[blackshoe]

The number of possible bridge deals is fairly easy to figure out, or look up. How many possible bidding sequences are there? If we call the former D and the latter B, then the number of deals that have to be described (on average) by each bidding sequence is D/B. I suppose that number would at least give us an idea of how big the problem of designng a bidding system is.

[kenrexford]

That number would be large, but incredibly small.

What I mean is that completely designing a "bidding system" involves much more than simply charting out all possible sequences and defining these. Many variables come into play.

One simple example is in the relatively calm auction 2♦-2♥ (overcall). That is one auction start. However, my guess is that 2♥ has many possible meanings, depending upon the meaning that the opponents assign to a 2♦ call, whether strong with diamonds, weak with diamond, intermediate with diamonds, strong balanced, Flannery, Roman, Mini-Roman, Multi, Minors, Precision, or whatever. Then, extend the auction to 2♦-2♥-2♠-X. What "double" shows will depend not only on the meaning of 2♦, and hence the meaning of 2♥, but also on what Responder meant by 2♠. Toss in any number of passes (P-2♦-2♥-2♠-X, or P-P-2♦-2♥-2♠-X, or P-P-2♦-2♥-2♠-X), and more complexity develops. Maybe vulnerability affects the meaning, or even the form of scoring.

[inquiry]

An IBM scientist published a paper calculating the number of possible auctions at bridge... for easy reference, here is his findings..

The number of possible deals (well known number) is
53,644,737,765,488,792,839,237,440,000 or 5.4 x 10 raised to the 28th power

The number of possible auctions is
128,745,650,347,030,683,120,231,926,111,609,371,363,122,697,557 or
1.3 x 10 raised to the 47th power

He concluded that the number of possible auctions exceeds teh number of possible hands by 2.4 x 10 raised to 18th power to one.

See this study at... Number of Possible auctions at bridge

[kenberg]

inquiry, on May 30 2007, 02:06 PM, said:

An IBM scientist published a paper calculating the number of possible auctions at bridge... for easy reference, here is his findings..

The number of possible deals (well known number) is
53,644,737,765,488,792,839,237,440,000 or 5.4 x 10 raised to the 28th power

The number of possible auctions is
128,745,650,347,030,683,120,231,926,111,609,371,363,122,697,557 or
1.3 x 10 raised to the 47th power

He concluded that the number of possible auctions exceeds teh number of possible hands by 2.4 x 10 raised to 18th power to one.

See this study at... Number of Possible auctions at bridge

More auctions than deals came as a surprise. However, after a little thought:

Without reading the paper, you can easily verify that the number of auctions is substantially higher than the number of deals.
After a club is bid, there are at least ten ways for the auction to continue to 1D. Here are ten.
CD
CPD
CXD
CPPD
CXPD
CX XX D
CPPXD
CPPX XX D
CPPXPD
CPPX XX PD

I haven't checked carefully, there may well be more. But then, after reaching 1D, there are ten ways to get to 1H. And so on. These multiply.
1C-1D-1H-...-7NT takes 34 steps.

So a coarse underestimate is 10 to the 34th, already far more than the number of deals.

It's enough to make 10 to the 47th at least a bit plausible.

Perfectly useless, of course. Still, a surprise.

[EricK]

A more interesting, but unfortunately far less defined, question is how many useful bidding sequences are there.

Whatever meanings you assign to various bids that will automatically make certain future actions impossible. I guess that most bidding systems have roughly the same number of useful sequences (for some value of "roughly"!)

[Echognome]

kenberg, on May 30 2007, 11:48 AM, said:

More auctions than deals came as a surprise. However, after a little thought:

Without reading the paper, you can easily verify that the number of auctions is substantially higher than the number of deals.
After a club is bid, there are at least ten ways for the auction to continue to 1D. Here are ten.
CD
CPD
CXD
CPPD
CXPD
CX XX D
CPPXD
CPPX XX D
CPPXPD
CPPX XX PD

I haven't checked carefully, there may well be more. But then, after reaching 1D, there are ten ways to get to 1H. And so on. These multiply.
1C-1D-1H-...-7NT takes 34 steps.

So a coarse underestimate is 10 to the 34th, already far more than the number of deals.

It's enough to make 10 to the 47th at least a bit plausible.

Perfectly useless, of course. Still, a surprise.

You forgot a few:

CXPPD

CPPXPPD

CPPX XX PPD

CPPXPPXXD
CPPXPPXXPD
CPPXPPXXPPD

[jtfanclub]

Let's see...

2 ways to go from 1 club to X
2 ways to go from X to XX
3 ways to go from (anything) to 1D.

So leaving out all the passes:

1C 1D= 3 possiblities.
1C X 1D= 6 possibilities
1C X XX 1D= 12 possibilities

Total 21.

Repeat 33 more times, and I come to 4*21^34=3.6*10^45 power. That's the number of possible auctions that include EVERY bid from 1 club all the way to 7NT. Then you have all the auctions that jump right from 1 club to 1 heart, lots of possible passes and doubles in there....

[kenrexford]

EricK, on May 30 2007, 03:33 PM, said:

A more interesting, but unfortunately far less defined, question is how many useful bidding sequences are there.

Whatever meanings you assign to various bids that will automatically make certain future actions impossible. I guess that most bidding systems have roughly the same number of useful sequences (for some value of "roughly"!)

You obviously play a different game than I play. I cannot imagine any auction being "impossible," considering some auctions I have actually seen at the table.

[jdonn]

kenrexford, on May 30 2007, 04:59 PM, said:

EricK, on May 30 2007, 03:33 PM, said:

A more interesting, but unfortunately far less defined, question is how many useful bidding sequences are there.

Whatever meanings you assign to various bids that will automatically make certain future actions impossible. I guess that most bidding systems have roughly the same number of useful sequences (for some value of "roughly"!)

You obviously play a different game than I play. I cannot imagine any auction being "impossible," considering some auctions I have actually seen at the table.

1♣ p p 1♦ p p 7NT.

[barmar]

OK, so there are about 10^19 possible auctions for every possible deal. But how many of those auctions reach a plausible contract given the hands. E.g. I'd expect that 7NTXX has the most possible auctions, but the fewest hands make this a reasonable contract to bid. Conversely, there are only 4 auctions that lead to 1C, but many hands where this is an OK place to play.

Basically, for any bidding system to be useful, it needs a significant number of natural, non-forcing bids so that you can set the contract. This significantly constrains the design.

[kenrexford]

jdonn, on May 30 2007, 06:49 PM, said:

kenrexford, on May 30 2007, 04:59 PM, said:

EricK, on May 30 2007, 03:33 PM, said:

A more interesting, but unfortunately far less defined, question is how many useful bidding sequences are there.

Whatever meanings you assign to various bids that will automatically make certain future actions impossible. I guess that most bidding systems have roughly the same number of useful sequences (for some value of "roughly"!)

You obviously play a different game than I play. I cannot imagine any auction being "impossible," considering some auctions I have actually seen at the table.

1♣ p p 1♦ p p 7NT.

This one does seem a tad unusual, I must admit, but one might imagine a simple explanation. Partner opens 1♣ precision. You somehow think this has been doubled, and so you pass. Now, having realized the problem, you take a position.

I have actually experience a somewhat similar auction, in some respects.

Partner opened 1♣, a "standard" opening, insofar as it showed about SAYC opening strength. We did use a very short club, meaning that you opened 1♣ with any balanced hand that might otherwise be opened 1♦ (1♦ being reserved for unbalanced hands).

I responded 1NT after a pass, which showed 2-5 HCP's. The more vulnerable, the more minor cards that were expected. However, with that range, I essentially had a hand qualifying for a pass.

After a double of 1NT, partner leapt to 6♣ as his second call. As it was, a finesse was off. That left us down one doubled, vulnerable, for -200 against their cold 5♠ (-450). Had the finesse worked, we would have made 6♣ and would have set 5♠ one trick.

I'll bet that 1♣-P-1NT-X-6♣-X-P-P-P is not a frequently seen auction. The additional benefit of the calls all working on this hand was a definite bonus.

[kenberg]

At a concert tonight, my mind wandered during Sibelius.
The answer, in compact form, is

(4*22^35 - 1)/3

That is, raise 22 to the 35th power, multiply by 4, subtract 1, divide by 3.

You get 128745650347030683120231926111609371363122697557 just as Ben said.

It's not hard. 22^35=(1+21)^35, and the 21 is the number of possible fills between successive suit (or nt) bids. Etc.

If you want, for example, the number of auctions where no one bids at the seven level, just drop the exponent from 35 to 30.

[barmar]

Of course, if you include auctions where someone makes an insufficient bid and the next player accepts it, there are an infinite number: 1♣(1♣)1♣(1♣)....

[Fluffy]

competitive bidding is pointless, maybe a calculation for just offensive bidding calculating only 2 hands, not 4 might have a point.

[Jboling]

Fluffy, on May 31 2007, 04:03 AM, said:

competitive bidding is pointless, maybe a calculation for just offensive bidding calculating only 2 hands, not 4 might have a point.

Fluffys right, in practice you can't rely on the opponents. I found 68719476735 (6.87*10^10) auctions with the opponents silent.

The longest auction you can have is pass-1♣-1♦-...-7NT-pass, consisting of 37 bids. Then you can leave out one of the first 36 bids (you can't leave out the last pass), and get 36 auctions consisting of 36 steps. And then leave out two of the bids, and get 36!/34!/2!=630 auctions consisting of 35 steps. And so on, all the way to the 36 possible sequences where we only have two bids.

[BillHiggin]

Jari, on May 31 2007, 06:15 AM, said:

Fluffy, on May 31 2007, 04:03 AM, said:

competitive bidding is pointless, maybe a calculation for just offensive bidding calculating only 2 hands, not 4 might have a point.

Fluffys right, in practice you can't rely on the opponents. I found 68719476735 (6.87*10^10) auctions with the opponents silent.

The longest auction you can have is pass-1♣-1♦-...-7NT-pass, consisting of 37 bids. Then you can leave out one of the first 36 bids (you can't leave out the last pass), and get 36 auctions consisting of 36 steps. And then leave out two of the bids, and get 36!/34!/2!=630 auctions consisting of 35 steps. And so on, all the way to the 36 possible sequences where we only have two bids.

Number the bids 1..35 so that bid(1) = 1♣ and bid(35) = 7N

AuctionsWhichStopBelowBid(n) = 2^n - 1
Auctions which stop below 1♣ = 2^1 - 1 = 1 ; i.e. pass pass

AuctionsWhichContractForBid(n) = 2^n

AuctionsWhichContractForNoMoreThanBid(n) = 2^(n+1) -1
Max auctions = 2^36 - 1 ; the same number that Jari produced.

[kenberg]

Jari, on May 31 2007, 06:15 AM, said:

Fluffy, on May 31 2007, 04:03 AM, said:

competitive bidding is pointless, maybe a calculation for just offensive bidding calculating only 2 hands, not 4 might have a point.

Fluffys right, in practice you can't rely on the opponents. I found 68719476735 (6.87*10^10) auctions with the opponents silent.

The longest auction you can have is pass-1♣-1♦-...-7NT-pass, consisting of 37 bids. Then you can leave out one of the first 36 bids (you can't leave out the last pass), and get 36 auctions consisting of 36 steps. And then leave out two of the bids, and get 36!/34!/2!=630 auctions consisting of 35 steps. And so on, all the way to the 36 possible sequences where we only have two bids.

Added: Bill, you got up earlier than I did, I see. I'll leave this up anyway.

In short form, 2(2^35-1)+1=2^36-1
This is 68719476735
In any non-contested auction look at the sequence of bids. Assuming the hand is not a pass-out there are 2^35-1 possible sequences (the number of non-empty subsets of a set of 35 objects). Double it to allow for the possibility the pair's auction begins with a pass. Then add 1 for the pass-out.

I suppose this figure is marginally more relevant than the total number of legal auctions (not counting the insufficient bid auctions) but I doubt anyone really cares about either number.

But it's summer, life is slow, some of us have a weird sense of entertainment. I was surprised first by the large number of legal auctions and then by how easy it was to calculate it. The combinatorics used don't go past what used to be taught in high school. Now, however, everyone wants to stuff every child who has mastered the addition of fractions (and maybe some who haven't) into a calculus course so that the principal can boast about the number of advanced placement students. This leaves no time for casual musing about the number of bridge auctions.


Btw. 2^36 is 2^6 times (2^10)^3. Now 2^6=64 and 2^10 is about 1000 (1024, actually) so we easily estimate, without technology, that there are approximately 64,000,000,000 uncontested auctions. As notes, some of us have a weird sense of entertainment.


For anyone still reading: On Studio 60, where they do weird/sham news stories, they mentioned that a mathematician (Perelman) won a large prize (a million dollars, which he turned down) by showing that the Poincare conjecture, stating that a bunny is a sphere, is true. This is more or less a correct statement of the result. Well, mostly less. But it was a hoot to see it mentioned at all.

[kenrexford]

I just tried to use my other computer to calculate the number of different auctions that would actually occur on a single, specific deal in a real game with real people making what they believe to be intelligent decisions. My computer crashed.

[dogsbreath]

hi

Would someone care to mention to my random Indy partners that among this vast choice of possible calls the one they seem to forget or ignore is ....PASS.

Rgds Dog

[Free]

Fluffy, on May 31 2007, 10:03 AM, said:

competitive bidding is pointless, maybe a calculation for just offensive bidding calculating only 2 hands, not 4 might have a point.

That's just too easy... 35th Fibonacci number: 9227465. Well ok, you can also pass in 1st seat, so 14930352 if you want to be a pain in the ass

edit: I forgot all auctions where you can pass at lower levels... The fibonacci numbers are the possible auctions you can have to 7NT alone. So basically it should be a sum of all the fibonacci numbers F1 to F37 (pass in 1st seat and passout are also possible): 63245985