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Does anyone know the probability (e.g. 1 out of xx) of a partnership having a 13-card fit in any suit? My opps had one on BBO the other week (no, not a goulash game) ....
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13-card fit
#1
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Posted 2007-October-04, 11:32
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#2
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Posted 2007-October-04, 11:42
I think it is 4*(26!)(39!)/[(52!)(13!)] - 6* (26!)(26!)/[52!].
You can plug it into your calculator if you'd like to see the number. The second part is negligible, don't worry about it.
You can plug it into your calculator if you'd like to see the number. The second part is negligible, don't worry about it.
Please note: I am interested in boring, bog standard, 2/1.
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#3
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Posted 2007-October-04, 12:19
If I did the math right, that comes to 1 in 15,639, which feels about right.
#4
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Posted 2007-October-04, 12:33
Yea, I got the same but I ignored the negligible part 
Thanks Han !
Thanks Han !
♣♦♥♠ Philosophy consists very largely of one philosopher arguing that other philosophers are all jackasses. He usually proves it, and I should add that he also usually proves that he is one himself. H.L. Mencken. ♣♦♥♠
#5
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Posted 2007-October-04, 14:19
I've seen ti 3 times n my life, the last one it was a nightmare trying to force partner to lead another suit other than ours (didn't work)
#6
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Posted 2007-October-04, 14:57
Hannie, on Oct 4 2007, 12:42 PM, said:
I think it is 4*(26!)(39!)/[(52!)(13!)] - 6* (26!)(26!)/[52!].
You can plug it into your calculator if you'd like to see the number. The second part is negligible, don't worry about it.
You can plug it into your calculator if you'd like to see the number. The second part is negligible, don't worry about it.
Han, I think that's the probability that a particular pair (e.g., the opponents) will hold a 13-card fit. So the probability that either side would hold such a fit is twice that, or about one in 8,000:
(2 choose 1)(4 choose 1)(39 choose 13)/(52 choose 26)
neglecting the correction.
just plain Bill
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Posted 2007-October-04, 15:03
ralph23, on Oct 4 2007, 12:32 PM, said:
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Does anyone know the probability (e.g. 1 out of xx) of a partnership having a 13-card fit in any suit? My opps had one on BBO the other week (no, not a goulash game) ....
Does anyone know the probability (e.g. 1 out of xx) of a partnership having a 13-card fit in any suit? My opps had one on BBO the other week (no, not a goulash game) ....
I'm not sure, but I have witnessed just two in my life. On one, at a small tournament, the best result was obtained by the defenders against a 5♣ contract...on the 0=0 fit. One hand had cued its void a couple of times and the other figured it had to be to play (the opps having been quite silent given their fit). The other resulted in a slam making when opening leader led his 9 card suit and gave a ruff and sluff.
'one of the great markers of the advance of human kindness is the howls you will hear from the Men of God' Johann Hari
#8
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Posted 2007-October-04, 15:47
I get 4*(26!)(39!)/[(52!)(13!)]=38/580027
This seems more likely than 1/15,639: Cancel the 19 and the 38 from 39! with the same numbers in 52!, leaving another occurrence of 19 in 26! that doesn't cancel.
Since two of you got a different answer and the two of you agree, this worries me. Still, I am at a loss to see where the third 19 goes.
Wouldn't be the first time I have been wrong.
Numerically the answers are close. Added: I see, you rounded a bit to make the numerator 1. Fair enuf. The answer just looked so exact I took it literally.
The correction term is very very small (of course)
This seems more likely than 1/15,639: Cancel the 19 and the 38 from 39! with the same numbers in 52!, leaving another occurrence of 19 in 26! that doesn't cancel.
Since two of you got a different answer and the two of you agree, this worries me. Still, I am at a loss to see where the third 19 goes.
Wouldn't be the first time I have been wrong.
Numerically the answers are close. Added: I see, you rounded a bit to make the numerator 1. Fair enuf. The answer just looked so exact I took it literally.
The correction term is very very small (of course)
Ken
#9
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Posted 2007-October-04, 16:17
Han's formula is the chance that a particular partnership holds a 13-card fit. The chance that one or the other has a 13-card fit is slightly less than twice that. The "neglible" term that Kenberg forgot corrects for the possibility of a double fit.
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Posted 2007-October-05, 14:43
Ralph, why don't you post puzzles on the beginner's forum anymore??
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