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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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