded multiple times, but it remains true and demonstrable that the chance of the Mets beating both teams in one race is greater than the chance of them beating one team times the chance of them beating the other.

If you disagree with my math, or the philosophy behind the math, I understand, but please do not again write that you have no idea what I am asserting.

[Frayed Knot]

I am sorry you do not have the slightest idea what I am asserting. -- So am I

I do not know why you have printed the eight daily possibilities, so there! -- "So there" ... Really?

OE: My "two possibilities," which I thought was clear, refers to the two season outcomes of (1) the Mets finishing ahead of the Diamondbacks, and (2) the Mets finishing ahead of the Reds. These are related, as a single Mets performance affects both outcomes. -- So I see now that you are talking about end of season outcomes not game outcomes although it's the game outcomes that determine the season outcomes so I'm not sure why the distinction. And you keep repeating how a single Met outcomes affects "both" outcomes but the outcome of every game and every final record of the concerned teams affects all others so, again, I don't know what the point is.

The bottom line is that there are a finite number of possibilities involving these fifteen NYM/CIN/ARZ games and, by my calculations, more than half of the eventual outcomes wind up with the Mets on the outside looking in. Now the possibilities of the Reds being the odd team out and the Snakes being left behind (for a second straight season) are also more than half. But this just means that because the Mets are currently ahead they have a better shot than EITHER Cincy or Zona, that doesn't add up to them having a better chance than the other two combined.

Even if you disagree, and disagree strongly, it is easy enough to understand that thesis, having been stated several times. -- I'm still not sure what the thesis even is. You began by implying that my data was wrong because I wasn't accounting for the idea of 'outcomes being not independently produced'. Yet every win and every loss by each of the three teams is independently produced and the season outcomes for each team will be as a result of those games.

[Fman99]

A bunch of dead mathematicians all read this from heaven and got boners.

The mathematicians in hell did, too, I think. But they could just have boners because hell is so metal