[benlessard]

Let's do a thought experiment. Two super-computer AI are bidding millions of hands after the start 1S-2Nt (2Nt is GF with 4 trumps support forcing to 4S, not suitable for splinter) . There is no opponents for the bidding & opener is going to play all the hands (so no rightsiding, no worry about lead directing X, no worry about leaking information to the defense and you cannot stop before 4S). The computers are programmed to reach the best contract & any bid they do over 4H is to play, so they have between 3C and 4H to give or ask information for finding what is the best contract. Computers like to work with one and zeros so thats what we are going to do to. A bit is one "one or zero" a byte (or octet) is 8 bit.

If a bid is made its one if its bypassed its a zero.

2NT...

3C-3D-3H-3S-4H = 11110001 (four bids followed by 3NT,4C,4D that are bypassed,

3C-3H-3NT-4D-signoff = 10101010 (the last zero = 4H is bypassed)

3S-3NT-4D-4H = 00011011

a Bytes/octet or 8bit can carry values from 0 (00000000) to 255 (11111111) so in my example there is 255 bidding sequence that allow you to get from 3C to 4H (not counting all the final contracts possible). Any number of bidding steps can be seen as a binary string so 6 step is 5 2exp6 for 2x2x2x2x2x2 = 64 possible sequences.

Computers have perfect memories so they can easily encode the meaning of each bytes (bidding sequence) and they will not have bidding misunderstanding. For them all the sequences have the same complexity and no sequences will have unclear meanings. In short they will not "think" in term of one bid they will think in term of one byte even if they are not in control of all the bids.

What is very important to see is that bidding the first bid 3C (1???????) or bypassing it (0???????) represent 128-128 sequences or 50%-50% of the total number of sequences. The computers will of course need to maximize the use of all the possible sequence so they will spread the meaning of the bids according to the total bidding space. They cannot afford to have sequence that do not exist or don't have meanings, they also need need to try to make each bytes as frequent and as significative than the others.

However note that the goal is not to sort bridge hands, the goal is to bid to the best contract, so some hands frequencies will not represent their "bidding" values/space, for example two minimum hands balanced hand facing each other is a frequent occurence, however since the final contract will often be 4S they are not valuable when the goal is reaching the best contract. So the bidding space will match the bidding value of a hand not its intrinsic frequency.

My 2 main hypothesis are

1- That the computer will maximize the bidding space by using the responses frequency that would look like...

3C=50% (1???????)

3D = 25% (01??????)

3H - 1/8 (001?????)

3S = 1/16 (0001????)

1/32, 1/64, 1/128

2- The computer will start by showing shapes and finish by showing extras or not.

IRL it make sense to show strenght early, if both are weakish you stop leaking information and quickly signoff, however this method is not optimal if you want to maximize the bidding space.

[Zelandakh]

I think bidding theory says that Fibonacci numbers are optimal. I make it that 3♣ should be about 39% and in general a 40-20-40 split is a decent guideline (but usually not achievable in practise).

Why is it that you think showing strength early is not optimal for bidding space? You can also mix and match as in the responses I have been using (3m = min/mid + shortage; 3M = min/mid without shortage). Even if you just had 3♣ min and others showing extras, I am not sure what you are losing whenn Opener shows their distibutional information on the next round.

[BillHiggin]

I see a complication related to:
Which computer makes the final decision? Answer - as the OP is presented, either might.

The final decision will be based on what that computer knows - its own holding and what it has learned about the other holding. Any information it has told about its own holding will have virtually drifted off to bit heaven! (well, it may have been used to modify the messages sent by the other).

The decision maker can be deduced from Ben's bytes simply by the parity of the byte - even parity means that the opener has made the final call and odd parity that responder has. But, that is known only after all the bids have been made.

To demonstrate the dilema, allow me to deviate from the original scenario and present this as two humans with a defined system might do it. They might decide to have opener describe his hand and responder to merely relay for more information (creating that Fibonacci like sequence that Zelandakh mentions). Because of the constraints on end of auction, we get the curious condition where a bid of 4H by responder at any point in the auction whould essentially say "now that I know all about your hand, you should place the final contract". Of course, since opener knows nothing other than the meaning of the original 2N call, this becomes essentially a transfer to 4S and so all the sequences ending in a 4H call by responder are equivalent to the same sequence where responder instead bids 4S himself.

The binary distribution of bids is appropriate if we wish to maximize the total information presented to the opponents (not the sort of priority we really want). Fibonacci distribution of first rebids by opener would be best if we wanted responder to be captain. If we want opener to be captain, then almost all his first rebids should be 3C (maybe he already knows enough to bid 7N!)

[helene_t]

 Zelandakh, on 2014-September-04, 04:55, said:

I think bidding theory says that Fibonacci numbers are optimal. I make it that 3♣ should be about 39% and in general a 40-20-40 split is a decent guideline (but usually not achievable in practise).

Fibonacci is optimal for relay systems. What Ben describes is optimal if the aim is to maximize mutual information exchange.

Edit: sorry, just noticed that Bill said the same.

Probably the best allocation of the captain's bids is that he usually makes the relay but sometimes makes some more specific asking bid. Then the optimal allocation of relayees bids will be somewhere between the binary splits and Fibonacci.

[benlessard]

Yes I know very well that symmetric bridge information (same amount of information going from A to B vs B to A) is not really optimal since picking the final contract is a one man job not a 2 man job. If bridge bidding was like at the end both players pick a contract and we make some sort of average between the 2 contracts than my guess is a pure 50-25-12.5 would be optimal.

Same for spiral scan.

If one player hold no cards and ask his partner for cards starting with A

1- no card A
2 card A but not B
3 A+B no C
4 A+b+c no D
5 ABCD

These responses will be very close to 50-25-12.5...

However captain is unlikey to have no cards so its possible scheme of 2 cards like

1- A or B but not both
2 no A, no B
3- 2 of the next 3 cards
4 3 of the next 4 cards
5 4 of the next 5 cards

are better since they take into account that responder is likely to have some cards.

captain next inquiries could be

1 asking for 2 next cards
2 Asking for 2 next cards but I need to know if you held A or B.

Anyway these are really interesting problems but very hard one imo. Sometimes you are close to the limit where you risk to bypassing your last making contract. I think in a "perfect" scanning methods they change according to your overall strenght to the lenght in your suits and according to the safety of the "landing strip". So irl they are not practical.

[mycroft]

Heh, another instance of the "I'm bored" call (more serious systems call it "Stop Signal") - although usually it's 4♦, not 4♥. "Hey pard, I'm bored; I'm going to pass whatever you bid."

[benlessard]

I think the proper term is "Polish end signal" and its 4D in the original version, ive suggest and im sure others did that 3S should also be a PES to 3Nt. This may wrongside 3NT however However its clear that cheapest call artifical are often multiplier of bidding sequence like 2NT lebenshol or xyz.

[Zelandakh]

I do not get all this end signal stuff. Say the last call was 4♦. Now we can transfer one bit of information by bidding either 4♥ or 4♠. 4♥ is clearly not an end signal - it provides information that partner can use to choose either 4♠ or something else.

Also, one of the things that playing relay methods has taught me is that the space requirements for them match the space requirements for natural. It is only that in natural there are typically win sequences (where we have extra space at the end) and lose sequences (where we run out of space and have to guess). Therefore I believe the 40-20-40 rule works just as well for bi-directional systems as for mono ones, particularly when we are talking about a codified system as here.

[mycroft]

Right. And the bit that may be the most useful at this point is "I've told you everything, you decide" for 4♥, and "I haven't told you everything yet, here's some more" for 4♠ and up.

[benlessard]

The polish end signal is when trumps is not known and relayer still want some options.

your in 3NT and opener showed a 5530.

4C is an ask/QP/RKC etc
4D is a puppet to 4H to stop somehwere
4M is something else (could be NF slam invitation) could be voids could be rkc 2,rkc 3 etc.

the 40-20-40 is because you count the next relay step as being part of the bidding space cost.

1step = cost 2
2nd step cost 3
3rd step cost 4
4th step cost 5

So this make the cheapest bid least efficient.
Not sure if this follow Fibo numbers im in a rush for time now.

[benlessard]

Ok playing a relay system the numbers should be

asssuming there is 9 steps. The frequency is inverse of the space cost of the bid.

cost = x+1 because the next relay is included in the cost.

So we get

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

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 doenst follow a Fibonacci sequence.

[yunling]

I believe the real distribution is something between Fibonacci and Geometric, but closer to Fibonacci.
The lower bids contains more hand types and leaves more space, so its nature is to ask more of partner's hand rather than show mine; the higher bids are just the opposite.
However, for each of the two person at the table, the information revealed to the other is not symmetric. After 1♠-2NT-3x, certainly responder knows more about opener's hand, instead of the other way around, so he has more desire to ask rather than show, which indicates that for him, the frequency of step +1 bid is always far above 50%(if it is 100% then we get Fibonacci).
Using the same logic we can conclude that opener will bid 3♣ less than 50% of the time.

[Zelandakh]

No Ben, the numbers for pure relays (Opener describing) are:-

4♠ = 1
4♥ = 1
4♦ = 1
4♣ = 2
3NT = 3
3♠ = 5
3♥ = 8
3♦ = 13
3♣ = 21

but I shifted these up a step in my earlier calculation to take account of it being bi-directional, hence 4♦ = 2 and so on. In symmetric the numbers for steps 3+ will always be equal to that of step 1 so your result of 27/18/55 is obviously wrong.