how many hearts
#1
Posted 2008-September-30, 19:14
(3♠)-Dbl-(p)-4♥
What is partner's median heart length for hands that would pass a 3NT or 4m response? (we're not counting power doubles here)
#2
Posted 2008-September-30, 20:11
I think that 4 is the most common number, but that 3 is more likely than 5. I would guess 3.75 on that basis.
Ricky
#3
Posted 2008-September-30, 21:11
I would say that typical doubles will respect 3NT from partner, and will respect 4m from partner unless offshape, advancer hit doublers doubleton, and having 5-card hearts or some other prepared response. Since I would say that less than 10% of takeout doubles will fulfill all of these conditions, you are looking at almost the same probability distribution as for the original double.
So, in other words, responder shouldn't count on opener to recover the heart suit very often, and should strain to bid hearts when could be right. But we already knew that.
#4
Posted 2008-October-01, 01:00
1444
1(345) I guess you can come up with 15(34) hands that would bid 4♥
2(344) but may be not marginal 2=3=4=4 hands
0(445)
0(355) but may be not 0=3=5=5
We have additional information:
o The preempter has something like 7 spades
o The preempter's partner is unlikely to have 3+ spades and heart shortness, since he didn't raise.
o We have our own hand which probably contains at least 4 hearts. But we also know how many spades we have, making it easier to zoom in on partner's spade length. (If we have 5 of them, it is not likely that partner has a 2344 hand, while if we have a spade void, we can almost play partner for a (3334) hand without a spade stop.)
I would say that the first 3 distributions are much more likely than the others:
1444: 4 hearts (12/3)
1(345): slightly less than 4 hearts (<12/3)
2(344): slightly more than 3.67 hearts (>11/3)
If we are crude and just average these 3 we get: (12+ '<12' + '>11')/9 which is approximately 35/9 = 3.888888, etc. Let's just call it 3.9.
Rik
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#5
Posted 2008-October-01, 01:47
#7
Posted 2008-October-01, 06:18
#8
Posted 2008-October-01, 06:28
#9
Posted 2008-October-01, 07:17
Edmunte1, on Oct 1 2008, 01:02 PM, said:
Doesn't that make the incorrect assumption that 3, 4 and 5-card suits are equally likely?
(My statement that it was 4 derived from experience, not mathematics.)
#10
Posted 2008-October-01, 08:13
gnasher, on Oct 1 2008, 02:17 PM, said:
Edmunte1, on Oct 1 2008, 01:02 PM, said:
Doesn't that make the incorrect assumption that 3, 4 and 5-card suits are equally likely?
(My statement that it was 4 derived from experience, not mathematics.)
If you assign probabilities to the event of observing 3, 4 and 5, then 4 is the median if p(3)<0.5 and p(5)<0.5. If either observation has more than 0.5 probability, it will be the median.
#11
Posted 2008-October-01, 08:19
gnasher, on Oct 1 2008, 08:17 AM, said:
Edmunte1, on Oct 1 2008, 01:02 PM, said:
Doesn't that make the incorrect assumption that 3, 4 and 5-card suits are equally likely?
(My statement that it was 4 derived from experience, not mathematics.)
Yes,
Median (3,3,4,4,4,4,4,4,5)=4 should be a better example in relation with our problem

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