[tournament-org] WhereIsTheBar

[Geoff Kaniuk]

[[ inline ]]

Geoff

33 Ashbury Close, Cambridge CB1 3RW 01223 710582

On 02/10/2018 12:57, Jenny Rofe-Radcliffe via tournament-org wrote:
> Hi,
> 
> To add to Toby's points, note that (as described by Alison) upsets do
> occur. One of our 1ds could certainly win the tournament, by beating the
> other 1d, a 1k and having an upset against the 5d. It would be grossly
> unfair to that 1d to advertise a McMahon tournament and then switching it
> around so that they don't have a chance to get an upset.
> 
[[
I agree that one should not change the rules.

But if the rule: "handicaps are allowed above the bar if the entry 
produces large gaps at the top" is published ahead of time, then surely 
there can be no problem?
]]

> I keep coming back to the question of setting the bar rigidly at a grade
> point that means we have an un-even number above the bar ... let's suppose
> we set this bar at 1k, so above the bar we have five players. Then
> inevitably one of the 2ks is drawn up in the first round ... which sort of
> functions as them being above the bar, but docking them a McMahon point for
> not being strong enough. This always seems a bit unfair to me. Clearly if
> you have two 2ks (or whatever level) about whom you know nothing, no GoR to
> judge from, etc., then I guess you might as well randomise by leaving them
> both below the bar. But I much prefer, if I have a differentiator, to put
> one of them above the bar to give an even number up there.
>
[[
I am not convinced by the argument that we need an even number of 
players above the bar.

A common rule is that for a 3 round event you should have between 4 and 
8 players.  The first is justified by the idea that the top player will 
play a round robin.  But this is false, as in round 2 some of the top 
losers might play winners from below.  The bar population is not isolated.

Again the 8 comes from the idea that the top play a 3 round Swiss or 
Knockout,so then you get a unique winner. Yes you do (with non-integral 
komi) but it is not a pure Swiss as the population keeps swelling.
]]

> This is obviously a bit fiddly and most practical if (a) you've used GoDraw
> quite a lot and (b) have someone else handling other aspects of the
> tournament organisation so you only have to deal with the draw at this
> awkward point when everyone is waiting for round one to begin. (I very much
> advocate strong-arming someone into making announcements for you, or
> alternatively coercing them into handling the draw for round one, at this
> stage of a tournament. It's no fun doing it solo; no one can multi-task
> that much.)
> 
[[
If you have an odd number so be it. You could of course arrange manually 
that the 5d is not the one to be drawn down. But sticking to the example 
there is no issue as everyone is below the bar and the whole population 
is even!
]]
> Thanks,
> 
> Jenny
> 
> 
> 
> On Tue, 2 Oct 2018 at 12:35, TobyManning via tournament-org <
> tournament-org at lists.britgo.org> wrote:
> 
>> Geoff:
>>
>> 1) Your probabilities are incorrect : 86.6 +(2*1.8) + (2*0.4) does not add
>> up to 100%.
>>
>> 2) You state: "We have agreed so far that setting the bar at 1k or 1d
>> produces an unfair tournament". *False.* The tournament is perfectly fair
>> - one player has a greater chance of winning than others (based on previous
>> results) but this does not mean it is a foregone conclusion. Heather Watson
>> and Serena Williams both enter Wimbledon and they have equal opportuities
>> to win - you don't say "Serena is so much stronger than Heather therefore
>> she should be handicapped to give Heather a chance" (or to ban Heather from
>> entering because she is considered to have no chance at all).
>>
>> 3) Please remember that the purpose of a Go Tournament is for people to
>> enjoy themselves - not to produce a "perfect" result. And the strongest
>> players - those affected by your analysis - would rather play even games
>> than handicap games. Until you can provide a consensus view to the
>> contrary, please stop this ridiculous idea of handicap games at the top of
>> the draw.
>>
>> Toby.
>>
>> On 02/10/2018 12:10, Geoff Kaniuk via tournament-org wrote:
>>
>>
>> There is overwhelming support for a bar at 1k with a spread from 2k to 1d,
>> and a few suggestions allowing handicap above the bar.
>>
>> It seems that we expect the 5d to win no matter where we set the bar, but
>> in order to run a McMahon tournament we are going to set the bar to 1k
>> anyway.
>>
>> One guide to setting the bar is that all players above the bar should have
>> a reasonable chance of winning the tournament. As stated we assume that all
>> players are correctly rated. This means we know the win probability for
>> each pairing.
>>
>> Let us suppose the bar is at 1k in our entry of:
>> 5d 1d 1d 1k 1k 2k 2k 3k 3k 4k 4k 5k .....
>>
>> Then given a plausible tournament, the top players have these chances of
>> winning all three games:
>>
>> PLAYER    OPPONENTS    PER ROUND PWIN      PROB WINNER
>> 1k    1k 1d 5d     0.50*0.40*0.02 =  0.40%
>> 1d    1k 1d 5d    0.60*0.50*0.06 =  1.8%
>> 5d    1k 1d 1d     0.98*0.94*0.94 = 86.6%
>>
>> I think this makes it clear that the 1k and 1d have effectively no chance
>> to win this tournament. Not much changes if you set the bar to 1d.
>>
>> Going back to basics, the McMahon system is designed to provide a fair
>> pairing at every round. It does this by assigning an initial MMS determined
>> by your grade. The fairness comes about because we assume grades are
>> realistic and always pair players on the same MMS where possible.
>>
>> The winner is the player who ends on the maximum MMS. If there is only one
>> strongest then he or she will have a massive advantage. Hence we have a bar
>> - and in the old days there were plenty of 4d and 5d around in the top
>> group so no problems.
>>
>> If an an even pairing is not possible, then it is common for players below
>> the bar to play with handicap - often MMS difference-1 but this can of
>> course be varied.
>>
>>
>> We have agreed so far that setting the bar at 1k or 1d produces an unfair
>> tournament. Therefore in order to continue play in the spirit of McMahon,
>> we should consider raising the bar to 5d. For the sake of clarity this is
>> equivalent to setting the bar at 6d.
>>
>> In the example given, setting the bar at 1k produces an odd number in the
>> bar, so one player chosen at random will have to play down.  The 5d
>> therefore might play a 2k in round 1.
>>
>> Setting the bar at 6d avoids this kind of problem. We pair for the
>> smallest MM difference so the 5d will play the 1d in round 1 with a
>> handicap of 3 stones by usual rules.
>>
>> This levels the playing field somewhat but calculations show that this
>> still favours the 5d and better would be: handicap is the straight MMS
>> difference.
>>
>> My conclusion is that in these anomalous cases setting the bar anywhere
>> without handicap goes against the basic principles of McMahon pairing.
>>
>> In looking at variants of setting the bar with handicap say in a
>> super-group just leads to complications (e.g. when top players who miss
>> rounds are supposed to be below the bar).
>>
>> The simplest and most elegant solution to this problem is just to set the
>> bar to 1 higher than the highest grade. Assign handicaps by straight MMS
>> difference where needed. Then let McMahon do its job without any further
>> interference.
>>
>> Geoff
>>
>>
>> 33 Ashbury Close, Cambridge CB1 3RW 01223 710582
>>
>> On 02/10/2018 00:21, Richard Wheeldon via tournament-org wrote:
>>
>>
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