The enclosed paper from John Manning should give you all the mathematics you need.

A further proof of switching 1/8 of the rounds (if further proof be needed) is my little example of an 8-table tournament, playing 1-board rounds in a Mitchell movement. The best pair in the world are sitting at N/S 1 - the other 15 pairs are all equally matched.

Obviously, the best pair in the world would win any tournament - but in which direction would you like to sit in order to come second, and does your choice depend on how many arrow-switches (if any) there will be?

A top on a board is 14.

N/S 1 score 14 x 8 = 112 match points (100%).

All other N/S pairs score only 6 per board (N/S 1 always have a top). 6 x 8 = 48 match points.

E/W score 8 match points per board except when they go to E/W 1, when they score 0. So, their score is 7 x 8 = 56 match points (50%).

So, you are better off sitting E/W than N/S.

N/S 1 still score 112 match points.

The other N/S pairs score 6 when they sit the same way as N/S 1, and 8 when they sit the other way. Note that they sit the other way on 2 of the 8 boards: one when they themselves arrow-switch - the other in whichever boards N/S 1 arrow-switch.

So, they score (6 x 6) + (2 x 8) = 52 match points.

Similar calculations apply for E/W. Except for the pair who arrow- switch against N/S 1 (we'll come back to them in a moment), they score 8 match points on 5 boards (sitting the opposite way to N/S 1), 6 match points on 2 boards and 1 board on which they score zero (when they play against N/S 1).

(8 x 5) + (6 x 2) + (0 x 1) = 52 match points.

Lo and behold - the same score as the N/S pairs! This is the *'proof'* of switching 1/8 of the rounds.

Of course, there was a very lucky E/W pair - the one who switched against N/S 1. They still score a zero on the switch round, but they score 8 match points on all the other rounds for a final score of 56 (this because they never played a board the same way as N/S 1).

I won't go into this in detail, but you can see that as 0 switches was unfair to N/S and that 1 switch equalled things out, then 2 switches will move things too far the wrong way and will actually favour the N/S pairs.

This was a simple example. In general terms, less than 1 switch in 8 favours the pairs sitting in the weakest direction and more than 1 switch in 8 favours the pairs sitting in the strongest direction. I hope this helps by way of background information to the enclosed masterpiece by Dr Manning.

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