[kenberg]

Free, on Jun 1 2007, 03:48 AM, said:

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

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]

fred, on Jun 3 2007, 08:40 AM, said:

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?

[kenberg]

We just do it to impress the chicks.