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Bridge Winners Profile for Paul Barden

Paul Barden
Paul Barden
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Member Since
Aug. 20, 2015
Last Seen
23 minutes ago
Member Type
Bridge Player
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United Kingdom of Great Britain and Northern Ireland

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The ruling and appeal that weren't
I don't think fast arrival applies when you go past 3NT on a hand where there may be no fit. Without the UI, partner's bidding is consistent with a 4414 19-count. Perhaps KQJx AQxx x AKxx. I want to play in 6S opposite that. With the UI, partner ...
The ruling and appeal that weren't
I agree that 4 should be natural and strong, 4414 or 4405. I don't think it's clear to pass it. Our diamond values are not altogether wasted, since the difficulty in 6 may be tricks not losers.
Patterns VI - Suit Combinations
Thanks Napoleon. You needn't look at specific distributions: the number of hands with a singleton diamond is simply: (number of possible singleton (or tripleton)) diamonds) * (number of ways to choose 10 (or 8) cards from the 18 major-suit cards outstanding) which is 4 * 18!/(10!8!) = 175,032 note ...
In the Well: Jeff Meckstroth
Jeff, I think it would be good for the game if top pairs published their systems. How can there be full disclosure if opponents don't have the opportunity to consider your system as a whole? Plus, it would help vugraph commentators, and it would be good for the game ...
Patterns VI - Suit Combinations
I applaud Napoleon's open-mindedness in testing his theory. 291,720 is correct for the number of hands East can have with two diamonds. However, the number of hands he can have with a singleton diamond is 175,032
Patterns VI - Suit Combinations
Do you think that once the club lengths are even in the example, it's possible for both defenders to have three odd-length suits?
Patterns VI - Suit Combinations
Somewhat to my surprise, I have reread "Patterns V". I have no idea how you can think that the 56.25% you just calculated is relevant to Kit's example hand. We know, because we can see declarer's hand and dummy's, that the defenders have combined odd length ...
Patterns VI - Suit Combinations
You're right Napoleon, I don't understand. Some way up this thread you wrote ".75*.75=.5625 is the math probability for the two patterns to coincide" - I took that to be where your 56.25% came from.
Patterns VI - Suit Combinations
Clubs are 2-2, and the defenders have 9 spades, 9 hearts, and 4 diamonds between them. Let's say that each defender has an "even pattern", i.e. three even-length suits, 3/4 of the time. Then the probability that they both have even patterns is 9/16 (56.25 ...
Patterns VI - Suit Combinations
No, I said if the only possibilities for RHO are singleton Q or doubleton QJ, it will be QJ 52.6% of the time. Kit said that if the only possibilities are singleton Q, singleton J or doubleton QJ, it will be QJ about 1/3 of the time. Folloing ...
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