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- How to play 6!H and win or lose the Bermuda Bowl
- If West has the ♥K, finessing the ♥10 succeeds when he has ♥KJ5 or ♥K5, ♥A and a heart back succeeds when he has ♥K95 or ♥K85. In addition, the latter play succeeds when West has ♥K5 (the ♥Q is not played), unless East can ruff a diamond return. It ...
- How to play 6!H and win or lose the Bermuda Bowl
- If defenders play this carding scheme consistently (but should they not randomize that, too, if the purpose is to make life more difficult for declarer?), West having ♥95 or ♥85 can be excluded. Then each alternative succeeds in 3 cases. If West has shown some length in the minors, he ...
- How to play 6!H and win or lose the Bermuda Bowl
- If the defenders also randomize adherence to this scheme, West would play the ♥5 about half the time from ♥K95 and ♥K85, and half the time from ♥95 and ♥85. Then, in each case West has three hearts once and two hearts twice making the best play a complete guess.
- How to play 6!H and win or lose the Bermuda Bowl
- If declarer is tired and did not notice which small heart West played, finessing the ♥10, then the ♥A, succeeds when West has ♥KJ9, ♥KJ8, ♥KJ5, ♥K9, ♥K8, ♥K5, ♥J9, ♥J8, or ♥J5. That's 9 cases. In other cases when the play succeeds, the alternative plays also succeed. Playing ...
- Applying Vacant Spaces to Distributional Hands
- Dave, I think your idea is useful, but I think you didn't apply it in the best way in this case. There are 7 cover cards outside spades, not 9, so the expected number of useful cover cards according to your method is 4*4/7 = 2.29. According ...
- How to calculate the odds
- 8 means the remaining cards (in addition to the 4 spades and the Ace) in West's hand.
- How to calculate the odds
- 4 spades out of 5, 8 remaining cards out of 20, excluding the already placed Ace. Addition: This gives C(5,4)xC(20,8) = 5x20!/(8!x12!).
- How to calculate the odds
- As always, there are different ways to do the calculations. Here is one focusing on the number of combinations of the various cases. Assume West has the Ace of interest. Then there are C(5,4)xC(20,8) combinations where West has 4 spades, and C(5,3)xC ...
- Another Question for Mathematicians--well...Statisticians
- I can repeat my calculations for 26 cards (2 hands) instead of 13 cards (1 hand). This gives Var(X) = 16/17 and Cov(X,Y) = -4/51, which for both is an increase by the factor 4/3. Thus, the result sqrt(1160/51) is obtained. Does Helene have ...
- Another Question for Mathematicians--well...Statisticians
- The honor card distribution is a multivariate hypergeometric distribution. The following expressions follow from formulas for that distribution (see, e.g., Wikipedia). The variance of the number of honor cards of a given rank X (J, Q, K or A) is Var(X) = 12/17. There is a negative correlation ...

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