A numbers game

[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]

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]

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]

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]

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]

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]

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]

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]

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]

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

[kenberg]

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 :P

 

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

I don't see how fibonacci numbers can be used. And since your answer is in disagreement with the answer Jari, Bill and I get, I doubt that it's right. I have been wrong before, possibly even Bill has been wrong before, but our argument is simple enough to make it seem unlikely this is one of the times. Also, check the simplified game below where it can be explicitly checked.

 

Take the 35 possible bids. (Where bid means some number of something, not a pass). Any non-empty subset of these can occur (and in only one way) in an auction that begins and ends with a bid (followed by a pass). There are 2^35-1 non-empty subsets. Double this, to include the auctions that begin with a pass and end with a bid. Then add in the pass-out. So we get 2(2^35-1)+1 = 2^36-1=68719476735

 

What we did was to first count all auctions such as 1H-3N-7C-P

then count all auctions such as P-2H-3N-5D-P

and then PPPP

 

It seems clear we have counted them all and counted none of them twice.

 

To check this, take an abbreviated game. Suppose we want the number of auctions ending at or before 1H. So the available bids are three (C,D,H) instead of 35 (C,D,...7NT)

 

Our argument would give us 2(2^3-1)+1=2^4-1=15 auctions. Here they are:

 

The seven starting with a bid:

CP

CDP,DP

CDHP, DHP, HP

 

The seven starting with a pass but having a bid

PCP

PCDP,PDP

PCDHP, PDHP, PHP

 

The one pass out

PPPP

 

The Fibonacci numbers begin 1,1,2,3,5,8...

Whether you regard 1,1,2 or 1,2,3 as the first three, you aren't going to get 15 (or 7) by summing them.

Ah. You do get 7 by summing 1,1,2,3. But you don't get 2^4-1 by summing 1,1,2,3,5. So summing F's seems to be out.

 

Counting the auctions with all four players allowed to call proceeds along similar lines, using nothing more than binomial coefficients, similar to how Jari did this restricted problem

[fred]

About 1 month ago I received an e-mail from Bill Gates that basically said:

 

I was on an airplane and I was bored so I tried to figure out the number of possible bidding sequences in bridge. I came up with the following 45-digit number. When I checked with the Official Encyclopedia of Bridge, I saw that my number was wrong by a factor of 4/9. Can you check my calculations please?

 

Not surprisingly, Bill's method of calculation was very elegant (I won't go into the details here), but I was pleased with myself for being able to spot 2 mechanical errors that he made. One of these errors (forgetting that an auction can start with 3 passes - not just 0, 1, or 2) added a factor of 4/3 to the final number. The other error had to do with the way Bill computed the number of possible sequences of Passes, DBLs, and RDBLs after the final call of the auction. Correcting this reduced the final number by a factor of 1/3.

 

When you multiply 4/3 by 1/3 you get 4/9 and the correct 45-digit number :)

 

Bill has an amazing ability to work with numbers in his head. We were once discussing how to compute bridge probabilities so I asked him as an excercise to try to figure out the odds of a 3-2 break. Naturally I expected him to do some serious scribbling on paper and, after several minutes, to come up with the correct 68%ish number.

 

What I got instead was a fraction made up of 2 smallish numbers (sorry I can't remember what they are) that described the odds of 3-2 vs. 4-1. This did not take Bill more than a minute and he did not require any paper at all! I had never heard of bridge probabilities being expressed in this way and I must admit I had my doubts. When I did the math myself (requiring paper!) I saw that he was right :)

 

Fred Gitelman

Bridge Base Inc.

www.bridgebase.com

[cherdano]

Bill has an amazing ability to work with numbers in his head. We were once discussing how to compute bridge probabilities so I asked him as an excercise to try to figure out the odds of a 3-2 break. Naturally I expected him to do some serious scribbling on paper and, after several minutes, to come up with the correct 68%ish number.

 

What I got instead was a fraction made up of 2 smallish numbers (sorry I can't remember what they are) that described the odds of 3-2 vs. 4-1. This did not take Bill more than a minute and he did not require any paper at all! I had never heard of bridge probabilities being expressed in this way and I must admit I had my doubts. When I did the math myself (requiring paper!) I saw that he was right :)

I suspect I know how Bill Gates came up with this number (that is, I know how most people with some math training would compute this number), but here is a way to get to this number that I believe is more intuitive for most bridge players:

 

We know there are 20 4-1 breaks, and 40 3-2 breaks. This would superficially imply the ratio of 3-2 vs 4-1 is 2:1 -- except as we all know every specific 3-2 break is slightly more likely than every specific 4-1 break. So how much more likely?

 

Well, we just need to find this out for two specific such breaks! Say we are missing KQ432, it is ok if we just know exactly how much more likely a K32-Q4 break is compared to a KQ32-4 break. Well this is easy using vacant space, after we deal out the "known" K32-4, then LHO has 10 spaces left to get the queen, versus RHO's 12, so it is 6:5 for the 3-2 break. So to get the relative probabilities right, we have to multiply the "40" above by 6, and the 20 by 5, which gives 240:100 or 12:5, or in other words a probability of 12/17 for the 3-2 break (once we know the suit doesn't split 5-0).

 

Will this help you at the bridge table? I doubt it, but maybe one day it may help you solve one of those evil suit combinations on the last page of the bridge world, when you don't know whether to play for 4 specific 3-2 breaks or 5 specific 4-1 breaks...

[kenberg]

For counting the hands, what I did and what I imagine most everyone else who counted them did, is as follows:

 

Say that a bid is a call of some number of somethings, and a call is anything else. We need to count the number of possible calls between successive bids.

There can be 0, 1,2 passes.

There can be a double by one of two people followed by 0,1 or 2 passes.

There can be a double by one of two people, later a redouble by 1 of two people, and then 0,1 or 2 passes.

 

The first can happen 3 ways, the next 2 times 3=6 ways, the last 2 times 2 times 3 =12 ways.

 

 

3+6+12=21

 

Similar counting shows there can be 7=21/3 sequences of calls after the final bid, and there obviously are 4 possible sequences before the first bid. If we take a specific set of k bids from the choices 1C through 7NT and ask for the number of auctions with exactly those bids we get 4X(21)^(k-1)X(21/3)=(4/3)(21)^k. There are B(35,k) subsets of a 35 element set, where B(35,k) is the binomial coefficient.

So there are (4/3)XB(35,k)X(21)^k sequences in which exactly k bids occur, except for the special case k=0.

 

If we sum B(35,k)(21)^k over all k>0 then the binomial theorem gives us the answer ((1+21)^(35))-1.

 

 

Multiplying by 4/3 and then adding 1 gives us the desired formula.

 

I did this during a concert, BG did it during a flight. I also forgot to divide the last 21 by 3 when doing it in my head. My long lost twin! Do I get to share in the company's profits?

[Guest Jlall]

my head is gonna explode :angry: you people are pretty smart lol.