THE BOOK--Playing The Percentages In Baseball

z is 3.3, as all players involved should be regulars.  x is probably between 5.0-5.3, as the 0.7 player, while above replacement level, should still be fairly available in the FA market.  y depends a lot on roster construction, I think, and how much usage the 1.7 player gets.

Of course, it depends on the roster construction of both teams, but this is what I’m going with:
x = 5.1
y = 4.6
z = 3.3

A 0.7 WAR player is a bench player or decent reliever. A 1.7 WAR player is a starter on a weak to good team, or strong role player on a great team. A 2.7 WAR player is as much of a regular as a 3.3 or 3.0 WAR player, so there doesn’t need to be differentiation between total WAR between 3.3 + 2.7 and 3.0 + 3.0.

Ok, how about these groups of pitchers.  I’m providing the numbers in terms of what a team’s W/L record would be, with those pitchers pitching.

(League average is 4.50 RA9 if you need that.)

Ready:

Team A:
15-9, 3.38 RA9, 216IP, 81RA
5-9, 6.07 RA9, 126IP, 85RA

Team B:
13-9, 3.68 RA9, 198IP, 81RA
7-9, 5.13 RA9, 144IP, 82RA

Team C:
11-9, 4.05 RA9, 180IP, 81RA
9-9, 4.50 RA9, 162IP, 81RA

In all cases, the pairs of pitchers are 20-18.  In all cases, the pairs of pitchers threw 342 IP.

Which of the three pairs of pitchers do you want?

Eric/2: do you mean to say y=4.2 ?

Question #3 is easy, I’d want team A.  Because I believe I can easily get rid of pitcher #2 and replace him with a pitcher that gives up fewer than 6 runs per game.

Concur with Rally.  However, if forced to run that pair of pitchers out there for the stated numbers, which one I’d want would probably depend on my offense/bullpen.  If I’m stronger in the former I’d probably want Team C, since both pitchers will give my superior offense a chance to win the game.  If I’m stronger in the bullpen, I’ll go with Team A and count on my bullpen to throw up blanks after the second pitcher gives up his 3-4 runs in 5ish innings.

Not sure this question can be answered ceteris paribus.

Rally: for the same cost, right?

Ok, if Team A is an easy one, then how about between Team B and Team C?

***

Kevin: the W/L numbers are there.  What does it matter about “chance to win”, when I’ve told you how many times the team will win?

5.0
4.1
3.3

I guess it wouldn’t matter if I’d read the question correctly.

you could argue 2 3 WAR players are better than 1 6 WAR player because with 2 players you’ve diversified some risk.  I think the real reason you might prefer the 1 6 WAR player is certainty, not scarcity.  You’re more sure that the 6 WAR guy will perform than you are that either of the 3 WAR guys will.

By actual wins, 6 WAR = 2 x 3.0 WAR.

However, there’s only a handful of 6 WAR players and a bunch of 3 WAR players.  If I want to win a title (build a WAR outlier), then I need to get some of the very high WAR players, because playing time is limited.  To win a title, I need what, 60 WAR?

So what we really care about is WAR/playing time.  A 1.0 WAR player is useless to a contender if they take 600 PA to get there; but is valuable if they can do that in 300 PA.  Do you see what I mean?  Playing time is a limited commodity.  I have only so many PA, so many innings to pitch.  If I’m constructing a contender, I need to cram WAR into as little PT as possible.

From the other direction: injury risk.

If each player has a 25% chance of being injured, here are the WAR distributions for 6 + 0 vs. 3+3.

6 WAR: 75% chance of 6 WAR, 25% chance of 0 WAR.
3 + 3 WAR: 56% chance of 6 war, 38% chance of 3 WAR, 6% chance of 0 WAR.

If I’m trying to win a title (create a WAR outlier season), I’d rather have all my eggs in few baskets.

Hang on, I’ll work up a quick model.

Okay, here’s a chart showing the likelihood of various WAR totals.

The mean is the same, but the likelihood of getting the max WAR changes dramatically.  If I’m trying to win a title, I’m willing to accept some worse seasons when injury strikes for the chance at getting a dominant season.

Here’s the results:

This seems like a very good reason to value top players’ WAR more than a bunch of mediocre players’ WAR.

seems like a replacement level issue, now that I think about it.  There’s really no amount of 1 WAR players that add up to a 6 WAR player because a bunch of 1 WAR players adds up to a last place team.  The numbers being thrown out imply a replacement level of 1 WAR or a little less.  Which makes sense, since unless you’re historically incompetent, you’ll be able to scrape together a 65 win team.

Please answer Tango/3, by placing teams A, B, C in order of preference.

And the examples of 5 players on one side and 10 players on the other are not applicable.  We’re not talking about roster management. (Unless of course you are implying the 5 missing guys are all 0 WAR full-time players.)

A, B, C.

@Tango

If I know those are the stats, after the fact, then I believe all 3 options are equal.

In my examples, I was implying the 5 missing guys were 0 WAR; this was purely an exercise at the extremity to emphasize the difference.  I was simply pointing out, that, given a set of players with the same mean performance, we ought to select the group that has higher variance, if we are trying to win the title.

I should have been clear: those are not after-the-fact stats, but true talent level stats.

DSM: agreed that you wanted the larger variance, all things equal.  What I am asking is what kind of difference in mean would you need to make this a break-even proposition?

JeffZ for example says 5+0.7 = 3+3, but that 3.3+2.7= 3+3.

Therefore, you still haven’t told me what your answer is to the question.  The question is one of degree, and not one of yes/no.

Well, it depends on where you are on the championship probability curve and what your team objectives are.

If I assume 15% probability of injury, a team with 54 WAR if healthy, and 18 players, and compare the case with X + 0.7 vs. 3.0 + 3.0, X comes out to be 5.2.  (I’m assuming the healthy team is a title contender with ~19% chance of winning the title, and progressively less with more injuries.)

In other words, the effect is probably not very pronounced, but it depends on some other variables.

@3

Well, if I’m forced to pitch those pitchers for that number of innings, then I’d definitely go with pair C.  Two reasons:  their average starts are about the same length, and that give my bullpen a little more consistency (just having to cover a couple innings every night, rather than alternating between one inning and four).  Second, they give up the fewest runs.  Now, I understand they give up about the same number every time, but I’d rather only have to overcome 162 runs for the opposition, rather than 166.

How reliable is my manager?

If I can count on my manager to always play my ace in the highest-leverage situations, I’d say A is probably the best option. The difference in win expectancy per run probably outweighs the difference in total runs.

But if my manager doesn’t believe in leverage, obviously C gives up the fewest total runs.

Also, what’s the handedness of the pitchers in question, and what’s their platoon split?

If they’re opposite-handed and have large platoon splits (and assuming the numbers given are true talent vs. a mixture of hitter handedness), the value of C would greatly increase.

I ran some tests to see what would happen if I had a lineup of 7 AL-average players and 2 others with different stats: both with 3 WAR, 2.7+3.3, etc.

And even without considering scarcity, the unbalanced lineup is better simply because you can bat your best player first and your wost player 9th.

I got the following estimated equivalencies for that hypothetical team:
3+3 =
2.7+3.29
1.7+4.23
0.7+5.07
0.0+5.72

Of course, those estimates don’t take scarcity into account. If I had to come up with a quick and dirty estimate based on the assumption that 0.0+5.3 for 3+3 was a fair trade, then I might try multiplying every discount by 2.5 (because the 0.7 discount is 2.5 times the .28 discount) and I’d get:

3+3=
2.7+3.275
1.7+4.125
0.7+4.725

So a rough guess might be x=4.7, y = 4.1, z=3.28.

But those numbers are screwed up by the fact that a 1.7-WAR guy bats 9th while a 2.7-WAR guy bats 2nd on this particular hypothetical team. Realistically, the difference between x and 5.3 is going to be smaller, the difference between x and y is going to be larger, and the difference between y and 4.3 is going to be larger.

So I’m going to say x=4.8, y=4.0 and z=3.28.

Not very scientific, I know, but any effect that could cause a 5.3-WAR player to be as valuable as 2 3-WAR players would have to have a very small effect size for the 2.7 case and in-between effect sizes for the in-between cases. The lineup effect, which has a very small effect size for the 2.7 case and in-between effect sizes for the in-between cases, seems like a reasonable proxy to me.

Anyway, if 3+3=5.3, I can’t see the true values of x/y/z being very far away from my estimate or Jeff Z’s estimate.

Bill: what about Tango/3?

I answered that in 21.

Basically it’s A/B/C in an ideal world, but C/B/A if the pitchers are put in at random times rather than at the ideal times.

Maybe B is the best overall balance, though; only 1 more run allowed than C and we can assume that most managers will put the best guy in more often than the worst guy in close games (even if they’re not consistent enough to do it every time).