Hanoi5 Posted February 22, 2008 Report Share Posted February 22, 2008 This is the set-up. 5 pairs are playing for three places in a National Team. They are to match with and against each other pair and the best three pairs will have the right to represent their country. These are gonna be 24-board matches divided in two, twelve boards each time. Usually, when there is a forfeit the team which was present receives 18 VP (or their average, or the complement average of the other team, whatever is best). As these matches are divided in two halves, what is the best way to give VP's if a pair wins (or loses) by a GREAT margin the first match and decides to miss the second (making up an excuse, of course)?. Quote Link to comment Share on other sites More sharing options...
ArtK78 Posted February 22, 2008 Report Share Posted February 22, 2008 Forfeits should result in disqualification of the forfeiting pair absent extenuating circumstances. There is virtually no justification for a forfeit in an event of this type. Quote Link to comment Share on other sites More sharing options...
jtfanclub Posted February 22, 2008 Report Share Posted February 22, 2008 Forfeits should result in disqualification of the forfeiting pair absent extenuating circumstances. There is virtually no justification for a forfeit in an event of this type. For this team, sure, they shouldn't be allowed to continue or get a 0, depending on how good their excuse is. But what about their opponents? If they get a max score, that may win them the tournament without doing anything. If they get a 18/30 they might be unable to win if this is a drunk-punching contest. Quote Link to comment Share on other sites More sharing options...
Hanoi5 Posted February 22, 2008 Author Report Share Posted February 22, 2008 The problem here is that we're not talking about teams but pairs. Let's say, pairs A, B, C, D and E are playing and so the first day we have: A-B A-C A-Dvs vs vsC-D B-D B-C E is bye. Each match 12 boards are played. And let's say that A-B got a very bad result against C-D. However, for the next round (or the next 12 boards) A doesn't show up. Doesn't B deserve better than 0? What if C is the one missing, wouldn't D like to have its result stand or a better result than 0? How would this be fairly solved? Quote Link to comment Share on other sites More sharing options...
hrothgar Posted February 22, 2008 Report Share Posted February 22, 2008 Hey there Comment 1: I don't find the example with missing pairs particularly interesting... If a pair doesn't show up for a session boot their sorry asses from the event. In theory, you're selecting your National team. If a pair can't get their ***** together for the event they have no excuse participating. I don't think that you want to send a pair of flakes to the World Championships Comment 2: What is interesting is trying to design appropriate conditions of contest to sellect the best three pairs out of five. Here's the catch... How are you going to determine board results? Think about it. Suppose that you have Pair A playing v Pair B andPair C playing V Pair D with Pair E sitting out Does the result on a board show that (Pair A is much better than pair B) or that (Pair C is much stronger than Pair D)??? There are what? 15 different ways that we can match up the 5 pairs... Will you have enough time to run 15 separate sessions of 24 boards each? In theory, you can use some statistical techniques like "Design of Experiments" to select an optimal subset of the total test matrix. I still think that you're going to need to run quite a few rounds to get anything approaching reliable results Quote Link to comment Share on other sites More sharing options...
Cascade Posted February 22, 2008 Report Share Posted February 22, 2008 If one pair doesn't show up then you are down to four pairs for a much better format. If another pair doesn't show up then you have your three pairs selected - perfect. I don't want someone on my team that doesn't show up because they got beaten or make up some excuse. The conditions of contest need to insist that pairs show up. Quote Link to comment Share on other sites More sharing options...
jtfanclub Posted February 22, 2008 Report Share Posted February 22, 2008 There are what? 15 different ways that we can match up the 5 pairs... Will you have enough time to run 15 separate sessions of 24 boards each? There are 10 possible pairings: A v B E v C A v C D v B A v D B v E A v E C v D B v C D v E As you can see, it's easily enough to do in 5 rounds. While I haven't done it, it's also not hard to have each pair oppose a pair once at the same table, play against them at the other table once (N vs. N), and team with them once (N vs. other table W). As much as anything can, this will be completely fair and give you the top 3 teams, assuming a sufficient number of boards per round. Quote Link to comment Share on other sites More sharing options...
hrothgar Posted February 22, 2008 Report Share Posted February 22, 2008 There are what? 15 different ways that we can match up the 5 pairs... Will you have enough time to run 15 separate sessions of 24 boards each? There are 10 possible pairings: A v B E v C A v C D v B A v D B v E A v E C v D B v C D v E As you can see, it's easily enough to do in 5 rounds. While I haven't done it, it's also not hard to have each pair oppose a pair once at the same table, play against them at the other table once (N vs. N), and team with them once (N vs. other table W). As much as anything can, this will be completely fair and give you the top 3 teams, assuming a sufficient number of boards per round. Funny that, I was able to find 15 possible combinations of pairs for the test matrix... Exclude Pair E (A x B) x (C x D)(A x C) x (B x D)(A x D) x (B x C) Exclude Pair D (A x B) x (C x E)(A x C) x (B x E)(A x E) x (B x C) Repeat excluding pairs, A, B, and then C for a total of 15 You're counting the number of unique pairs, rather than the number of possible COMBINATIONS of pairs. As I mentioned originally, the scoring system (presumably) compares the Board result at table 1 with than at table 2. Unfortunately, we have no information about the relative strength of any of the pairs. Assume for the moment that 6S gets bid at both tables. At Table 1, pair A makes 6S. At table 2, pair C goes down in 6 Spades? Does this mean that Pair A declares very well? Pair C is very bad? Pair B defends brilliantly? Pair D screwed the pooch? There's no way to tell... As I mentioned earlier, the right way to approach this problem is to look at a complete test matrix and then partioning it systematically using something like a Box-Behnken design... Quote Link to comment Share on other sites More sharing options...
han Posted February 22, 2008 Report Share Posted February 22, 2008 Richard, can you make an argument for having both (AxB)+(CxD) and (AxB)+(DxC). I think that is necessary for a truly fair competition, which would mean that you need 30 matches. Or rather, 15 matches, each cut in two. Quote Link to comment Share on other sites More sharing options...
hrothgar Posted February 22, 2008 Report Share Posted February 22, 2008 Richard, can you make an argument for having both (AxB)+(CxD) and (AxB)+(DxC). I think that is necessary for a truly fair competition, which would mean that you need 30 matches. Or rather, 15 matches, each cut in two. URK... You're right... The ordering matters (on occasion...) The whole problems with this trial is that you have such a small number of pairs: Lets assume that you had 10 or 30 or however many pairs competing. There's not way that you could exhaustively test all the various combinations. Here you have a small enough number of combinations that this is feasible, but enough that its still damn annoying... Quote Link to comment Share on other sites More sharing options...
Hanoi5 Posted February 22, 2008 Author Report Share Posted February 22, 2008 That's the number of players that are willing to represent our country, so they're the only ones playing. We're planning exactly the 15 matches described by hrothgar but twice as han pointed out. I was, however, asking about what to do in case one of the pairs missed a match or better still (or worse still) half a match. Quote Link to comment Share on other sites More sharing options...
hrothgar Posted February 22, 2008 Report Share Posted February 22, 2008 Wow... I'm genuinely impressed. From the sounds of it you're planning a total of 30 12 board matches. Thats an impressive number of boards Quote Link to comment Share on other sites More sharing options...
Hanoi5 Posted February 23, 2008 Author Report Share Posted February 23, 2008 It's a ten-day event, I'm not playing though... Quote Link to comment Share on other sites More sharing options...
han Posted February 23, 2008 Report Share Posted February 23, 2008 Richard, can you make an argument for having both (AxB)+(CxD) and (AxB)+(DxC). I think that is necessary for a truly fair competition, which would mean that you need 30 matches. Or rather, 15 matches, each cut in two. URK... You're right... The ordering matters (on occasion...) The whole problems with this trial is that you have such a small number of pairs: Lets assume that you had 10 or 30 or however many pairs competing. There's not way that you could exhaustively test all the various combinations. Here you have a small enough number of combinations that this is feasible, but enough that its still damn annoying... Consider the case that pair C always does well against pair D but not necessarily does better against other teams. If the pairing is AxB and CxD in round one then A has an advantage. But, if you split the matches in 2 as described in the initial post and switch the teams halfway (so A still plays against B but now A and D are a team and B and C are a team) then no pair ihas an advantage. It seems to me that you don't have a problem in this case, even if you don't play all possible matches. Just make sure that for the matches that are played, at one table the pairs switch directions halfway. Quote Link to comment Share on other sites More sharing options...
cherdano Posted February 23, 2008 Report Share Posted February 23, 2008 It's a ten-day event, I'm not playing though... Wow. I hope that the last pair won't be out of contention as long as possible, as otherwise it may really skew the result if it starts playing worse... If you play it as a double round-robin you could offer the pair in last position to drop out halfway. Quote Link to comment Share on other sites More sharing options...
Guest Jlall Posted February 23, 2008 Report Share Posted February 23, 2008 This seems horrible. Lets say pair A is pwning everyone. Pair B is doing well. Pairs C and D are fighting it out, and pair E is out of it. So basically only pairs C and D matter, assuming pair B doesnt get crushed. But pair A can get crushed. Now pair A can basically decide who gets in C, or D. This may be personal (friendship or spite). This may induce pair C to pay pair A some money to dump to them. etc etc. I don't like this format as collusion and dumping become so easy. Quote Link to comment Share on other sites More sharing options...
Hanoi5 Posted February 23, 2008 Author Report Share Posted February 23, 2008 You're being a little dramatic Mr. Lall. I don't think anyone is gonna be paying another pair to do bad or anything like that... My main fear is that a pair stopped playing or someone purposefully played badly, but I don't think anything can be done about it. Anyway, since you feel it is a bad way to decide the team , what would you propose? Quote Link to comment Share on other sites More sharing options...
gwnn Posted February 23, 2008 Report Share Posted February 23, 2008 may I ask what country this is? or is it some sort of a military secret? Quote Link to comment Share on other sites More sharing options...
Hanoi5 Posted February 23, 2008 Author Report Share Posted February 23, 2008 It's a military secret. Quote Link to comment Share on other sites More sharing options...
Codo Posted February 25, 2008 Report Share Posted February 25, 2008 I don´t have a solution for your original question, but why do you insist on 5 pairs?Isn´t it much easier to play a big imps across the field tournement with 6 pairs on three tables? And if you have no sixt pair, why can't you simple let some trainers/Foreigners/dummies play, who are not allowed to qualify? Quote Link to comment Share on other sites More sharing options...
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