[Stefan_O]

How many possible, different bidding-auctions are there?
(We are talking abt auctions that are legal according to bridge-rules, not only the ones that are sensible...)

Just discovered this calculation, and it's actually beyond staggering

Before you check the link below, I invite you to work out with pen/paper -- or just have a guess :

Assume North deals, and opens 7♥.
In how many different, legal ways (according to bridge-rules) can such an auction proceed before it completes?
[a] 38
[b] 388
[c] 3388
[d] infinity

http://tedmuller.us/...dgeAuctions.htm

[hrothgar]

Stefan_O, on 2016-August-14, 14:27, said:

How many possible, different bidding-auctions are there?
(We are talking abt auctions that are legal according to bridge-rules, not only the ones that are sensible...)

Just discovered this calculation, and it's actually beyond staggering

Before you check the link below, I invite you to work out with pen/paper -- or just have a guess :

Assume North deals, and opens 7♥.
In how many different, legal ways (according to bridge-rules) can such an auction proceed before it completes?
[a] 38
[b] 388
[c] 3388
[d] infinity

http://tedmuller.us/...dgeAuctions.htm

Just think how many more possibilities there are once we start allowing for insufficient bids...

[kenrexford]

When you add in definitions, it also adds.

For example, 2D-P-P-P is one auction. However, there are likely several hundred variations on what Opener and Responder will have in this one auction.

[barmar]

kenrexford, on 2016-August-14, 18:37, said:

When you add in definitions, it also adds.

Since the calculation doesn't care whether the auctions are meaningful or sensible, definitions are irrelevant.

[kenrexford]

barmar, on 2016-August-16, 08:33, said:

Since the calculation doesn't care whether the auctions are meaningful or sensible, definitions are irrelevant.

Sounds like some of my partnerships.

[helene_t]

3388 sounds about right

[Jinksy]

I remember reading that there are more possible auctions than possible 52-card distributions.

If you design your auction (cooperatively with the opps) to just describe every card in your hands ('twas said), you will always have described the last cards by the 6♦ bid.

[barmar]

Jinksy, on 2016-August-23, 10:16, said:

If you design your auction (cooperatively with the opps) to just describe every card in your hands ('twas said), you will always have described the last cards by the 6♦ bid.

So you have a choice between describing your hands completely and getting to a reasonable contract.

[The_Badger]

Stefan_O, on 2016-August-14, 14:27, said:

How many possible, different bidding-auctions are there?
(We are talking abt auctions that are legal according to bridge-rules, not only the ones that are sensible...)

Just discovered this calculation, and it's actually beyond staggering

Before you check the link below, I invite you to work out with pen/paper -- or just have a guess :

Assume North deals, and opens 7♥.
In how many different, legal ways (according to bridge-rules) can such an auction proceed before it completes?
[a] 38
[b] 388
[c] 3388
[d] infinity

http://tedmuller.us/...dgeAuctions.htm

hi Stefan_O,

I love Ted Muller's and Richard Pavlicek's (rpbridge.net) bridge sites

Actually answer [d] is incorrect. ( I'm having one of my pedantic days - apologies ) It is an finite number, but a very. very, actually very, very, very large one.

It anyone wants some light relief from our lovely card game, it's worth logging into both Ted's and Richard's sites. Great fun!

[masse24]

The_Badger, on 2016-August-26, 00:37, said:

hi Stefan_O,

Actually answer [d] is incorrect. ( I'm having one of my pedantic days - apologies ) It is an finite number, but a very. very, actually very, very, very large one.

I think you misread the question. [c] 3388 is hardly a very, very, very large number.

[The_Badger]

masse24, on 2016-August-26, 05:58, said:

I think you misread the question. [c] 3388 is hardly a very, very, very large number.

Apologies everyone. I was so excited by the number that represented the number of possible auctions that I read the question incorrectly. (Must do better next time!)

[661_Pete]

Had me worried for a moment! I thought the OP's question was "how many possible auctions are there (starting from scratch)" - and I was thinking, something preposterous like Graham's Number - that is, until I read the OP more carefully!

Starting from 7♥, doesn't look overly difficult. Let's consider the cases where 7♥ - with all its doubling options - is the final contract. There are seven cases: passed, two ways of being doubled (by E or W), then four ways of being doubled and redoubled.

Now for each of these seven cases, there are three ways in which the next bid of 7♠ can occur. Thereafter there are, once again, seven pass/double/redouble combos. That makes (7 x 3 x 7) possible ways in which 7♠ becomes the contract.

Also there are an equal number of cases where the next bid is 7NT (bypassing 7♠). Another (7 x 3 x 7)

Finally, for each 7♠ case, there are another 3 x 7 cases where the final outcome is 7NT. (7 x 3 x 7 x 3 x 7) in all.

So, to tot up:
7 +
(7 x 3 x 7) +
(7 x 3 x 7) +
(7 x 3 x 7 x 3 x 7)

On my calculator this comes to...... 3388 ..... bingo!

[Vampyr]

LOL also misread the OP question.

[661_Pete]

In fact, looking into the problem a bit further (perhaps I should drop Bridge and take up Maths.... ) I've figured out a fairly straightforward iterative formula for deriving the number of possible auctions A_n after an opening bid "n" where n is the number of possible bids higher than the opening bid.

A₀ = 7
A_n = 7 + 21(A₀ + A₁ + ..... + A_n-1)

Using this formula, I get:
7NT: A = 7
7♠: A = 154
7♥: A = 3388
7♦: A = 74536
7♣: A = 1639792
etc.

I didn't try extending the process all the way down to 1♣, but at a rough estimate, I'd say that for 1♣ A = about 10⁴⁶ or around ten billion billion billion billion billion......

[edit]I wrote the above before looking at the Ted Muller link in the OP (honest! ). Seems my guesstimate was out by a factor of about ten - but at least in the right ball-park!

[Vampyr]

661_Pete, on 2016-August-27, 02:42, said:

I didn't try extending the process all the way down to 1♣, but at a rough estimate, I'd say that for 1♣ A = about 10⁴⁶ or around ten who billion billion billion billion......

[edit]I wrote the above before looking at the Ted Muller link in the OP (honest! ). Seems my guesstimate was out by a factor of about ten - but at least in the right ball-park!

How do you get a factor of 10? The number in the link was 2.4 X 10¹⁸.

[Stefan_O]

Vampyr, on 2016-August-27, 09:41, said:

How do you get a factor of 10? The number in the link was 2.4 X 10¹⁸.

Look again, Vamp...

It is: 1.29 * 10⁴⁷

[Stefan_O]

Yes, I choose 7♥ in the intro question, because one would think "intuition" could at least give you some estimate in such a simple case.

I would just guessed it could be no more than "like a few dozens or so"... but looking into the math details, it turns out it multiplies much faster than one (I) would believe.

[barmar]

661_Pete, on 2016-August-27, 02:42, said:

In fact, looking into the problem a bit further (perhaps I should drop Bridge and take up Maths.... ) I've figured out a fairly straightforward iterative formula for deriving the number of possible auctions A_n after an opening bid "n" where n is the number of possible bids higher than the opening bid.

A₀ = 7
A_n = 7 + 21(A₀ + A₁ + ..... + A_n-1)

Isn't that essentially the same formula as in the article that the OP linked to? And it includes the grand total for starting from 1♣.

[661_Pete]

barmar, on 2016-August-27, 20:16, said:

Isn't that essentially the same formula as in the article that the OP linked to? And it includes the grand total for starting from 1♣.

Yes it is. Sorry, I hadn't noticed the link in the OP before I started number crunching. But I figured it out for myself: cross my heart, I haven't plagiarised Ted Muller! Honest!

[barmar]

661_Pete, on 2016-August-28, 04:05, said:

Yes it is. Sorry, I hadn't noticed the link in the OP before I started number crunching. But I figured it out for myself: cross my heart, I haven't plagiarised Ted Muller! Honest!

I suppose there's something satisfying about the confirmation that your calculations were correct.