[nige1]

dburn, on Dec 27 2008, 05:27 AM, said:

Quick quiz: a player opens 7NT as dealer and claims before seeing the opening lead. What is his most likely distribution?

cherdano, on Dec 27 2008, 06:41 AM, said:

Nice! Must be

Spoiler

7321

(hidden).

My guess is (hidden)

Spoiler

9211 because
AKQxxxxxxx A A A = 10C7 = 10*9*8/3/2 = 120 (or 120 * 4 = 480)
AKQJxxxxx AK A A = 9C5 = 9*8*7*6/4/3/2 = 126 (or 126 * 4 * 3 = lots)
AKQJTxxx AK AK A = 8C3 = 8*7*6/3/2 = 56 (or 56 * 4 * 3 = fewer)
AKQJT9x AKQ AK A = 7C1 = 7 (or 7 * 4! = 168)
AKQJT9 AKQJ AK A etc = 1 (or 4! = 24)

[MFA]

Spoiler

9310

[cherdano]

Fluffy, on Dec 27 2008, 07:00 AM, said:

So, is it 8311 or 8221?

Both are equally likely. (The odds I gave above are still wrong by a factor of 4, btw.)

[MFA]

MFA, on Dec 27 2008, 10:08 AM, said:

Spoiler

9310

lol
[Note to self: Don't play poker and try to reply to tricky dburn posts at the same time.]

[Fluffy]

lol MFA

[fred]

I am going to go out on a limb here and suggest 9-2-1-1.

Fred Gitelman
Bridge Base Inc.
www.bridgebase.com

[dburn]

We have a winner - more than one, in fact, since that was also Nigel's suggestion (and perhaps cherdano's second suggestion, although the figures he gives do not accord with my own calculations).

There are 3,812 hands in total on which one could open 7NT and claim before the lead. Of these, 1,512 have 9-2-1-1 distribution, since the nine-card suit must consist of AKQJ and five of the remaining nine cards. There are, as Nigel says, 126 ways of selecting five cards from nine, and there are 12 possible 9-2-1-1 patterns (here I use the Bridge World notation in which 9-2-1-1 means any nine-card suit, any doubleton and any two singletons; 9=2=1=1 would mean nine spades, two hearts and one card in each minor). 126 * 12 = 1,512.

The next most common 7NT openers have 8-3-1-1 and 8-2-2-1 pattern (the eight-card suit must be AKQJ10 and three of the remaining eight small cards; there are 56 ways of choosing three cards from eight; twelve ways to hold each distribution; 56 * 12 = 672). Then comes 10-1-1-1 (480 of those), and 7-3-2-1 (168 of those).

The chance of holding one of the 9-2-1-1 shapes is 1,512/635,013,559,600, or 0.00000024% (give or take). The total chance of being able to open 7NT and claim is a more encouraging 0.0000006%. Suggest not holding your breath while waiting for one.

[marie__L]

Very clear !
And very useful !

[RMB1]

dburn, on Dec 27 2008, 08:46 PM, said:

There are 3,812 hands in total on which one could open 7NT and claim before the lead.

Under the heading "Perfect Bridge Hand" the Official Encyclopedia of Bridge gives the figure as 3756.

By my calculation, this breaks down as follows.

Shape nCr x Hands
10111 120  4  480
9211  126 12 1512
8311   56 12  672
8222   56 12  672
7411    7 12   84
7321    7 24  168
7222    7  4   28
6511    1 12   12
6421    1 24   24
6331    1 12   12
6322    1 12   12
5521    1 12   12
5431    1 24   24
5422    1 12   12
5332    1 12   12
4441    1  4    4
4432    1 12   12
4333    1  4    4
Total        3756
nCr = binomial n choose (13-n), where n>6 is the length of the longest suit
x   = number of shapes of a given pattern

Robin

[dburn]

Sorry - the Encyclopedia is quite right. For some reason I thought that there were rather more 7-2-2-2 shapes than there actually are.

I am abashed, though not especially surprised, to discover that when I set this as a problem in the bulletins for the 1988 Junior World Championships I had calculated the correct answer. Twenty years later, I managed the wrong one. Corpse evangelists nothing.

[qwery_hi]

Free, on Dec 27 2008, 04:42 AM, said:

You only need 11HCP for 7NT (by South):

---

Since the original question was what are the chances of a player bidding and making 7nt, in a club at mps I guess, the answer of 37+ hcp hands is probably not far off the mark.