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I think I solved this hand in about 8 minutes; I'm sure it was under 10. It was the fastest time on that hand, and one of the fastest by anyone on any hand.

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This is sad news indeed. I saw Harold during the Nationals in Las Vegas in July and he seemed fine.

Harold was an integral part of my group of bridge friends while I was going to MIT, though he never lived in Boston and never went to MIT. He and Mike Gurwitz met at the partnership desk for the Flight C Pairs (!) at the Eastern States Regional in New York City in 1967 or so. They hit it off and Harold became a regular with us.

The first time Harold came to Boston to visit, he got directions (from a local) to take the subway to “Pack Street Station”. Eventually he arrived at “Park Street Station” and everyone got off. The conductor arrived at Harold's seat. Conductor: “This is the end of the line.” Harold: “But I'm going to Pack Street.” Conductor: “You're there.”

Harold claimed that he was such a presence at my dorm that one time we came in and the doorman said “Hi Harold” to him and “Who are you?” to me.

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I agree. My best hope is to be third. Aces are my only controls (!), so making them the start card should max out my chances.

I also look for unusual fantan hands while playing bridge. And to show you that it's (slightly) more than a mindless distraction, one of my bidding principles is to “overbid with a good fantan hand”, which normally is loaded with fillers.

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This particular accusation is ludicrous. The only possibilities are stiff ace and AJ tight. The (alleged) hesitation is clearly meaningless, as the defender will ALWAYS win the ace with either holding.

Declarer is basically insisting that the defender MUST give him a reliable tell. If the defender had won the ace “in tempo”, would declarer have complained had he then finessed and lost to AJ tight?

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In 1969 or 1970 Ken was playing against George Boehm (Augie's father) in a New York tournament and spotted the word “lebensohl” on George's convention card. The following conversation (approximately) took place.

KL: What's that? GB: You should know, you invented it. (Describes convention) KL: No I didn't. (Describes vaguely similar treatment in which 2NT is TAKEOUT and double is penalty, which is what he DID play) GB (crestfallen): I just wrote a Bridge World article about it with that title. KL: You also misspelled my name. GB (brightening): Then do you mind if I keep calling it that and spelling it that way, starting with a small “l”? KL: Fine.

And that's why the lebensohl convention is misnamed and misspelled after Ken Lebensold.

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If you're going to defend double-dummy, I'm going to play double-dummy.

In your ending North must pitch TWO clubs. But after ONE club pitch, I exit in clubs instead of cashing the last spade. Now there is no squeeze. Dummy saves three hearts and two diamonds, while declarer saves 1=2=2=0. The defense is endplayed.

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And I surely would have led a club without the double, based on my only other experience of this type.

In the 1998 Rosenblum semifinals I opened 3♦, first seat favorable, on Qxx-xx-Qxxxxx-xx (vile, but I had a feeling). LHO, Versace, bid 4♦ and we passed the tray. In tempo it came back with 7♦ by my partner, Sidney Lazard, and 7♥ by my RHO, Lauria, then all pass. I knew that Sidney would have doubled with a potential casher outside of diamonds, so I led a diamond, and Sidney's ace lived. He had AKJx of diamonds and out. We'd have gone for 1700 (down 7). We won that battle but, alas, we lost the match.

On the recent deal I would have applied the same logic to my lead.

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I'm not understanding the math. Let's assume that the only possible outcomes are making and down one, and further that the same number of tricks are taken at each table. (If one table makes and the other goes down, the IMP swing from that will dwarf the extra swing from the redouble.)

Let's look at the scores for 4♠ “singled”, doubled and redoubled, making or down one: Making: +620 +790 +1080 Down 1: -100 -200 -400

If singled vs. doubled, the swings are +5 imps and -3 imps. If singled vs. redoubled, the swings are +10 imps and -7 imps. Here, the redouble has 5 to 4 imp odds in favor.

If doubled vs. redoubled, the swings are +7 imps and -5 imps. Here, the redouble has straight 7 to 5 imp odds in favor, as the board will otherwise be a push.

While those are good odds they are nowhere close to what is being suggested.

Note that when it comes to redoubling, this is one the few areas of bridge scoring where the odds are better when the declaring side is NOT vulnerable. The gains from making stay the same, while the losses from going down are smaller, since the comparable raw scores for down one not vulnerable (singled, doubled, redoubled) are -50, -100 and -200. In the singled vs. redoubled case, the redouble now has 5 to 2 imp odds in favor. If doubled vs. redoubled, it's 7 to 3 in favor.

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Yes. By the way, Vanderbilt and Spingold use the same numbering scheme of 30-board sets. Thus, Board 30 is none vul, East dealer, and Board 31 is none vul, North dealer (like Board 1).

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And to clarify even further, this was actually board 29 of a 30-board set. In this event the dealers and vulnerabilities recycle starting with Board 1 in each set of 30 boards. The opening bidder was North, the dealer. The spade hand was East, in second seat. Craig is right about that. Don't make authoritative assumptions without all the facts.

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Dummy actually had KQ9 of spades. At my table LHO shifted to the spade TEN, covered and ducked. Then he continued with the spade eight when he won the club ace. Maybe everyone watching thought it was automatic to play for the (winning) strip-squeeze when LHO let go of his last spade, but I was sweating bullets before I did that, because if LHO had double-crossed me by discarding this way with jack-third of diamonds (initially), then I would have gone down in 2NT for a near-certain loss.

Adding to my concern was that this was the next-to-last board of the segment (out of 14) and I had been having a terrible set. I was greatly relieved to get it right, securing a vital point near the end.

This is a classic game-theoretical position, with psychological overtones. Both declarer and LHO have choices in the ending. Note that RHO contributed by ducking the spade ace twice. If he had won the ace at either turn, the diamond finesse is then declarer's only chance for the overtrick, and it's free. On that basis perhaps I should have taken the finesse, as the actual defense was the only one to give me a losing option when the finesse was winning.

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Kit is correct. In the original problem, before declarer plays a card, the chance that East has exactly one of the two missing key cards is about 2/3.

Above I had a hard time seeing the fallacy in David Burn's argument using 22 yellow cards, two red cards, and two green cards. Now I see it. We are instructed to deal West both red cards, to deal East one (random) green card, and then to deal the rest of the cards randomly. It is true that with this method, the chance that East gets the other green card is approximately 1/2. However, that's not how to count all the cases. To do so we must perform the calculation twice, once each for the possibility of East receiving one specific green card or the other. However, when we then deal out the remaining cards and count cases, we will DOUBLE COUNT the cases where East gets both green cards. Thus, there are only three “pools” of cases, one where East gets both green cards and two more for East getting exactly one of the green cards.

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I ignored the possibility that West has three aces for the same reason as in the original problem: No opening bid.

I maintain that the original problem and my restatement are logically comparable. Each statement of the problem assumes that there are three significant high-card holdings, of which West is known to hold at most two. On opening lead he shows you one of the holdings. In the original it is a specific holding. In my restatement it could be any of the three holdings. How does that change the subsequent calculation?

David Burn's reformulation of “restricted-choice” when East is KNOWN to have at least one of the significant cards is on point. Indeed, in his example of a nine-card fit missing king-queen, declarer makes a clear error by not playing the ace on the first round when he knows that at least one royal is offside.

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Reluctantly I find myself agreeing with David. Reluctantly, not because I have any problem with David personally, but because his solution is highly non-intuitive for me.

Here's how I convinced myself. Let's restate the problem in simpler terms. Suppose we are missing three aces. Further suppose that we know that West always leads an ace when he has one, choosing randomly when he has more than one. On the dealing there is an equal probability of West having either one ace or two aces. We will see the lead of any specific ace exactly one-third of the time. Therefore, when West leads an ace there is a 50% chance that he has no other ace, which means there is a 50% chance East has BOTH other aces.

In turn, this implies that in the original problem the chance that East holds both the ♠K and ♥A is approximately 50%, rather than the approximately one-third chance produced by the standard restricted-choice argument.

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Billy was a teammate when I won my first national (Men's BAM, Spring 1980), which happened to be Billy's last national win. Several years later I moved to Chicago to trade options, which allowed me to see Billy regularly on the options floor as well as at the bridge table, plus occasionally on the golf course. He was a delight in any venue, a class act all the way.

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I echo Barry's sentiments in all respects, especially the nice guy part. I first met Anders many years ago when he played in the Cavendish Pairs in New York several times (and won it once). More recently I corresponded with him on matters of esoteric analysis.

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I have arrived. In addition to Chuck Berry's Memphis (I love the line “with hurry-home drops on her cheek that trickled from her eye”) and Dylan's Stuck Inside of Mobile (“it all seems so well timed”) I heard Big River (Grateful Dead version) and Dylan's Highway 61 Revisited. Many more songs available.

Bart Bramley

Bart Bramley

Harold was an integral part of my group of bridge friends while I was going to MIT, though he never lived in Boston and never went to MIT. He and Mike Gurwitz met at the partnership desk for the Flight C Pairs (!) at the Eastern States Regional in New York City in 1967 or so. They hit it off and Harold became a regular with us.

The first time Harold came to Boston to visit, he got directions (from a local) to take the subway to “Pack Street Station”. Eventually he arrived at “Park Street Station” and everyone got off. The conductor arrived at Harold's seat.

Conductor: “This is the end of the line.”

Harold: “But I'm going to Pack Street.”

Conductor: “You're there.”

Harold claimed that he was such a presence at my dorm that one time we came in and the doorman said “Hi Harold” to him and “Who are you?” to me.

He was a natural at everything he did.

Condolences to Janis.

Bart Bramley

I also look for unusual fantan hands while playing bridge. And to show you that it's (slightly) more than a mindless distraction, one of my bidding principles is to “overbid with a good fantan hand”, which normally is loaded with fillers.

Bart Bramley

Declarer is basically insisting that the defender MUST give him a reliable tell. If the defender had won the ace “in tempo”, would declarer have complained had he then finessed and lost to AJ tight?

Bart Bramley

KL: What's that?

GB: You should know, you invented it. (Describes convention)

KL: No I didn't. (Describes vaguely similar treatment in which 2NT is TAKEOUT and double is penalty, which is what he DID play)

GB (crestfallen): I just wrote a Bridge World article about it with that title.

KL: You also misspelled my name.

GB (brightening): Then do you mind if I keep calling it that and spelling it that way, starting with a small “l”?

KL: Fine.

And that's why the lebensohl convention is misnamed and misspelled after Ken Lebensold.

Bart Bramley

In your ending North must pitch TWO clubs. But after ONE club pitch, I exit in clubs instead of cashing the last spade. Now there is no squeeze. Dummy saves three hearts and two diamonds, while declarer saves 1=2=2=0. The defense is endplayed.

Bart Bramley

In the 1998 Rosenblum semifinals I opened 3♦, first seat favorable, on Qxx-xx-Qxxxxx-xx (vile, but I had a feeling). LHO, Versace, bid 4♦ and we passed the tray. In tempo it came back with 7♦ by my partner, Sidney Lazard, and 7♥ by my RHO, Lauria, then all pass. I knew that Sidney would have doubled with a potential casher outside of diamonds, so I led a diamond, and Sidney's ace lived. He had AKJx of diamonds and out. We'd have gone for 1700 (down 7). We won that battle but, alas, we lost the match.

On the recent deal I would have applied the same logic to my lead.

Bart Bramley

2. F***in' Up - Neil Young

Bart Bramley

Let's look at the scores for 4♠ “singled”, doubled and redoubled, making or down one:

Making: +620 +790 +1080

Down 1: -100 -200 -400

If singled vs. doubled, the swings are +5 imps and -3 imps.

If singled vs. redoubled, the swings are +10 imps and -7 imps.

Here, the redouble has 5 to 4 imp odds in favor.

If doubled vs. redoubled, the swings are +7 imps and -5 imps.

Here, the redouble has straight 7 to 5 imp odds in favor, as the board will otherwise be a push.

While those are good odds they are nowhere close to what is being suggested.

Note that when it comes to redoubling, this is one the few areas of bridge scoring where the odds are better when the declaring side is NOT vulnerable. The gains from making stay the same, while the losses from going down are smaller, since the comparable raw scores for down one not vulnerable (singled, doubled, redoubled) are -50, -100 and -200. In the singled vs. redoubled case, the redouble now has 5 to 2 imp odds in favor. If doubled vs. redoubled, it's 7 to 3 in favor.

Bart Bramley

Bart Bramley

Don't make authoritative assumptions without all the facts.

Bart Bramley

Adding to my concern was that this was the next-to-last board of the segment (out of 14) and I had been having a terrible set. I was greatly relieved to get it right, securing a vital point near the end.

This is a classic game-theoretical position, with psychological overtones. Both declarer and LHO have choices in the ending. Note that RHO contributed by ducking the spade ace twice. If he had won the ace at either turn, the diamond finesse is then declarer's only chance for the overtrick, and it's free. On that basis perhaps I should have taken the finesse, as the actual defense was the only one to give me a losing option when the finesse was winning.

Edited to correct a suit designation.

Bart Bramley

Bart Bramley

Above I had a hard time seeing the fallacy in David Burn's argument using 22 yellow cards, two red cards, and two green cards. Now I see it. We are instructed to deal West both red cards, to deal East one (random) green card, and then to deal the rest of the cards randomly. It is true that with this method, the chance that East gets the other green card is approximately 1/2. However, that's not how to count all the cases. To do so we must perform the calculation twice, once each for the possibility of East receiving one specific green card or the other. However, when we then deal out the remaining cards and count cases, we will DOUBLE COUNT the cases where East gets both green cards. Thus, there are only three “pools” of cases, one where East gets both green cards and two more for East getting exactly one of the green cards.

Kit and others were right all along. My bad.

Bart Bramley

I maintain that the original problem and my restatement are logically comparable. Each statement of the problem assumes that there are three significant high-card holdings, of which West is known to hold at most two. On opening lead he shows you one of the holdings. In the original it is a specific holding. In my restatement it could be any of the three holdings. How does that change the subsequent calculation?

David Burn's reformulation of “restricted-choice” when East is KNOWN to have at least one of the significant cards is on point. Indeed, in his example of a nine-card fit missing king-queen, declarer makes a clear error by not playing the ace on the first round when he knows that at least one royal is offside.

Bart Bramley

Here's how I convinced myself. Let's restate the problem in simpler terms. Suppose we are missing three aces. Further suppose that we know that West always leads an ace when he has one, choosing randomly when he has more than one. On the dealing there is an equal probability of West having either one ace or two aces. We will see the lead of any specific ace exactly one-third of the time. Therefore, when West leads an ace there is a 50% chance that he has no other ace, which means there is a 50% chance East has BOTH other aces.

In turn, this implies that in the original problem the chance that East holds both the ♠K and ♥A is approximately 50%, rather than the approximately one-third chance produced by the standard restricted-choice argument.

Bart Bramley

Bart Bramley

I will miss his warmth and his keen intellect.

Bart Bramley

Bart Bramley

And congrats to Peter and Judi! We all waited a long time for this.