The last post discussed the two alternatives for the Consolation bracket needed for an upcoming tournament. But in addition to a Consolation, my friend also needs a Last Chance bracket. Third brackets are almost always rather difficult brackets to build, and they depend on the second brackets they’re attached to in the same way that second brackets relate to first brackets.
My friend wants to pay two places again in the consolation. That means that the first version of the Consolation bracket, 48Consol22Xv3, spins off these drops: {16, 8, 8, 6, 4, 2}, and alternate version, 48ConsulAlt, had these: {8, 12, 10, 8, 4, 2}.
Here’s a Last Chance for the first drops vector:
16, 8 > 12
12, 8 > 10
10, 6 > 8
8, 4 > 6
6, 2 > 4 > 2 > 1
yielding a bracket in the form FG.H.I.J.K.|.|: 48 last chance. The last chance itself has 7 rounds, but its first round depends on four rounds in the brackets above, so it requires 11 rounds to play out. The worst thing about the structure is the double drop from the F and G rounds.
Here’s a structure for a last chance linked to the alternative Consolation:
8, 8 > 8
8, 4 > 6
6, 10 > 8
8, 8 > 8
8, 4 > 6
6, 2 > 4 > 2 > 1
This bracket is FG.G.H.I.J.K.|.|: 48LCAlt. This last chance has 8 rounds, but it can start after only three rounds above, so it’s essentially the same length as the other one, which my simulations confirms.
The squirrelly feature of this bracket is that the G drops are divided between the first two rounds. Eight of the G’s drop to the first Last Chance round (the N round, if you’re keeping track), but four lucky G’s drop one round further along.
It could be argued that this is less squirrelly than the other pattern, in which all of the G’s drop a round too early. But that’s a fine point that may be lost on many players. Somehow, splitting the G drops just seems to look fishier.
So, there are the alternatives for my friend’s 48 tourney brackets. It’s now up to him to decide which set will garner fewer complaints.
tourneygeek.com 
Inspired by Metzgerism’s response to my comment on the Friedman-Zieliński table, I worked out another possible solution for this problem: Put all of the “Byes” in the same quadrant, effectively putting a “Bye” in the Semifinals. Let’s have WD1 vs WD2 in the Semi (E1) with WD3 getting the “bye” and advancing to the final (F1). Now it may happen that, if WD3 loses F1, they’ll have the same record as the loser of E1 (both 4-1). In that case, there would be a match for 2nd between the 2 (G1= LE1 vs LF1).
If this was an activity where there is a sublevel of scoring, e.g. sets or points, a Round-Robin between the 3 quad winners would be preferable. But even if there was no subscore that could be incorporated in tiebreaks, this could still be accomplished by having WD3 face off against LE1 first, and then WE1, regardless of result. If WD3 wins F1, then G1 is a straight-up Final. If, however, WD3 Loses F1, then it’s still playing for 1st if it wins, but drops to the Consolation if it loses. The concern there is that the loser of E1 becomes limited to 2nd, and could be dropped to the Consolation if all 3 contenders have a 5-1 record in the end. (Unless you have the Loser of G1 drop to the Consolation in an all-5-1 scenario, ensuring that the winner of F1 is “in the money”, i.e. 2nd at worst.)
The F-Z table recommends a recharge round if the winner of E1 loses the Finals. (“A [WD1] vs B [WD2], Winner vs C [WD3] twice, C loses ties”) But it’s focused on 1st and not 2nd. Still, it would set the bar for the Championship at 6 Wins (6-1 if WE1 loses F1 (the 1st match against WD3) but wins the Recharge, 6-0 otherwise) and the bar for 2nd at 5 Wins (5-2 if WD3 is 6-0, 5-1 otherwise), eliminating the vagaries of my “quasi-round-robin”: if WD3 wins F1, then G1 becomes a “true Final” whereas if WD3 loses F1, G1 becomes a fight for 2nd.
What this does is give all of the drops in the consolation bracket an equal record: the 24 A-drops are 0-1, the 12 B-drops are 1-1, the 6 C-drops are 2-1, and the 3 D-Drops are 3-1. The format would be A.B.|.C.|.D.[E/F/G].|.| , so you would get uneven records starting in round K, becoming worse in Round M. But that’s the nature of this beast. The drops vector would be:
24 > 12
12 > 12 > 6
6 > 6 > 3
3 > 3
1 > 2 > 1 (Whichever entrant faces the 1 would be a luck draw, though this drop only creates a rematch if at least 2 of the 3 from the Consolation are from the “1”‘s “quadrant”.)
Now I’ll admit that it still makes the “Last Chance” screwy, though not as much as the method inspired by your friend’s concern. There are 2 rounds that take multiple drops (P takes HI (12 drops each, total 24) and V takes MN (3+2=5)), but no rounds have their drops split. The format is HI.|.J.K.|.L.MN.|.| and the drops vector is:
12
12 > 12 > 6
6 > 6
6 > 6 > 3
3 > 3
3
2 > 4 > 2 > 1
I then thought about shifting the consolation, and I’ll cover that in my next comment, since I believe that it contains a potential exception to the “don’t split drops” rule.
LikeLike
When I tried to shift the consolation in the tourney format I proposed in my prior comment, I got a very screwy mess in the Last Chance; odd-even combinations everywhere. After trying it unsplit, I considered splitting the drops from the Consolation Semis (now M) due to that round receiving the loser of the quasi-Round-Robin, but that would end with having 3 contenders and only 2 payable places. The loser of the “quasi-Round-Robin” in my main bracket has recourse in the Consolation; losing here allows no recourse.
However, by that principle, if I split the drops from the J round instead, segregating the 6 matches that feature a C drop from the 3 that don’t, then I could have the latter drop to the R round while the former would drop to the S round. (Alternately, you could have the 6 C drops play each other, but that wouldn’t be fair in-and-of-itself, Ottawa Bonspiel aside, let alone dropping the losers of those matches to the R round with the HI side dropping to the S.)
From here, I have 2 options: If I have the K drops join the S round, then that’s the only round with multiple drops (the I drops start in Q instead of P) and I remove 1 round from the Last Chance. Effectively, the J drops from matches featuring a C drop get a “bye”. However, if I let the K drops go to the T round, then we’re back to V taking both L & M (pre-shift M & N) and the Last Chance still has 9 Rounds, though, again, it doesn’t have to wait for the I drops.
Consolation: A.B.C.D.|.[E/F/G].|
24 > 12
12 > 12
6 > 9
3 > 6 > 3
1 > 2 > 1
Last Chance J.JK version: H.I.J.JK.L.M.|.|
12 > 6
12 > 9
3 (HI) > 6
6 (CHI)
6 > 9
3 > 6
2 > 4 > 2 > 1
Last Chance LM version: H.I.J.J.K.|.LM.|.|
12 > 6
12 > 9
3 (HI) > 6
6 (CHI) > 6
6 > 6 > 3
3
2 > 4 > 2 > 1
Screwy, yes. But how screwy depends on whether a match with a higher received drop matters more than one without. The key question is “is facing a (e.g.) C drop more “difficult” than facing a fellow winner from the Loser’s Bracket?” i.e. Does it matter where the “1” in a 2-1 record occurs when playing in a Bracket?
C Drop dropping to J: W,W,L
A Drop making it to J: L,W,W
B Drop making it to J: W,L,W
LikeLike