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A couple of problems for the stats geeks. These all ACTUALLY happens in my last two sessions of hand-shuffled club play.

 

What are the odds of holding a 30 HCP hand?

 

[hv=s=sakthakqdakxcaqj]133|100|[/hv]

 

What are the odds of a player bidding and making 7NT.

 

Part 1: On two consecutive boards

Part 2: Three times in a nine board stretch

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[hv=s=sakthakqdakxcaqj]133|100|A couple of problems for the stats geeks. These all ACTUALLY happens in my last two sessions of hand-shuffled club play.

What are the odds of holding a 30 HCP hand?

What are the odds of a player bidding and making 7NT.

Part 1: On two consecutive boards

Part 2: Three times in a nine board stretch

[/hv]
  • Richard Pavlicek's magnificent site says that the probability of dealing a hand with exactly 30 HCP is 0.000219848534%.
  • Tyler might obtain rough answers to his other questions from analysis of BBO hand-records or from GIB simulations.

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A couple of problems for the stats geeks. These all ACTUALLY happens in my last two sessions of hand-shuffled club play.

 

What are the odds of holding a 30 HCP hand?

 

Dealer: ?????
Vul: ????
Scoring: Unknown
AKT
AKQ
AKx
AQJ
 

 

What are the odds of a player bidding and making 7NT.

 

Part 1: On two consecutive boards

Part 2: Three times in a nine board stretch

Assuming perfect defense, 7nt requires 37, 38, 39 or 40 points. There are hands with these points which do not make 7nt, but they would be a minority.

 

Hence

 

Part 1: making 7nt on two consecutive boards = having 37+ hcp in the two combined hands on two consecutive boards

 

Part 2: similar to above.

 

Using the site that nige1 mentioned, the above can be calculated. The answers will not be accurate, but they may be more accurate than the simulations.

 

EDIT - I realized that some hands with less than 37 hcp may make 7nt if the opening leader has no ace to cash. The above solution is only a stab at the answer

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Assuming perfect defense, 7nt requires 37, 38, 39 or 40 points. There are hands with these points which do not make 7nt, but they would be a minority.

1). 7N requires 13 tricks. Read some of Victor Mollo's books to get good examples of how little in HCP you really need.

 

2). I would be willing to bet that the majority of hands that make 7N do not have 37 HCP. Running suits are far more common. Unfortunately, that makes the calculation harder, as the base assumptions are, well, wrong. Double unfortunately, I don't have a better way of calculating other than just have the computer do 100000 random deals and double dummy analyzing to see how many make 7N, then picking a selection of those deals and seeing how many are actually worth bidding.

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Assuming perfect defense, 7nt requires 37, 38, 39 or 40 points. There are hands with these points which do not make 7nt, but they would be a minority.

Uh, no! Where did you get this from?

Axx

AKQxx

Axx

xx

 

KQxxx

xxx

xx

AKx

 

As an example.where you have 13 tricks provided both Ms break 3-2

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Dealer: ?????
Vul: ????
Scoring: Unknown
[space]
8765432
765432
[space]
KQJT98765432
[space]
[space]
K
[space]
AKQJT9
AKQJT98
[space]
A
[space]
[space]
AQJT98765432

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

Provided West isn't dealer, when he can pre-empt a 7N "sacrifice" for a grand-slam swing :)
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Quick quiz: a player opens 7NT as dealer and claims before seeing the opening lead. What is his most likely distribution?

6332

Not this either, since although hands with 14 cards are quite likely on BBO, in real life they are less probable. Neither is 6-3-2-2 the answer, but feel free to try again.

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I'm running out of the house for four days in Chicago so I don't have time for a "real" answer.

 

I will simply note that I suspect that this is one of those questions where a simulation would work quite well... If I had a bit more time, I'd code it up.

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I'm running out of the house for four days in Chicago so I don't have time for a "real" answer.

 

I will simply note that I suspect that this is one of those questions where a simulation would work quite well...  If I had a bit more time, I'd code it up.

I am not quite sure what is meant by "one of those questions" - four questions have so far been asked in this thread:

 

[1] What is the chance of holding a hand with 30 hcp?

[2] What is the chance that a player will bid and make 7NT on two consecutive deals?

[3] What is the chance that a player will bid and make 7NT three times in nine deals?

[4] What is the most likely distribution of a player who opens 7NT and claims before seeing the opening lead?

 

Of these, [1] and [4] can be calculated precisely. [2] and [3] can neither be calculated nor simulated at all, since we do not know who this player is.

 

Perhaps one should restate them as "what is the chance that a player who always bids to the optimum contract and plays double-dummy will bid and make 7NT...". In that case, precise calculation would be possible in principle but exceptionally tedious in practice; a simulation would certainly be a better approach.

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

Nice! Must be

7321

(hidden).

If it must be, then it would be. But it mustn't, so it isn't.

Silly me to consider AKQJxxx a solid suit. Anyway, I was off by a mere factor of 10,.5 and who cares about the difference between 0.000001481% and 0.000000141%.

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I'm running out of the house for four days in Chicago so I don't have time for a "real" answer.

 

I will simply note that I suspect that this is one of those questions where a simulation would work quite well... If I had a bit more time, I'd code it up.

Make sure you run enough samples, though :)

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

7321

(hidden).

My guess is (hidden)

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)

 

 

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