Fairest Movement (choice of)
My primary focus in my choice of movement is a movement which I believe to be the fairest and I believe that to be a movement where (as far as possible) each pair play all other pairs and all other boards. When I have an odd number of tables I play a Mitchell with a N/S winner and an E/W winner. Occasionally (out of curiosity) I look at other Bridge Clubs to see what they do and I see that a club local to me with e.g. 5 / 7 tables play a Mitchell with E/W pairs numbered +1 and an arrow switch on the last table.
Standard Mitchell with a N/S - E/W two winner movement
OR
Mitchell with E/W+1 and Arrow switch one winner movement
Which is ‘fairest’ and why?
Comments
In terms of absolute balance, the standard mitchell is fairer. The two winner solution will give you an almost perfectly balanced field but, inevitably, two winners. Arrow switches aren't perfect and unless you're able to make it an all play all howell you lose a bit of the balance.
Sometimes this isn't the only consideration, with 5 tables a standard Mitchell will be 5 rounds of 5, and I prefer to play a slightly less balanced movement with more rounds. I'm quite confident inflicting slightly complex movements on my players though. Maybe the neighbouring club just likes having a single winner movement. Sounds like they might be playing Hesitation Mitchells which would let them play 6 fours or 8 threes.
I can't see how a Mitchell arrow switch movement produces any kind of 'fair' result . . . how can your produce a fair result for an E/W pair against a N/S pair when they are playing different boards against different other pairs?
For a full answer see https://www.ebu.co.uk/documents/media/bridge-movements-the-maths.pdf
To answer your original question, it depends what it is compared with. If you have 11 tables it is fairer to have a hesitation Mitchell playing 12x2-board rounds than to have a two-winner Mitchell playing 8x3-board rounds. In the first one you play all of the boards and 12 other pairs, in the second you play 24/33 boards and 8 other pairs.
There must be a compromise between fair movements and various other factors: easy to operate, not to confusing for the players, some players not wanting/able to move, short rounds, ...
Re: For a full answer see https://www.ebu.co.uk/documents/media/bridge-movements-the-maths.pdf
I have read this doc and it is as clear as mud (and I've got a bachelor's degree in maths). His approach presumes that all players are of equal ability (in my experience the more expereinced players tend to sit N/S). He is clear that some movemnets are less fair than others. He clearly states that that and arrow switch is necessary to make the Mitchell single winner fairer but clear on why.
Why is it neccessary to arrow switch - what does doing it / not do9ng it add / take away from the fairness of the rfesult?
In my experience, it is the less mobile players that tend to sit N/S.
If your club actually is one where the NS pairs are all very experienced and the EW pairs all relative novices, and you have a Two - winner Mitchell, it is much harder for a NS pair to win, even though it all works fine for the NGS.
The obvious answer is that for a Mitchell movement, you can't have a single winning pair unless you do introduce an arrow-switch. I don't believe that there is a mathematically perfect movement, but all the generally recognised movements are, in my opinion, sufficiently "fair", particularly if it is club sessions that concern you.
Also, as Robin says.
Barrie Partridge - CTD for Bridge Club Live
Let's start off by defining a couple of concepts.
Competition: Two pairs are in competition on a board if the better one does, the worse the other does.
Co-operation: Two pairs are in co-operation on a board if the better one does, the better it is for the other pair.
Example:
Pair 1 Vs Pair 2: NS +400 : director changes this to NS +430
Pair 3 Vs Pair 4: NS +420
Pair 3 is competing against pair 1: Initially they have gained a matchpoint, but when the director changes the score then they get no matchpoints. (and vice versa)
Pair 2 is competing against pair 4: Initially they would have gained a matchpont but have now lost it
Pair 2 is co-operating with pair 3: Inititally they (and pair 3) would have got 1 matchpont: now BOTH get nil.
Pair 4 is co-operating with pair 1: Initially they (and pair 1) would have got no matchpoints: now BOTH get one.
So: let us consider two pairs in a 2-winner Mitchell - one NS, one East west. To make the maths easier we will assume that each round consists of one board and that one matchpoint is at stake each time it is played - like in the ACBL. To give a specific example we will assume an 8-table Mitchell.
Each board is played 8 times so the top is 7 (you beat 7 other pairs).
So: when the EW and the NS play AGAINST each other they are competing for 7 matchpoints.
On the other seven rounds, however, the two pairs are co-operating: (Each of them is hoping that the other will get a good score on the board). Co-operation is the opposite of competition, so on each of those boards the level of competition is -1.
So the net level of competion is (1X 7) - (7 X 1) = 0: EW and NS pairs aren't competing so we must have two winners.
Hopefully it is clear that the competition between each pair in the same direction is 8 (rounds) X 1 (matchpoint) = 8
Now suppose we arrowswitch one round.
First of all we have to work out what the ideal level of competition between all pairs should be. To do that we note that the total number of matchpoints at stake is 56 (8 rounds of 7) and the total number of opponents is 15 (there are 16 pairs and none of them want you to win). So ideally each pair should compete for 56/15 matchpoints against all the others. This of course is not possible. This is just under four.
Now consider two pairs in the same direction. For six boards they are competing in the same direction but for 2 boards (remember the arrow-switch applies at two tables) they are co-operating so the net level of competition has been reduced from 8 to 4 (8 - 2 rounds not competing - 2 rounds of co-operation): There is still too much competition between each direction (4 compared to 56/15) and to reduce it slightly we would have to arrowswitch slightly more often. i.e. just over an eighth of the rounds.
Going to a 7 table Mitchell we get: 42 / 13 = 3 and a bit, and the actual level is reduced from 7 to 3 - which is a fraction too low - this is because too many boards (in theory) have been arrowswitched.
Going to a 9 table Mitchell we get 72/17 = 4 and a bit and the actual level is reduced from 9 to 5 - not enough: we need to arrowswitch some more boards. (If that were possible)
Obviously since there is too much (or too little) competition between some pairs there must be too little (or too much) competition between other pairs. We can work out the variance (or SD) from the mean by simple statistics. A perfectly fair movement would have 0 variance, in real life we must mimimise this variance to ensure that the movement is as fair as possible.
All the mathematics does is generalise this situation and we come up with the answer that just over an eighth of the boards should be arrowswitched to minimse the variance of competition in each direction.
Note that the ability of the players is irrelevent in terms of a single-round arrowswitched. Once more than one round is arrowswitched the level of competition varies as for some competing pairs only one round is arrowswitched uniquely (the other round involves BOTH pairs arrowswitching) - so the generalised maths deals with two pairs that are some distance apart. In this circumstance you want to be a pair (in the other direction) that arrowswitches only one round with the strongest pair in the room..
Something that's worth noting more clearly is that which pairs play which other pairs is only half of what makes a movement fair – which directions they are sitting in is also very important. It is, for example, possible to construct an all-play-all movement that is very unfair because some pairs of pairs usually sit in the same direction as each other, and other pairs of pairs are normally sitting in the opposite direction. (In fact, BBO actually uses such a movement for its "Howell" games, which are not actually Howells – it's sufficiently unbalanced that some pairs of pairs have negative competition between each other, i.e. one pair doing better tends to increase the other's score.)
For example, a naive movement for a barometer howell is to have 1..N start EW at tables 1..N and pairs N+1 .. 2N sit NS at tables N down to 1. Pair 2N are stationary and pairs 1..2N-1 move in a loop. Over 2N-1 round every pair meets every other pair, but the balance is very bad. Any moving pair usually sits in the same direction as pairs that have a pair number (numerically) close to the given pair, and in the opposite direction to pairs half-way round the loop.
The balance can be greatly improved with arrow-switches in the original layout of pairs (e.g. high numbers start EW at some specified tables) and for some values of N (and some number theory) the movement can be made perfectly balanced.