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A minor Fibonacci issue symmetry vs probability

#1 User is offline   pilun 

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Posted 2022-October-22, 03:11

8 is a Fibonacci number. If you have a group of 8 hands, you can resolve them all within 5 steps. (3+2+1+1+1)
The question is. Should you?

Take the example of 3-suiters with both minors. There are eight of those.
This is a logical Fibonacci way to group them:

4-0-4-5, 4-0-5-4, 5-0-4-4 (so 3 hands with a heart void)
4-1-4-4, 1-4-4-4 (2 of those)
0-4-4-5
0-4-5-4
0-5-4-4

If 2♥ is the bid to show 3-suited with both minors, you can get them all out by 3♠, zooming with three shapes.
All good and easy enough to remember. Is it efficient?

4441 hands are more common. If we look at hands short in hearts, 4-1-4-4 is twice as common as the three heart void hands put together.
With the scheme above, 4-1-4-4 comes out at 3♥, 1-4-4-4 comes out 3♠ with zooming.

One asymmetric way to group them is like this

2N = 4 spades (4 hands)
3♣ = 1-4-4-4
3♦ = 0-4-4-5
3♥ = 0-4-5-4
3♠+ = 0-5-4-4

then
3♦ = 4-1-4-4
3♥ = 4-0-4-5
3♠ = 4-0-5-4
3NT = 5-0-4-4

So one hand finishes high. In fact all the heart void hands are a step higher. (There are 2 zooming shapes, not 3)
To compensate, the 4441s come out earlier, which has to be good.

1-4-4-4 at 3♣
4-1-4-4 at 3♦

That's a gain of two steps for both the 4441s. Seems a better deal; getting the more common hands out earlier.

For us, there is another reason for an asymmetric split. Two of our major openings deny four cards in the other major.
As a consequence, when describer shows a 3-suiter, there are only four shapes to show.
Four is not a Fibonacci number so there is no gain in splitting. Just show them in a line, naturally starting with the 4441.
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#2 User is offline   DinDIP 

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Posted 2022-November-05, 01:44

KK Relay (Woolsey and McCallum) decided to show shapes symmetrically. To achieve that they
(a) use the cheapest step (2♠) after teller shows a three-suiter with both majors (1♣-1♥-1♠-1N-2♣-2♦) to cancel the message of a three-suiter and replace it with a 6+card semi-solid major; and
(b) use 3♣ to show a 4441 with a singleton A, K or Q (so that other 4441s are always shown by 3♦)

I think symmetry is a big plus but one step can matter so I'm not keen on losing space.
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