Log in

entries friends calendar profile Previous Previous
Mildly Interesting Blog
I made a movie.

3 comments or Leave a comment
3rd, none

KT82 AK 32 K6543
QJ765 J32 54 A87


Opps cash two diamonds and exit a heart. Now what?

I slipped a spade through, cashed the other high heart, CK, C to the ace, ruffed a heart, and threw in the opp who started with SAx and a doubleton club. Making four.

The overcaller held:

Ax xxxx KQxxx Jx

Should he have won the first spade to avoid the endplay? I think so (I did lead a spade off dummy, away from the king...) but it didn't hurt to play for that small an error.

1 comment or Leave a comment
I'm used to seeing people butcher the Monty Hall problem. In fact, the only person who gets it right (besides me) is Monty Hall. He did a great interview in the New York Times a few years ago. Now a Canadian, bridge playing mathematician is bridge-blogging about it and I just can't stand it any more. So, here is a breakdown of the usual problem statement, the most common answers, and the correct answer.

BTW, this is almost entirely not about bridge. The common analogy of MH to restricted choice works well because, in a bridge setting, all the usual assumptions are met. In short, the usual answer there (finesse when you see an honor, it's about 2-1 in your favor) is correct. If that's all you're interested in, you don't need to read any further.

The Usual Setup

You go on Let's Make a Deal and Monty Hall shows you three doors.  He says there is a goat behind each of the two losing options, and some prize you want (e.g. a car) behind the other door.  He tells you to pick a door.  You pick a door and he opens a different door, revealing a goat.  Then he asks whether you'd prefer to switch to the third door, or stick with your original choice.  What do you do, and why?

The Usual Answers
  • Marilyn Vos Savant, with the world's highest recorded IQ, wrote in her Sunday column that you should switch.  She said you had one chance in three of picking correctly at first, and nothing has changed, since Monty could always show you a goat, regardless of which door you picked originally.  She was wrong.
  • A bunch of mathematicians (professors, PhD's, etc.) wrote in to say various mathy and nasty things (e.g. You haven't learned the math they dole out in eighth grade).  They believe that there are two doors remaining, and each has a different prize, so it's a straight 50-50 chance whether to switch or not.  Needless to say, they were hella wrong.

Monty Hall's Answer
  • He said that if you picked a goat, he could just let you have the goat.  If you picked a car, he could offer to let you swap.  Therefore, you shouldn't swap.  I'm not going to argue with Monty.


People make excuses for why certain answers are right, even after you demonstrate they are completely wrong.  In particular, when you point out a direct contradiction, they claim it's a linguistic issue and you know what they meant.  Fine.  Except if they said something other than what they meant, and what they said was wrong, it's fair game.  The real problem I have with this sort of imprecision is that they say one thing meaning another, then add a third thing that makes perfect, intuitive sense, given what they said, but not given what they "meant".  That leads to fuzzy thinking.  I expect better from the smartest woman in the world.  And I demand better from mathematicians.

Unstated Assumptions

In a way, Marilyn was right, given certain (false) assumptions.  The mathematicians were entirely correct, given different (false) assumptions.  Monty is awesome, obviously.  Let's look at various assumptions people have made, but not stated:

  1. Each person playing the game is always shown the goody behind a door other than the one chosen, and is always offered the opportunity to swap the first door for the third door (Note: I didn't mean the first door for the third door, I meant the prizes behind those doors, but you "knew what I meant".  Isn't that confusing?  I'll try to avoid that from now on...)
  2. Monty knows where the goats are
  3. Monty doesn't want you to win a car
  4. You want to win a car (I have to admit, I like goats, though not in that way...)
  5. You can neither hear nor smell the goats specifically enough to locate them
  6. Monty picks a door, on purpose, that hides a goat (as opposed to using some random mechanism independent of his knowledge and goals)
  7. Monty picks a door via some random mechanism, independent of his knowledge and goals.  In particular, that random mechanism is equally likely to pick either door you hadn't chosen

Marilyn assumed, without stating, numbers 1, 2, 3, 4, 5 and 6.  The mathematicians (note: these are the ones who disagreed with her, not every single mathematician in the world) assumed, without stating, 7.  When pressed, they might admit that they had also assumed 1, 3, 4 and 5, but those were not necessary for their argument.  FWIW, Monty seemed to believe 2 and 3.  The fact that Marilyn made SO MANY more untested assumptions makes me lean towards the mathematicians (or, really, their argument.  I'm doing it again...), but their one assumption is so grossly unreasonable they come off looking extremely foolish.  I don't mind pointing this out, since I don't know any of their names.

The Answer

Should you switch doors?  Of course not!  If you want a car, and don't know what's behind each door, and believe that Monty doesn't want you to have a car, and believe Monty has free will, the bulk of the time he offers a swap its not in your best interest.  If, however, you grant all the assumptions of the mathematicians (or don't grant assumption #4), what you do doesn't matter.  If you grant Marilyn's reasonable (but incorrect) assumptions, you should swap.

  1. Some believe that the problem statement implies you will always be offered the option to swap.  Even when I discuss this, and THEN ask the victim to state the problem in their own words, they never, ever state this assumption (assumption number 1)
  2. When anyone offers to prove you're wrong about, essentially, anything by showing you the results of a computer simulation, you should be extremely wary.  In particular, if a mathematician offers to play a simulation game with you, and tells you to BRING MONEY, what they mean is that they'll write a computer program (or pull tickets out of a box, or flip coins, or whatever) to demonstrate the results of their unstated assumptions.  It's easy to write a simulator, in good faith, that proves your point, as long as you don't question all your assumptions
  3. That's why restricted choice is 2:1, wihtout any further (non-bridge) assumptions.  When you have ATxxxx opposite Kxx, and it goes K-x-x-quack, you always get to decide whether to finesse or play for the drop.  Monty can't make your decision for you
A Girl Named Florida

There's a bridge blog about this type of math puzzle and how it applies to different bridge situations, by the Canadian, bridge playing mathematician I mentioned earlier.  I didn't even make it to the bridge part of the blog.  Here's the relevant part.  BTW, it looks like he found this problem in a probability book (one I haven't read) so I hope this isn't too much like a game of telephone, where my version bears no resemblance to the original formulation.  The blogger writes (with my interruptions):

Suppose that a couple have produced 2 naturally conceived children. What are the chances they are both girls? We assume that at the time of conception a boy is as likely to result as a girl. The event is mathematically equivalent to tossing a coin. First we present a false argument that was common in centuries past which goes as follows. There are 3 equally probable states: 2 boys, 2 girls and a boy and a girl. The chance of producing 2 girls is 1 in 3? That is wrong because the probability of a given outcome of a series of random events is proportional to the number of ways in which that outcome could have been produced. One must take into account the birth orders, of which there are 4: boy-boy, boy-girl, girl-boy, and girl-girl. The chance of producing 2 girls is 1 in 4.

Pascal would have got it right, as would most bridge players who are asked, given that 2 finesses are to be taken, what are the chances both succeed? There are 4 equally likely possible outcomes of the play, for two of which one finesse wins and the other loses. The chance of both finesses succeeding is 1 in 4.

I could pick nits here (boys are more likely to be born than girls, coin tosses are biased and serially correlated, and birth order is irrelevant (it's a proxy for conditional probability)) but the math and conclusion are sound, and the assumptions are even stated!  Oh, the probability of two finesses succeeding is either 24% or 26% (depending on whether you finesse through the same hand both times).

Next we ask if one child is known to be a girl what are the chances the other is also a girl? It would be wrong to argue that given that one is a girl doesn’t affect the odds the other is a boy, so the chances of 2 girls should be 50%. The correct argument is that the birth order of boy-boy has been removed from consideration, so there are 3 possible sequences remaining leaving the chances of 2 girls at 1 in 3. Similarly, if we are assured that at least one finesse wins, the chance of the other also succeeding is 1 in 3.

Call me a nit picker, but this is COMPLETELY false.  Grab the nearest five year old you can, and pose the question.  You have two children.  ONE is a girl.  Then, what must the other one be?  The kid will get it right, two kids, one girl, therefore one boy.  Yes, he meant at least one is a girl, of course, but that's TOTALLY different a paragraph from now.

Next we ask, what are the chances of 2 girls given one of them is named Florida? There are those who would argue that whether the girl was named Florida, or Jane, or Laura should make no difference to the odds that their other child is a boy. Although the name Florida is unusual, there is no causal effect at work. Consider the problem statistically and imagine going through US census data looking for all parents with 2 children one of whom is named Florida. Can we expect to find that the other child is a more likely also to be a girl? It doesn’t make sense that we should.

Now my head explodes.   Here are my issues:
  1. Those who argue the name makes no difference are correct.
  2. There might be no causal effect, but he hasn't demonstrated that.  Kid named Florida gets teased, the parents don't want another girl, and so on.  Ok, maybe that's a stretch
  3. Even granted there's no CAUSATION, that doesn't mean there's no CORRELATION.  The self selected group of people who name a girl Florida might be less likely to want girls (e.g. the ratio of boys to girls among third children in China is almost 3-1, not that I assert they name any of them Florida)
  4. Remember when I complained that he said one of them was a girl?  Now he looks for families where that girl (that one girl) is named Florida).  Sound reasonable?  Try restating it correctly:  There are two children, at least one is a girl, and SHE is named Florida.  That's doesn't parse.  One is a girl and named Florida parses fine, but it isn't what he means.
  5. He assumes that two siblings can't have the same name.  That's false (see Michael Jackson's two sons, or George Foreman's five sons) but irrelevant, as it happens

Although there seems to be no causal link between the name and the probability of 2 girls, the argument doesn’t solve the Girl-Florida problem as posed. The correct solution is obtained by incorporating the information that a daughter is named Florida, condition FL, into the possible sequences of births. The possibilities are the following 4: boy- FL, FL-boy, FL-girl, and girl-FL, in half of which the other child is a girl. This corresponds to our intuitive feeling that it doesn’t matter whether the girl was named Florida, or Jane, or Laura, the chances are 50-50 the other child is also a girl. What does matter is that the naming of one child changes the odds for a second girl from 1 out of 3 to 1 out of 2.

AAAARGH!  This is the exact same INCORRECT argument he disdainfully claimed was common in centuries past.  Again, birth order has NOTHING to do with anything.  He just stated by fiat that four possibilities are equally likely, when that's not true given the selection criteria.  In practice, if you pick some normal name (e.g. Joan) there's a pretty good chance the probability of two girls will be about 50%.  That's not because anyone named the child (note, it applies equally to second children named Joan) but because of the way you select your population.  I just wish people would state their assumptions clearly.

And if you don't believe me, I'll write a simulation to prove it to you.

Bring money...

Current Location: United States, San Francisco
Current Mood: annoyed annoyed
Current Music: Jackie - Scott Walker

4 comments or Leave a comment
We played an eight board swiss match against four WBF ranked players (Woman Master, World Life Master, and two Grand Life Masters. Note that these aren't namby pamby titles like ACBL Life Master, since three of them have won a Rosenblum, two won a Bermuda bowl, etc.). I declared six of the eight boards, and had a GLM in my seat at the other table, so I hoped that we'd win big and I'd get a nice blog post. I think there was one IMP decided by card play (I was defending that board) and everything else happened in the bidding. I declared a slam carefully, when I might have gone down, but they weren't in slam at the other table. I lost 11 IMPs in 3N down two when my counterpart made an overtrick, but that all due to the opening lead. GLM on my left led SQ from SQTx (?) and my teammate led a club up to declarer's KJT8x, so that was a two trick difference (the overtrick was gravy). I was so annoyed when I heard their auction. I held A9 K98x Jx KJT8x, and the uncontested auction at the other table was 1D-1H, 1S-3N.

REALLY? I mean, why can't pard hold Kxxx A Axxx Axxx (slam is good), or KQxx A Axxx Axxx or Kxxx A Axxx AQxx (slam is excellent) or even KQxx A Axxx AQxx (grand is excellent. I realize pard might jump over 1H, but try getting to 7C now...).

Anyway, that was super unjust, but we still won 14/20 VPs.

BTW, in another match we had an aggressive auction to 6S:

ATxx Kx KQ98x Ax
KQJx AJTx J7 Qxx

The beer (D7) made me do it:


1N was 15-17. 3D was natural, game forcing. 4C was a control bid for spades (with both minors pard would start with 2S not 2C). 4N was rolling.

LHO led CJ and what's your line? Keep in mind she had already led from the club king against SEVEN DIAMONDS this round. I know they say to lead clubs against slams, but that's ridiculous.

Anyway, I thought about the dummy for a minute then won CA, SK, and tabled HJ. She covered and I won, drew trumps (they broke), cashed hearts pitching my club loser, and led DJ, with the DT dropping eventually.

FOR A PUSH! I don't know how they got to 6S, but they played it the other direction and got a heart lead. (N.B. Opening leads from three small worked very, very poorly all day).

Current Mood: content content

Leave a comment
You hold AT62 Q87 KJ2 432 and open 1N (10-12) at matchpoints. The opponents bid naturally:

1N-2S-P-3D, P-P-X

Your partnership agreement is to lead trumps against doubled partscores, so you lead a small diamond. Dummy tracks KQ873 A93 864 T7 and pard wins DA, then plays another diamond and you draw a third round of trump.

Just for fun, I'll tell you that pard started with 954 J652 A7 KJ65 and deep finesse thinks you can hold them to eight tricks. What do you play to trick four and why?

I haven't run this through a double dummy analyzer, but it looks like the quickest way to set the contract is to duck a diamond to declarer's singleton jack!

Current Mood: amused amused

1 comment or Leave a comment
9) 3rd, favorable
ATxxxx Q9xx AKx -


1C may be as few as two cards, with a weak NT
1N was 13-15 (not a really good 15)
2D was artificial, game forcing checkback
2H showed 4H, did not deny 3S
5C was exclusion RKC (pard showed 2 without)
Now what?

I wasn't ABSOLUTELY positive pard would take 6H as a signoff, but I was almost sure. Still, I bid 6S in case pard had something like KJx AJTx Qxx Qxx, since I can afford to lose a heart finesse as long as it doesn't result in a spade ruff! Not to mention, pard might have KJx Axxx xx Axxx and RHO might have KHJTxx!

In practice, pard had Kxx AJxx xx AQxx, and I ruffed two clubs along the way (trumps were 2-2) to see if I could drop CK and make seven with HK onside long.

Timed properly, if LHO has CK you can make seven anyway:

On a trump lead draw trumps, DA, DK, ruff, CA, C ruff, lead to HJ, C ruff, run trumps for the club heart squeeze.

4) 1st, both
Qxxx AKxxxx x Kx


2N is 4+ hearts, game forcing (not my favorite treatment)
3D is a singleton or void in diamonds, with or without extras (not my favorite treatment)
3H is a punt

We play 3S here as a non-specific, non-serious slam try. 3N would be a "serious" slam try. How do you evaluate this hand? Do you want partner to cooperate in a slam auction without any extra values, even with good controls? Or do you want to back pedal with your 12 HCP?

Since SAK and a minor ace makes slam worthwhile to wonderful pretty much regardless of partner's shape, I think you should make a serious slam try. We control bid first and second round controls economically, so 4C is just right. Pard, with AK Jxxx QJ9 Axxx would sign off over a wimpy 3S try.

31) 1st, unfavorable
AKJxx x KQx QTxx


2C is game forcing
3H is a singleton or void in hearts, club support, significant extra values (3C would have shown modest extras, usually four card support)
4C is natural (I think Redwood here has negative utility, though most of the Diamond LM's I know love it)
4D over 4C would have been RKC (I'm not a Luddite, I just don't like Redwood) so 4S is a control bid for clubs, neutral about diamond control

So, what is 4N and what's your call?

Pard could have bid RKC with 4D over 3H, so the RKC boat has sailed. It's either a diamond control bid, or it's rolling, too strong to back pedal with 5C, basically demanding slam when pard has diamond control. I prefer that treatment. What are the odds pard wants to show diamond control on this auction, but can't afford to pass 5C?

Pard had Qx AJx xx AKxxxx, so we had about 16 winners after the DA lead.

Current Location: United States, California, San Francisco
Current Mood: tired tired

Leave a comment
1) Both, IMPs, 1st seat
Ax xxxx Axx AT9x

1C-1H, 2H-2S, ?

2S is natural (think KQxx), trying for game. I bid 4H, pard went down. What do you think?
If he had something like Kxxx Axxxx xx xx I want to be in game, red at IMPs, since it's over 37.5%, and he should be better than that. I like aces...

2) None, IMPs, 3rd
Ax ATx AJ A98xxx

Pard opens 1N (14+ to 17), opps silent. What's your plan?

We play Walsh relay, so we bid:

1N-2D, 2H-2S, 2N-3C, 3N-4N, 6C

2D was, ostensibly, a transfer to hearts. 2S forced 2N. 3C showed a single suited slam
try with 6+ clubs, but not a great suit. 3N denied interest. 4N was quantitative, asking for
any non-min (4S would have asked for a max). Pard had KQTxx KJx Qx KTx.

I got a heart lead, and led up to CT, to secure my contract. LHO had singleton CJ so I
made seven. At the other table, they bid 1S-2C, 2N-4N, 6N and got a diamond lead,
then misguessed clubs for down one.

I'd waited at least 15 years for a walsh relay auction, and it was nice to reach the right contract for the right reasons. If pard had a max with bad clubs, like KQJ KQxx KQxx xx I'd be happy staying low, though that would be a pretty deep position. I definitely didn't want to force to slam if pard had a min and no interest in clubs.

Current Mood: happy happy

2 comments or Leave a comment
A9753 AQ7 AKT9 2
KQ4 K964 QJ6 J83

1C-(1S)-1N, P-3N

Down 1, -100, win three IMPs!

At the other table 6S went down two. Granovetter got what he deserved for overcalling 1S on JT862 82 87 QT65!

Rubber bridge at the Mayfair my butt. I wouldn't be surprised if Ira Rubin tracked him down tonight and (expletives deleted).

Current Mood: cheerful cheerful

Leave a comment
There are lots of people who play SO SLOWLY you don't know what they're thinking about. Ever. Other people think about valid bridge problems. You can take some of that offline, though. Say you know RHO has 5 spades, 3 hearts, and 2 clubs. 13-(5+3+2) = 3, so he has 3 diamonds. Easy, right? Everybody knows that. But not everybody memorizes it. If you memorize all the distributions, then you don't need to do math. If you're slow at arithmetic, that can speed up your game a lot. Next time, just say 5...3...2...THREE, without the math.

So, memorization is great. Sometimes. The problem memorization, as with anything, is overdoing it. If you memorize something that doesn't always apply (like 2nd hand low, say) you give yourself a mental block. That's a bad idea, though arguably still better than playing super slowly.

I gave myself a mental block to pitch a loser on a loser. A few months ago I built a winner and pitched my loser on the loser but I should have pitched a potential winner on the loser, then pitched my loser on the established winner. It was SUPER costly when the cards hit the 1% chance where it mattered. And my partner expressed his...discontent.

It came up again this weekend, and I got it right. Yay me, for overcoming my learning disability. BTW, I wrote up this deal using bridgebase's deal viewer. Pretty neat, huh?


4 comments or Leave a comment
In the August, 2007 ACBL bulletin, Marty had an article about this card combination:


There's a similar one in the 2001 Encyclopedia of contract bridge (AJ9xx opposite xxxx) but that's a little different since you can't pick up KQT7 onside with the actual holding.

He says "if East-West clubs split 2-2, your play doesn't matter." That means he'll play the ace on either the first or second trick. He then says that when RHO plays the 7 he'll play the ace, because QT7 and KT7 are twice as likely as KQ7. He finishes by saying you should finesse the jack if RHO plays the T instead of the 7.

Ok, fine. Not to mention this was in the intermediate players section, and he specified that RHO is a "normal" intermediate player. Still, against an expert, is it right to play the ace? If RHO will play the T from T7 (and why wouldn't he?) you can finesse the J and then cash the ace (remember, he claimed all doubletons), but now a T from QT7 or KT7 picks up a second trick for RHO.

The encyclopedia gives different probabilities of picking up the suit for one loser, and different lines of play, based on whether RHO will falsecard the T from those three holdings. Well done, ACBL! It also gives a probability of picking up the suit when RHO will split from KQ7. That's pretty optimistic.

If you're not sure you'd find this falsecard, as RHO, at the table, just remember it the way I do: play upside down count.

Current Mood: sleepy sleepy
Current Music: I Fought the Law - The Clash

Leave a comment