Join Bridge Winners
All comments by Kit Woolsey
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Stripe-tail ape: John Lowenthal. Snail: Roger Stern. Seal: Jeff Rubens.
6 hours ago
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But covering could cost from Kxxxxx. Why couldn't declarer's hand be: J9xxxx xxx xx A9. So if the bot knew its partner didn't have the ace it would be technically correct to duck. As far as I can tell, the reason for the cover was that the bot was afraid partner had the ace.
20 hours ago
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If that is the case then passing is a LA IMO (i.e. you weren't taking the pass then pull route to show weaker hand). Since the UI (I'm assuming that such exists) suggests bidding, I would not allow a pull on this hand.

Saying you are “barred” by the UI is not accurate. If bidding on were clear, then you may go ahead and do so. I don't think it is clear.
20 hours ago
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It might be very relevant whether pass followed by pull to 5 is stronger or weaker than an immediate 5 call in my partnership.

In other words, you imposed a pass of 4 on me. I need to know what my alternative action would have shown.
21 hours ago
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Of course the bot would cover. After all, partner might have the ace!

No competent human player could cover for that reason, since the human would know that partner doesn't have the ace. The day the bots can perform that sort of reasoning is probably the day the best players in the world will no longer be human, as has happened in other mental games.
March 27
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I will discard two clubs from dummy on the trumps. I can still take the club finesse through West if I so choose.

East has to come down to 9 cards. If he guards the hearts and keeps 3 clubs (it will be very difficult if not impossible for him to come down to Qx of clubs this early in the hand), he will have to come down to a doubleton diamond.

I will then cash AK of diamonds, AK of hearts, ruff a heart, and look around.

If it appears that only West guards diamonds and East guards hearts, I lead my last trump – discarding dummy's diamond if West keeps his diamond guard. This effects a double squeeze – whoever has the queen of clubs will be squeezed.

If it appears that West guards hearts (I will find that out if East discards a heart from a 3-card holding), I will lead my last trump discarding dummy's heart when West holds his heart guard. If West also keeps the diamond guard he comes down to a stiff club, so the finesse must work. If East keeps the diamond guard, nobody guards the clubs. I will have to read the position. This is in essence like a guard squeeze.

If I judge that East is keeping both red-suit guards and discarded down to a doubleton club, I will simply lead a club to the 10.

Some card-reading will be necessary. But it will be very difficult for the defenders to conceal the position.
March 27
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Andy,

Of course I will answer the question. As I said:

When the clubs are 2-2:

If East has played an honor on the first round of diamonds (from either a stiff honor or QJ doubleton), the probability that he has QJ doubleton is about 33.3%. Paul's 35.7% is more accurate since he is properly taking vacant spaces into account. I was just giving an approximation.

If you believe otherwise, I suggest you look closely at what Paul has written. He understands the concepts, and explains them very well.

If you still don't understand, I suggest you run the simulation that I propose. The results of that should remove any doubts from your mind.
March 27
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There is a pro which isn't mentioned. It is the catching of improperly entered scores.

If the results at other tables are shown, a player who has just played the board can (and will) look at these other results and an apparently impossible result will jump out at him. He can then call the director, who can check with the players who played the board earlier to find out what really happened and correct the score if it is wrong. Otherwise such scoring errors would likely to go unnoticed, since players generally don't pay much attention to the scores of others unless they have just played the board.
March 27
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From a straight mathematical point of view, if we assume that the clubs are 6-1 (the proper assumption to make since otherwise the hand is cold), I believe that tilts the normal odds enough to make the 3-0 spade split more likely than the specific 2-1 spade split.

However, we have to consider West's cover of the king of clubs. If he doesn't cover you do not have a convenient way back to your hand. Therefore, you would certainly lay down the ace of spades. West can see all this, since the actual layout is the only relevant layout from his point of view. The only reason he would ever cover is that he has the stiff king of spades and is giving you a chance to go wrong. Therefore, a spade to the ace is clear.
March 27
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Paul,

Thank you for supplying a very accurate and easy to understand explanation which illustrates the fallacy in Napoleon's reasoning.
March 27
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I could see bidding 2 with something like KQJ109 x KQJ10 xxx which is why I voted usually rather than always. But opener is definitely expected to pass unless he is either quite strong or has a good heart fit.
March 27
Kit Woolsey edited this comment March 27
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Without a club shift, you can keep both 3-3 diamond split and squeeze alive (risking the 5-1 spade split) as your analysis shows.

With a club shift, you play it the same way: spade to hand, cash ace of hearts, and run the spades. If West discards anything but a heart, assume that he has been squeezed and play accordingly.

If West discards only hearts (as he will on the actual deal) you have a straight guess as far as I can see.
March 26
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It is hard to see this hand making if the hearts aren't 2-2. So, I'll ruff a club at trick 2 and lead the queen of hearts. Assuming the hearts split, I will have the 10 of hearts entry to dummy for at least 2 clubs, and the diamond finesse will be #10 (4 hearts, 3 clubs, 2 diamonds, 1 spade). If the opponents are unable to make me commit to the diamond finesse before I get to dummy, I will have the additional chance that the jack of clubs comes down in 4 rounds.
March 25
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Napoleon says:

I will play for the drop if the fall in clubs is 2-2 and finesse the 10 if the fall is odd.

This is flat out wrong. The finesse is still close to a 2 to 1 favorite if the clubs are 2-2.

I will not attempt to give a mathematical demonstration of this, since the only math Napoleon knows is his own. However, I will suggest this very simple simulation:

1) Fix the N-S hands exactly.

2) Randomly deal out 10,000 (or some large number) of E-W hands.

3) Throw out hands where the E-W clubs are other than 2-2.

4) Throw out hands where East has something other than stiff queen, stiff jack, or QJ doubleton of diamonds.

5) Of the remaining hands (which are the hands which fit the required conditions), count how many times the finesse will win and how many times the drop will win.

I predict that the finesse will win close to 2/3 of the time. Napoleon clearly predicts otherwise.

I do not have the programming skills to do this. But I am sure there are BW readers out there for whom programming this simulation would be trivial.
March 25
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I don't think so. Having that extra trump is worth a lot on hands such as this. It would be unlikely that spades will take as many tricks as diamonds.
March 25
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Thank you for finally giving an unambiguous answer.
March 25
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Takeout double
March 25
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Thanks. Fixed.
March 25
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You write a lot of words. But you still haven't answered my very direct question. Is it so difficult to do so?
March 24
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Napoleon.

Frances is correct. Please answer my question without any elaboration about patterns.

I will repeat the question:

a) If the clubs are 2-2, how will you play the diamonds

b) If the clubs are 3-1, how will you play the diamonds.
March 24
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