[benlessard]

If we take as example

1C--1S (strong, GF bal)
1NT--?? (ask...)

Here any bids is available to show a bal hands GF. So basically responder will show his hand and opener will just bid the next step to know more.

Some say that the frequencies of 2C,2D,2H,2S should follow Fibonacci sequence but its not what I get.

Bidding the 1st step will cost 2 space not one because partner will bid the cheapest bid to know more.

Bidding the 3rd step will cost 4 spaces. As an example

2C-2D-2H-2S-??
2H-2S-??

so bidding the first step followed by the first step again is equivalent in space consumption than to bid the 3rd space.

The frequencies should be the inverse of the space cost of the bid. For example to consume 7 spaces (you bid 3C and partner will relay at 3D) you can..

bid step 6 (3C-3D)
4+1,3+2,2+3,1+4 (2S-2NT-3C-3D, 2C-2D-3C-3D)
2+1+1,1+2+1,1+1+2. (2D-2H-2S-2Nt-3C-3D) etc..

all these sequences should have equivalent frequencies because they consume equivalent space.

So the cost of a bid is cost = x+1 because the next relay is included in the cost.

So we get

2+3+4+5+6+7+8+9+10 ...

For simplicity lets stop at 10 and find the lowest denominator.
2/2 + 2/3 + 2/4 + 2/5 +2/6 +2/7 +2/8 +2/9 + 2/10

So the first step bid cost 2 should twice as frequent than step 3 (cost 4) and three times more likely than step 5 (cost 6)
a lowest denominator would be 420

420 + 280 + 210 + 168 + 140 + 120 + 105 + 88 = 1531

27%, 18% , 13.7% , 11% , 9% , 7,8% 6.85% , 5.75% = 100%

So this series doenst follow a Fibonacci sequence.

Im not good at math enough to be sure, could soemone tell me where the frequencies of bids in relays = Fibonacci started ?

[awm]

Suppose 3NT+ are never shape relays, and you want shape completely described by 3NT. If the first relay is 3♥, partner can show two shapes by 3NT (one by bidding 3♠, after which there are no more shape relays, the other by bidding 3NT). If the first relay is 3♦, partner can show three shapes (by bidding 3♥, 3♠, 3NT.. note that if partner bids 3♥ you cannot usefully relay again because you're not supposed to go past 3NT). If the first relay is X, then partner's next call will be one of X+1, X+2, etc. This gives you two possibilities:

1. Partner's next call is X+1, so you can resolve all the shapes that you could resolve when the first relay was X+2.
2. Partner's next call is one of X+2, X+3, ... etc which is exactly the same set of calls he could make if the first relay had been X+1, so you can resolve that many shapes.

In total the number of shapes I can resolve when the first relay is X is the sum of the number I can resolve with the first relay being X+1 and X+2. Continuing from above:

First relay...

3♥ = 2 shapes
3♦ = 3 shapes
3♣ = 5 shapes
2NT = 8 shapes
2♠ = 13 shapes

and so forth; the Fibonacci numbers. If I wanted shape to end at 4♣ (say) instead of 3NT it just shifts things a step.

[benlessard]

ok thanks a lot, if Ive understood correctly in the end there is always 3 branch, 3H-3S-3NT is a sequence but it cannot really carry information in theory. Since after 3H a 3S relay would "force" 3NT. However in practice there is a slight difference since the branch that end at -2 have twice the value I think. For a balanced hand bidding 3H will allow opener to bid 3S last train.

EX if S is trumps.

ending in 6S or ending in 6H have the same value. since opener need to guess 6S or 7S right now
over 6D opener cannot really relay anymore but he can do a final last train by bidding 6H.

But anyway your explanation is clear and convincing and im sure my maths was wrong.

[Cascade]

benlessard, on 2014-September-06, 20:04, said:

If we take as example

1C--1S (strong, GF bal)
1NT--?? (ask...)

Here any bids is available to show a bal hands GF. So basically responder will show his hand and opener will just bid the next step to know more.

Some say that the frequencies of 2C,2D,2H,2S should follow Fibonacci sequence but its not what I get.

Bidding the 1st step will cost 2 space not one because partner will bid the cheapest bid to know more.

Bidding the 3rd step will cost 4 spaces. As an example

2C-2D-2H-2S-??
2H-2S-??

so bidding the first step followed by the first step again is equivalent in space consumption than to bid the 3rd space.

The frequencies should be the inverse of the space cost of the bid. For example to consume 7 spaces (you bid 3C and partner will relay at 3D) you can..

bid step 6 (3C-3D)
4+1,3+2,2+3,1+4 (2S-2NT-3C-3D, 2C-2D-3C-3D)
2+1+1,1+2+1,1+1+2. (2D-2H-2S-2Nt-3C-3D) etc..

all these sequences should have equivalent frequencies because they consume equivalent space.

So the cost of a bid is cost = x+1 because the next relay is included in the cost.

So we get

2+3+4+5+6+7+8+9+10 ...

For simplicity lets stop at 10 and find the lowest denominator.
2/2 + 2/3 + 2/4 + 2/5 +2/6 +2/7 +2/8 +2/9 + 2/10

So the first step bid cost 2 should twice as frequent than step 3 (cost 4) and three times more likely than step 5 (cost 6)
a lowest denominator would be 420

420 + 280 + 210 + 168 + 140 + 120 + 105 + 88 = 1531

27%, 18% , 13.7% , 11% , 9% , 7,8% 6.85% , 5.75% = 100%

So this series doenst follow a Fibonacci sequence.

Im not good at math enough to be sure, could soemone tell me where the frequencies of bids in relays = Fibonacci started ?

I always thought the Fibonacci numbers came in from the number of shapes that you could show starting at a particular bid and below some other bid like 3NT. I have always worked these out starting from 3NT and working backwards like so:


3NT 1 shape
3♠ 1 shape
3♥ 1 shape (3♠ relay and only one step left)
3♦ 2 shapes = 3NT + 3♠
3♣ 3 shapes = 3NT + 3♠ + 3♥
2NT 5 shapes
2♠ 8 shapes
2♥ 13 shapes
2♦ 21 shapes
2♣ 34 shapes
1NT 55 shapes
1♠ 89 shapes
1♥ 144 shapes
1♦ 233 shapes
1♣ 377 shapes (for the forcing pass players)

This shows the number of shapes you can have at each level and those frequencies follow a Fibonacci sequence.

[Zelandakh]

Exactly! Good explanations from both Adam and Wayne. Just to add in relation to your other thread that in adjusting this to a bi-directional system, the 3rd from last step can now be said to hold 2 hand patterns - the Last Train as you put it - so a reasonable guideline to an ideal for pattern frequencies here would be 1-1-2-3-5-8, etc. The difference here is that the number for the first step does not exactly equal the number for steps 3+ but is 1 hand type more. In practise this makes little enough difference to the frequencies as to be negligible. Just think or the ratios for 1st step:2nd step:others as 40:20:40 and you will not go far wrong.

[benlessard]

Zelandakh, on 2014-September-07, 06:55, said:

Exactly! Good explanations from both Adam and Wayne. Just to add in relation to your other thread that in adjusting this to a bi-directional system, the 3rd from last step can now be said to hold 2 hand patterns - the Last Train as you put it - so a reasonable guideline to an ideal for pattern frequencies here would be 1-1-2-3-5-8, etc. The difference here is that the number for the first step does not exactly equal the number for steps 3+ but is 1 hand type more. In practise this makes little enough difference to the frequencies as to be negligible. Just think or the ratios for 1st step:2nd step:others as 40:20:40 and you will not go far wrong.

In bi-directionnal the binary model is working ok, its the +1 cost (realy doesnt give any information) that make it a Fibo model here. However a more precise model would be counting the bids that end in 3H (LTTC or -2) since they are a quite different than those who end at -1 or 0 (3S,3NT) ive used 0.001 to count them

stating point
3nt = 0 (forced to pass)
3S = 0 (forced to bid 3NT)
3H = 2 (3S or 3Nt)
3D = 3.001 (3H,3S,3NT) but over 3H opener could bid 3S lttc
3C = 5.001
2NT = 8.002
2S = 13.003
2H = 21.005
2D = 34.008 (you have 8 sequence that lead you to 3H, but 13 each for 3S&3NT)

For practical purpose I think the -2 are worth almost twice the value of the 3S/3NT endings. If we count them as 2pts we get

1NT =89.021 = 110
2C = 55.013 = 68
2D = 34.008 = 42
2H = 21.005 = 26
2S = 13.003 = 16
2NT = 8.002 = 10
3C = 5.001 = 6
3D = 3.001 = 4
3H = 2

This mean that of all the 110 sequences-pts there is 34 sequences that end in 3S,34 that end in 3NT and 21 that end in 3H (but they are worth 2 pts) for 110 pts

Of the 110 sequence pts-
42 are under 2C-2D-?? for 38%
26 are under 2D-2H-?? for 23.6%
16 are under 2H-2S-?? for 14.5%
10 are 2S-2NT-?? for 9.1%
6 are 2NT-3C-?? for 5.5%
4 are 3C-3D-?? for 3.6%
2 are for 3D-3H-3S/3NT for 1.8%
2 are for 3H direct =1.8%
2 are for 3S/3NT direct = 1.8%

[nennen]

I found a recent article in Bridge Winners which gives
a very good and completely general explanation of the relé-answer methods and their relationship with Fibonacci sequence. Worth a "coup d'oeuil"

http://bridgewinners...nacci-grouping/