A question that's been bugging me for a while and I'm not sure how to approach the solution.
RHO is playing something fairly normal for America.
a) 2+ card club
b) 3+ card club
You hold
1) 5 clubs
2) 6 clubs
How much does your holding actually impact the frequency of opener holding less than 4 clubs.
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Estimating suit length RHO opens 1C and you have lots of them
#2
Posted 2013-August-05, 02:17
Combinations maths will tell you. I can remember using a ZX81 to work out the suit splits but would need to google how to do it now.
Intuitively I'd say that even if you had as few as one club, compared with a void, the chance of RHO having a club suit is reduced. So obviously going from 5 to 6 will also reduce the chance.
Intuitively I'd say that even if you had as few as one club, compared with a void, the chance of RHO having a club suit is reduced. So obviously going from 5 to 6 will also reduce the chance.
#3
Posted 2013-August-05, 05:44
wanoff, on 2013-August-05, 02:17, said:
Combinations maths will tell you. I can remember using a ZX81 to work out the suit splits but would need to google how to do it now.
Intuitively I'd say that even if you had as few as one club, compared with a void, the chance of RHO having a club suit is reduced. So obviously going from 5 to 6 will also reduce the chance.
Intuitively I'd say that even if you had as few as one club, compared with a void, the chance of RHO having a club suit is reduced. So obviously going from 5 to 6 will also reduce the chance.
I don't think it's anything like that simple, the point being it's not "how many hands have 3 clubs rather than 4", but it's "how many of those hands are actually opened 1♣ rather than 1♦ or 1N", it may not be sensibly analytically solvable and require a sim.
I didn't say so in the original post, my reason for asking was to get more of a feel whether to overcall 1N or pass with 1N overcall type strength and clubs when a short club is opened as this will often allow me to X 1N after a 1♣-1?-1N auction showing the hand fairly precisely.
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