[1eyedjack]

Sorry if I am being lazy. I know how to work it out from first principles (I think)but I expect that there is a tool out there somewhere that will do it in a twinkling.

In a vacuum, what is the likelihood that a suit AKQ5432 (in one hand) will run for 7 tricks with no losers? By "in a vacuum" I mean that you only know your hand, there has been no bidding, and partner could have anything from void support to 6 card support, including J(x) or x(x) as possibilities.

Thanks (I will also try to work it out by hand but it would be nice to have corroboration).

[MickyB]

One-third of the time, pard will have the jack, then you are nigh on 100% - stiff jack and a 5-0 break the one time the suit won't run (assuming you have an entry?)

The other two-thirds of the time, you require neither oppo to have four cards in the suit.

So -

Give one opponent the jack. The chance of that oppo having four cards in the suit is

12/38*11/37*10/36*5!/(2!*3!) = 26.1%

The chance of the other oppo having four cards in the suit is

13/38*12/37*11/36*10/35*5!/4! = 4.8%

So I make it in the region of 21% to not run.

[1eyedjack]

Hmm - I manually made it about 14% that it will not run. I worked it out by a different method, so not sure which is right or where I go wrong.

Where n = number of cards support that partner has, and P(n) is the probability of that event, I calculate:

n ........ p(n)
0 ....... 0.070566
1 ...... 0.262102
2 ....... 0.357412
3+ ..... 0.30992

From P(n) = (C(33,13-n).C(6,n))/C(39,13)
(except for n>2 I just took 1 minus the total of the above).

Where n = 0, the suit runs only on a 3-3 break (0.355 x 0.070566)

Where n = 1 then that 1 = J on proportion 1/6 of the time in which case only a 5-0 break beats you, so you make 0.26102 * (1/6) * (1-0.039)

Where n = 1, then that 1 is not J on proportion 5/6 of the time, in which case you need a 3-2 break, so you make 0.26102 * (5/6) * 0.678

Where n = 2 then that includes J on proportion 1/3 of the time in which case nothing can beat you, so you make 0.357412 * (1/3)

Where n = 2 then that excludes J on proportion 2/3 of the time in which case only a 4-0 break can beat you, so you make 0.357412 * (2/3) * (1-0.096)

Where n >2 then nothing can beat you, so you make 0.30992

Add the whole lot up and I get to 86% success rate.

[wclass___]

MickyB, on Aug 3 2009, 07:34 AM, said:

12/38*11/37*10/36*5!/(2!*3!) = 26.1%

I think it should be
12/42*11/41*10/40*5!/(2!*3!) = 19,16%

Quote

13/38*12/37*11/36*10/35*5!/4! = 4.8%

and
13/43*12/42*11/41*10/40*5!/4! = 2,89%

[MickyB]

1eyedjack, on Aug 3 2009, 02:16 PM, said:

Hmm - I manually made it about 14% that it will not run.  I worked it out be a different method, so not sure which is right or where I go wrong.

Your answer looks right to me - thought mine seemed a touch high.

[MickyB]

wclass___, on Aug 3 2009, 02:44 PM, said:

MickyB, on Aug 3 2009, 07:34 AM, said:

12/38*11/37*10/36*5!/(2!*3!) = 26.1%

I think it should be
12/42*11/41*10/40*5!/(2!*3!) = 19,16%

Quote

13/38*12/37*11/36*10/35*5!/4! = 4.8%

and
13/43*12/42*11/41*10/40*5!/4! = 2,89%

Where are you getting those numbers - 43, 42, 41, 40 - from?

I was looking it at from the point of view of -

give one hand AKQxxxx precisely

give one of the opponents the jack

now the probability of that opponent receiving the next diamond is 12/38 - there are 38 vacant spaces in the other three hands, and the hand which we are assuming has the jack has twelve of them.

[helene_t]

There are 3^6= 729 ways the six remaining card can be distributed across three hands. One opp has four cards in 120 of those, with a 1/6 probability that p has the Jack. This makes the probability of the suit not running 100/729 or 14%. The combinations with the cards spread equally among opps are more likely, but OTOH one opp could have five cards (24 combinations).

Jack's calculations look correct so 86% it is.

[wclass___]

MickyB, on Aug 3 2009, 10:09 AM, said:

now the probability of that opponent receiving the next diamond is 12/38 - there are 38 vacant spaces in the other three hands, and the hand which we are assuming has the jack has twelve of them.

Not 38, but 44. S-6;W-12;N-13;E-13

But now i realised that also 42,41.. are not correct.

Lets examine probability of JT98

Probability=is valid/all events.

For valid i know that East has JT98 and North + East has remaining two clubs.
valid = (39!)/(30!*9!)
all = (44!)/(32!*12!)

=(32*31*12*11*10)/(44*43*42*41*40) and muliply with (5!)/(2!*3!) to get probability of Jxxx

For J-T987 it should be
Valid = (39!)/(30!*9!)
All = (44!)/(31!*13!)
=(31*13*12*11*10)/(44*43*42*41*40) and multiply wiht 5.

I hope that now it is correct.

However i doubt it as i can't get even close to 86%

[nigel_k]

An alternative to Jack's method. The remaining cards in the suit split as follows:

A. 6-0-0 0.158%
B. 5-1-0 3.077%
C. 4-2-0 10.256%
D. 4-1-1 11.111%
E. 3-3-0 7.521%
F. 3-2-1 53.332%
G. 2-2-2 14.545%

Of these you will run the suit:

A. One third
B. One third
C. One third plus (two sixths of half the others)
D. One third plus (one sixth of the others)
E,F,G: All

I make this a total of 85.973%, i.e. 86%

Of course, if partner knows to pull 3NT to 4 of a minor when he has a void your chances are even better

[bb79]

I wrote a matlab code if anyone cares... nchoosek dees combination operation C(n,k). I calculated failure rates and subtracted from 1 at the end.

%7 sp , 6 other known, .. 6sp, 33 other remaining
allcombs=nchoosek(39,13)*nchoosek(26,13);

% partner has J, one opp has xxxxx
jpard=2*nchoosek(33,12)*nchoosek(21,8)/allcombs  
% 0.001709360842488

%one opp has jxxx,9 other,
valid1=2*nchoosek(5,3)*nchoosek(33,9)*nchoosek(26,13)/allcombs
%0.094964491249321

%one opp has xxxx, 9 other, other opp has J
valid2=2*nchoosek(5,4)*nchoosek(33,9)*nchoosek(25,12)/allcombs
%0.023741122812330

%one opp has jxxxx,8 other,
valid3=2*nchoosek(5,4)*nchoosek(33,8)*nchoosek(26,13)/allcombs
%0.017093608424878

%one opp has xxxxx,8 other, other opp has J
valid4=2*nchoosek(5,5)*nchoosek(33,8)*nchoosek(25,12)/allcombs
%0.001709360842488

%one opp has jxxxxx
valid5=2*nchoosek(33,7)*nchoosek(26,13)/allcombs
%0.001051914364608


jopp=(valid1+valid2+valid3+valid4+valid5)
%0.138560497693625


success= 1-(jpard+jopp)

gives 0.859730141463888

[matmat]

bb79, on Aug 3 2009, 04:42 PM, said:

I wrote a matlab code if anyone cares... nchoosek dees combination operation C(n,k). I calculated failure rates and subtracted from 1 at the end.

%7 sp , 6 other known, .. 6sp, 33 other remaining
allcombs=nchoosek(39,13)*nchoosek(26,13);

% partner has J, one opp has xxxxx
jpard=2*nchoosek(33,12)*nchoosek(21,8)/allcombs  
% 0.001709360842488

%one opp has jxxx,9 other,
valid1=2*nchoosek(5,3)*nchoosek(33,9)*nchoosek(26,13)/allcombs
%0.094964491249321

%one opp has xxxx, 9 other, other opp has J
valid2=2*nchoosek(5,4)*nchoosek(33,9)*nchoosek(25,12)/allcombs
%0.023741122812330

%one opp has jxxxx,8 other,
valid3=2*nchoosek(5,4)*nchoosek(33,8)*nchoosek(26,13)/allcombs
%0.017093608424878

%one opp has xxxxx,8 other, other opp has J
valid4=2*nchoosek(5,5)*nchoosek(33,8)*nchoosek(25,12)/allcombs
%0.001709360842488

%one opp has jxxxxx
valid5=2*nchoosek(33,7)*nchoosek(26,13)/allcombs
%0.001051914364608


jopp=(valid1+valid2+valid3+valid4+valid5)
%0.138560497693625


success= 1-(jpard+jopp)

gives 0.859730141463888

Nerd.

[bb79]

matmat, on Aug 3 2009, 05:49 PM, said:

bb79, on Aug 3 2009, 04:42 PM, said:

I wrote a matlab code....

Nerd.

Thanks for the feedback.
I may have written something complicated but I was simply trying to help mickyb and wclass finding their mistake!?