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- Feb. 12, 2016
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- Stig Holmquist's bidding problem: AKxxxxx x AKxx x
- Would you also open 1 S if the hand had the Q of S? I'm trying to question the traditional wisdom that a 2C opening bid should be based on 22+ HCP. The 2C opening bid should be based on trick taking potential, rather than HCP. Having more quick ...
- Stig Holmquist's bidding problem: AKxxxxx x AKxx x
- Reply to Jack Spear. There are six ways the six cards can be split, if we disregard the very rare 6-0-0. Of these the 3-2-1 accounts for 553%. So you think this is the only one you need to calculate the mean for all six splits. I listed the frequencies ...
- Stig Holmquist's bidding problem: AKxxxxx x AKxx x
- I tried to respond but was informed there was a 400 error and it could not be printed. I don't know what that means and much less how to fix it. I then sent a 2nd short letter and was told: I admire your optimism. Then all disappeared. Is ...
- Stig Holmquist's bidding problem: AKxxxxx x AKxx x
- How would you do this calculation at the table?
- Stig Holmquist's bidding problem: AKxxxxx x AKxx x
- What facts/probabiliyies do you base your claim on?
- Stig Holmquist's bidding problem: AKxxxxx x AKxx x
- If you subject this hand to post mortem analysis, you'll discover some interesting facts/probabilities. One test would be to generate this hand as N with the 'Play Bridge Hands Generators" on the internet and then do 10 000 deals. Tabulating the statistics would be prohibitive.. So why not ...
- Stig Holmquist's bidding problem: AKxxxxx x AKxx x
- My partner held this hand and opened 2C. I held : Q6-QJ752-J7-Q762 and responded 2D. Partner then bid 2S and I rebid 4S. He made 5S. When holding 14 HCP there is 50% chance the pair has 23HCP. So why the pessimism?
- Stig Holmquist's bidding problem: 32 AKT4 AQJ5 765
- Does anybody here add 1 pt for each 4+-card suit with 3 honor cards as advocated be Bergen?
- Probability of 15 HCP
- Imagine having tested 100 000 deals and obtained the frequencies for all possible counts. That might include 30 with 20 HCP. Now connect all frequencies and form a smooth curve, like a continuous distribution. Calculate the mean and the std.dev. and don't be surprised if the data fit ...
- Probability of 15 HCP
- I'm trying to establish a frequency polygon for the HCP responder will have when opener has a specified set of 17 HCP. I'm a numbers freak. The 15 count is not significant. I picked it because it seemed odd in my testing 100 deals. I'm wondering if ...

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