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Madison - I'm close to a method for a 100% Swiss
Posted By: neilkaz In Response To: Madison - Thank you (Bill)
Date: Friday, 5 September 2008, at 4:21 a.m.
I've been emailing Bill Minser and Sam back and forth while thinking about it for several days. I just went thru all possible occurances of player's records with Steve at 9-0, three at 7-2 and I think it was 10 at 6-3 wasn't it? I also assumed that Brad had defeated Steve in rd 9 and then there'd have been an 8-1, four at 7-2, and nine at 6-3 and went thru all possible outcomes again continuing to have all players with fewer than 4 losses play 11 rounds.
In all cases (I'll recheck in morning) it seems that there were 6 players remaining with a record of 8-3 or better after 11 rounds. This seems really good to me, since paying 6 players without a conso seems really fair to me when there's 47 in the event. So, it would seem that a record of 8-3 should guarantee cashing and that if you don't manage 8-3 you're getting no prize. A slightly smaller event might only have 5 getting 8-3 or better and getting paid and a slightly larger event would likely have 7 getting paid.
Sometimes the T ends by the 11th round. If Steve had hypothetically gone 11-0 last weekend, everyone (5) left would have been 8-3 it seems. Another rare possibility seems to be one 9-2 winner and five at 8-3. But usually you get two at 9-2 and four at 8-3 after 11 rounds. Sometimes one at 10-1 and one at 9-2 and 4 at 8-3 occurs. This means that the 9-2 must beat the 10-1 twice so there's the possibility of a 13th round. A 13th round is not a serious issue, when one notes that the main went 12 rounds and the conso went 13 rounds last weekend and when we keep ALL matches in the Swiss 9 pts.
After every outcome of taking this weekend's possibilities out to 11 rounds, there were no more than 2 players with better records than 8-3. Therefore the 8-3's stop playing and share third thru 6th place, while the two remaining finish the T.
Lets say that we had a bigger turnout..70ish like in Indy sometimes. Then there could maybe be 3 with 9-2 and 5 with 8-3. All 8 are guaranteed prizes but all play rd 12. Two 9-2's play and one becomes 10-2. The 3rd 9-2 plays a 8-3 and if he win's he's also 10-2 and plays one last match with the other 10-2 to determine the winner. Then the three 9-3's share 3rd/5th while the three 8-4's (noting they went 8-3 and get a prize) share 6th/8th (each getting 4.89% of the pot under my current proposed payout scheme). (Note that the last place finishers(3/4 conso) in a normal T paying 8 places would get 1/24 of the post or 4.17% so this seems reasonable.) Note also that the winner gets 1/3 of the pot under my proposed scheme for paying 8 spots just like he does in the typical 8,4,2,2,4,2,1,1 scheme.(Indy 2006 for example) But what if the 9-2 lost to the 8-3 ? Then the lone 10-2 wins the tourney bcuz everyone else has 3 or more losses. The five 9-3 share 2nd/6th and the two 8-4's share 7/8th. Again noting this is for a 70 player event where we likely get 8 cashers (no conso !!) at 8-3 or better after 11 rounds. However, the tough issue to swallow is the fact that there's no for sure finals match if three players have equal and top records near the end of the event. For chess players like myself used to big Swisses this is not an issue. It shouldn't be an issue for BG players either and won't happen most of the time under this format. Once in a while there will be three players at 8-2 and a larger number like seven at 7-3 in a 50ish player field...so what..play rd 11 and if only 1 person remains at 9-2 he wins the event ! This is Swiss remember.
Unfortunately, there's the thorny issue of giving someone a bye into the prize money if there's an odd number of players remaining after round 10 (all those with less than 4 losses). This is not as big an issue as the bye on Sunday was, since, this player is only getting a bye into a share of what will most often be a split of 3rd thru 6th place money, but still it is unfortunate.
Obviously if a rd 11 bye is needed, it is going to one of the 7-3's and cannot go to anyone who had a bye previously, of course. I'll discuss methods to try to deal more fairly with byes than simply awarding a free point in a later post.
So how would I pay the 6 prize winners in this 47 player event with no conso ? I'll make no changes based on whether the winning was 11-2, 11-0, 10-2 or what ever. I also feel that with so many matches being needed to place, that the winner should get somewhat less than twice what 2nd place gets. Thus I came up with the following payout sequence.
16,9,7,5,4,3,(2),(2) etc ..the (2)'s are to take care of places beyond 6th in case the event is larger. If the event is smaller or only 5 players are 8-3 or better, then the first 3 is dropped and the winner receives 16/41 of the pot
Now the winner is getting 16/44 of the pot or 36.4% which is equal to the 4/11 (36.4%) he usually gets when 6 places are would be paid in a normal elimination event(assuming no LC). 2nd place is getting 20% of the pot and the players tied for 3rd thru 6th get to split 19/44 of the pot 4 ways each getting 10.8%. If there's just a 1st place and 5 tie at 8-3, the winner gets the same thing of course, while the other 5 share 2nd thru 6th.each getting 12.7% of the pot, and also the ABT pts. This doesn't present an issue for trophies, since it really is only necessary to give a trophy to the winner. The event now saves money on trophies and perhaps can even pump a bit of that back into the prize pool :).
Tell me what you think of my payout scheme for paying 6 players when there's no conso. Note that there's always ties for the lowest places. Often there's a clear 2nd place, sometimes a tie for 2nd 3rd also, as well as a few players tied for the next lowest spot.
Tell me what you think of no conso but still paying the same ammount of players, all from the main and noting that just like last weekend it takes 4 losses to be out of the Swiss-elim. Guaranteeing a prize for those 8-3 seems to pay the proper percentage of players. Those getting that 8-3 prize won't have to play anymore unless the event is large enough that there's more than 2 players with better records (ie two or fewer losses and thereby still eligible to win the entire T) than 8-3 and I think that takes considerably more players in the event than we have had in Madison. Those at 8-3 could also stop playing as if somehow there were 4 players with equal better records (but again this will take a HUGE event)
So here you have my proposed method for a 100% Swiss event without conso. Once again 8-3 wins a prize, and 4 losses mean you're out and your 3rd loss means that you cannot win the event. The winner will be the last guy with fewer than 3 losses, requiring 11 to 13 rounds with player numbers typical to the 50 or so Madison gets. This is no longer than the events have always been and we'll keep all matches to 9 pts. With the really poor 24 player Advanced turnout I'd have to take a look at the drawsheet and records, but a similar scheme could be done, noting that now we'd want to pay only 3 players if possible. I'd also change the payout scheme a bit for 3 places..perhaps 15,10,5 ratio noting that sometimes there could be a tie for 2nd/3rd.
Tomorrow I will try to address byes and I will most certainly desire your inputs. Note that byes are only needed if there's an odd number of players remaining and there always give to one of the players with the most losses and never given twice to anyone, so some of the time there may be noone in the money who had a bye..but yuck..that possibility of giving a 7-3 a bye into the money split for what is often 3rd-6th.
... neikaz ...
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