I have in front of me an anthology of bridge (as in the card game) essays entitled *For Experts Only*, edited by Pamela and Matthew Granovetter. Essay number six was written by Phil Martin, and is entitled “The Monty Hall Trap.” Sounds interesting, but I am most definitely not an expert at bridge. In fact, I know nothing about the game beyond the basic rules. So I was hoping there was someone out there who could explain to me what Martin has in mind. Below the fold I have transcribed a lengthy excerpt from the article, starting at the beginning. All italics are in the original.

“Behind one of these three doors,” shouts Monty Hall, “is the grand prize, worth one hundred thousand dollars! It's all yours -- if you pick the right door.”“I'll take door number one,” you say.

“Let's see what's behind door number -- No! Wait a minute!” says Monty Hall. “Before we look, I'll offer you

twenty thousand dollars, sight unseen, for whatever's behind door number one.”“No! No!” shouts the audience.

“Of course not,” you say. “Even assuming the booby prizes are worth nothing, the expected value of my choice is thirty-three and a third thousand dollars. Why should I take twenty thousand?”

“All right,” says Monty Hall. “But before we see what you've won, let's take a look behind

door number two!”Door number two opens to reveal one of the booby prizes, a date in the National Open Pairs with Phil Martin. You and the audience breathe a sigh of relief.

“I'll give you one last chance,”says Monty Hall. &kdquo;You can have

fortythousand dollars for what's behind door number one.”“No! No!” shouts the audience.

“Sure,” you say.

If Monty Hall had chosen a

randomdoor to open, you would calculate that you now had a 50-50 shot at the grand prize and would refuse the $40,000. But he didn't. Showman that he is, he intentionally showed you a booby prize to heighten the suspense. Because you already knew that at least one of the other two doors held a booby prize, you have learned nothing. You still have the same one chance in three that you started with.This scenario exemplifies a common probability trap; treating biased information as random. Whenever the information itself has a direct bearing on whether or not you receive it, you must be careful to take that into account.

Here the trap is easy to spot. But the same trap can crop up more subtly in a bridge setting. Let's make up a deal:

NorthSpades: A, 5

Hearts: 8, 7, 5

Diamonds: J, 5, 3

Clubs: K, J, 7, 4, 2

SouthSpade: K, 7, 2

Heart: A, 6, 4, 2

Diamonds: A, 7

Clubs: A, 10, 5, 3

South: 1 NT North: 3 NT

Problem One.West leads a low spade. You duck in both hands; East continues spades. It appears from the carding that spades are 5-3. What is the percentage play to run the club suit?

Problem 2.What is the percentage play if it appears spades are 4-4?

Problem 3.Suppose, instead of a spade, West leads from a broken four card heart suit. Now what is the percentage play in clubs?

Solutuion 1.Some players would reason this way: East began with three spades to his partner's five. That leaves East with 10 unknown cards; West with eight. So East is five to four to hold the club queen.This is falling for the Monty Hall trap. If West had led a

randomsuit and that suit hadhappenedto split 5-3, the reasoning would be valid. But that's not what happened. West, with malice aforethought, chose to lead hislongestsuit. Is it any surprise that he has more spades than his partner?Suppose your opponents at the other table somehow reach three notrump from the North hand. East leads a red suit and (surprise!) he has more cards in that suit than West. Is your opponent supposed to finesse against

Westfor the club queen, while you finesse againstEast?No, like the booby prize behind door number two, the relative distirbution of the spade suit is biased information. You knew ahead of time that West was apt to be longer in whatever suit he led. “Discovering” what you already knew cannot change the odds.

How then do you determine the percentage play? Given that spades is West's longest suit, the expected spade break is roughly four and a half - three and a half. West rates to have one more spade than East. So the actual 5-3 break is only one card away from expectation. It is equivalent to a

randomsuit's breaking four-three.That means you can counter the bias by pretending that East has only

oneextra unknown card instead of two. You cash rthe club king and lead toward the ace. East follows. Now he haszeroextra unknown cards. So it's a toss-up. The finesse and the drop are equally likely to fail (or to succeed, for those with positive attitudes.)If you're not too convinced, you can work it out the hard way. Calculate the frequency of all of West's possible patterns assuming he has one, two or three clubs (4-0 breaks are irrelevant) and no suit longer than five cards. If the pattern includes a second five-card suit, divide that frequency by two, because half the time West would lead the other five card suit. Then compute the odds for each club play.

Martin then goes on the provide a table listing the results of these calculations. There's a bit more to the article beyond this, but I shall stop here for now.

I'm afraid I can't make heads or tails of this once the bridge jargon starts. If anyone can offer some help I'd greatly appreciate it.

I should also mention that this essay was originally published before the Marilyn vos Savant controversy about the Monty Hall problem. In the anthologized version Martin has a footnote indicating that in light of the controversy he can no longer describe the problem as easy.

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Try Dr. Bradley Lehman, a mile or two to your west in Dayton.

http://www-personal.umich.edu/~bpl/hdbc.htm

I don't know all the bridge jargon, but I can get through most of this.

The basic point is that N/S need to guess who has the Qc - E or W?

Two rounds of spades were played, starting with W. Based on how W/E played their spades, N/S have deduced that W started with 5 spades and E with 3. That means W has 8 non-spades, while E has 10 non-spades. Naively, that suggests that the odds of E holding the Qc rather than W are 10:8 (=5:4).

The problem is that W did't lead spades at random. He did it because he had 5 to start with. It's like Monty picking a door because he knows it's not the prize.

If W hadn't had 5 spades, he would have led a different suit where he did have at least 5 cards.* So, the fact that he had 5 spades is biased information.

To help see that, I imagine taking a deck, giving N & S exactly the cards they have above, then dividing the remaining 26 cards randomly between W & E. That way, it's obvious that the Qc has a 50:50 chance of being in either hand. Even so, W will (usually) be able to lead from a suit where he has 5+ cards. If our knowledge of bridge tells us that's what he'd want to do, then the fact that he did so here doesn't help us.

*Given the cards N/S have, it's unlikely W would have no suits of 5+ cards, but it's possible. I don't know the odds.

I'm no expert, but I've played a bit of competitive bridge.

First, there appears to be a mistake in your transcription of the info provided. "South: 1 NT North: 1 NT" doesn't make sense, since every subsequent bid must be higher than the previous. I suspect that North bid 3 NT, which is a game bid and also forecloses any further bidding by South. It's also justified by North's high card point count and distribution. Part of what causes the dilemma is that South opened with a No Trump distribution, but with only 15 high card points (18 is desirable for the 1 NT bid.). Making 2 NT is not a problem, so fairly obviously the contract reached is 3 NT.

So, if South is to win a contract of 3 NT (the "book" of six tricks, plus 3 more, with no suit as trump), it's immediately obvious that he only has six sure tricks and thus his only chance of making the contract is by winning 5 tricks in the Club suit.

He is faced with trying to make the best determination of the distribution and placement of the remaining four clubs. If they are split 2 and 2, there is no problem. If they are split 3-1 with the queen in the hand with three, then he must initially choose which way is the most favorable to make a finesse, since he is capable of doing it both ways. (If you need an explanation of finessing, ask.)

Aside from the fairly intuitive assumption that West has 5 spades, there is no information to indicate distribution of the other suits other than pure probability. Although, with East returning spades, there may be an inference that East has no specific agenda of his own, which probably rules out any very long suit there. It may also be fair to make the assumption that neither hand holds KQJ of hearts.

If, after the first two spade tricks, South cashes the Ace of clubs, he has essentially already made an assumption that either the Queen lies in West's hand or the cards are split 2-2, although I can't see logical grounds for that. That move means he can only finesse toward the King.

After that, the math mavens can take over. I hope this has been somewhat helpful.

chezjake:

I doubt N bid 3NT, based on the paragraph that begins "Suppose your opponents at the other table..." It sounds like he's describing another table at duplicate, with a different N/S pair where N did raise to 3NT, and E led at trick 1. I don't actually play, so I'm not sure, but does that make sense? Maybe the original problem had N pass?

An interesting thought occured to me after my first post, as I was driving home. This problem is essentially the inverse of the Monty Hall problem.

Monty's biased action seems useless in locating the prize, but in fact it helps a lot. Here, West's biased action seems helpful in locating the Qc, but in fact it isn't!

Also, I was going to dispute Martin on this:

I started to argue that the odds do change in the Monty Hall case; that they go from equal probabilities to 2:1 in favor of the 3rd door. But I realized that's wrong.

At first, there's a 1/3 chance that 'our' door wins. There's a 2/3 chance that one of the remaining doors wins while the other loses. After Monty opens a losing door, there's still a 1/3 chance that our door wins, and a 2/3 chance that one of the other two wins. We just know which of the other two can't be it.

I agree this is only an interesting problem if NS are in 3 NT (that means they need to take nine tricks out of 13, you always add six to the amount bid).

chezjake:

"Part of what causes the dilemma is that South opened with a No Trump distribution, but with only 15 high card points (18 is desirable for the 1 NT bid.)."

It's not unusual for a partnership to play 15-17 high-card points for NoTrump, and also not unusual for a person to think that having three aces and a king is worth an extra point even if they are playing 16-18. For those who aren't familiar with high-card points, you ususally count 4 for an ace, 3 for king, 2 for a queen, and 1 for a jack, but some people feel an ace is worth sighly more than 4 and a king a little more than 3.

qetzal:

"It sounds like he's describing another table at duplicate, with a different N/S pair where N did raise to 3NT, and E led at trick 1. I don't actually play, so I'm not sure, but does that make sense? Maybe the original problem had N pass?"

The first person of a partnership to bid a denomination plays the hand. So, perhaps some other South, thinking as chezjake does and playing that you can open a four card heart suit, opens the bidding with one heart, and North replies 1 NT. The final contract will still be 3 NT, but this time North will play the hand.

I think knowing the spade split 5 (with West) vs. 3 (with East), as opposed to 4-4, is useful iinformation that you did not have at the beginning of the problem, and so does change the possibility of West holding three clubs to the queen. If you finesse, you finesse through East. The article is probably right that the odds are about even on which works and which fails.

There appear to be several misstatements in Mr. Martin's quote that confuse the problem:

"It appears from the carding that spades are 5-3"

"'Discovering' what you already knew cannot change the odds"

These are inconsistent, depending on what "appears from the carding" means. Since he proceeds to reason assuming q 5-3 split, I have to conclude that the first statement means just what it appears to mean. Then the second is irrelevant since that's not what happened.

Knowing that W will lead the long suit results in some

a prioriprobability that the split is 5-3, but the actual play will change that probability. After the initial lead, S knows the possible splits (those that sum to 8), and after E's return lead, some of the splits are eliminated (any that don't have E with at least two spades)."biased information"

I am (well, used to be several decades ago) relatively savvy in probability theory and information theory. I don't recall ever encountering this concept. Without a definition, it isn't clear what it means - and therefore, what the argument using it means.

"This is falling for the Monty Hall trap."

By analogy, it seems that falling for the MH trap would be to assume that since the Qc can be in either E or W, the probability is 1/2 for each. Noting that the play changes the relative probabilities seems to be having learned not to fall into the trap.

It isn't immediately obvious to me what knowing additional information about the spade split does to the statistics of the club split, which is the essence of the problem (ie, should the Qc finesse be targeted at E or W, both of which are doable). Eliminating the splits where E has 0 or 1 spade increases the probabilities of the rest, in particular those which cause the finesse to fail. But I have no intention of following my own advice and working out the sample space to find out!

- (Lazy) Charles

In that last part, replace "in particular those which cause the finesse to fail" with "which in turn changes the probabilities of the club distribution". I was in a hurry and mixed up clubs and spades. I think with that change, the rest is OK, or at least no more likely to be wrong than it was without that offending phrase in there.

- Charles

Charles:

I think Martin is arguing the reverse. Falling for the MH trap in this case means concluding E is more likely to hold the Qc, when the probability is really 1/2 for each. Thinking that the spades play changes the relative probabilities IS the trap.

I'm almost sure that's Martin's argument, and I'm pretty sure he's correct.

How about it Jason? Are you able to make sense of this now?

First, thanks to everyone for the remarkably clear explanations. They've helped a lot.

Also, there was indeed an error in my transcription, and North did bid 3 NT. Sorry about that. I'll go back and correct it as soon as I post this comment.

qetzal- You're explanation made a lot of sense. Thanks!

chezjake- Just to be sure I follow what you're saying: The six sure tricks you're referring to are the four aces and the two kings. Since there is no trump suit, North/South can be sure to win any trick in which they play an ace, and since they have all the aces they can be certain of winning with their kings as well.

Now, clubs are the only suit in which North/South are strong. Except for their aces and kings, they have only low cards in every other suit. That is why they have to win five tricks in clubs; the two sure tricks with the ace and the king, and three others. That is why the location of the queen of clubs is so important. How am I doing?

I'm afraid I will have to have it explained to me what a finesse is. If you wouldn't mind, perhaps you can also explain the term “ruff.”

Charles-

I feel your pain with regard to Martin's writing! His explanation of the Monty Hall problem leaves a lot to be desired. I don't think he intended “biased information” as a technical term in mathematics. If he did intend it that way I can only say that I have never heard the term.

I think all he had in mind was that in updating your probability assessments in the light of new information, you need to be aware of whether the new information was chosen at random from equally likely possibilities, or whether there was some bias among the pieces of information you might have received.

In the Monty Hall problem, the distinction is between Monty opening a door at random and revealing a goat by chance, or Monty being forced to open a door concealing a goat. This distinction is relevant to how you update your probabilities.

In Martin's bridge example, if I'm understanding it correctly, we are given information (via East/West's play) about the distribution of spades, and are tyring to use that to draw an inference about the location of the clubs. If our information had come from learning something about the locations of randomly chosen non-clubs, then we would be justified in drawing the 5:4 inference described above. But since we were not equally likely to receive information about any of the four suits, we must adjust our reasoning accordingly. Or something like that.

Anyway, thanks again for the help. I'll go back and read the rest of the essay and see if I can make sense of it inlight of what you all have said. If only there was some way to relate the Monty Hall problem to chess, then I'd be much more in my element!

qetzal-

Your last comment crossed mine in the ether. I think you are right as to Martin's intention. But I will have to mull it over a bit more before deciding if he is right.

Sorry I got here late, but things seem to be under control. In case you are still wondering, to "ruff" in bridge is the same as to "trump", which means it does not play any part in this example (in NT there is no trump suit). "Finesse" means to avoid losing to a high card in one of your opponent's hands by means such as leading through it to a hand (either your hand or the dummy) which has cards above and below that card. For example, if you knew (or assumed) that West had the Qxx (queen and two small cards) of clubs, you would take the second spade lead in your hand (South) and lead a small club toward the dummy. If West played low, you would play the jack, then drop his queen with the ace and king. (If he plays the queen you take it with the king.) If you knew West had three clubs but didn't know whether the queen was one of them, you would play the ace from your hand first, in case East had the singleton queen.

With cards shown, you can run the club suit against a 3-1 or 1-3 division of the outstanding clubs (providing you know which side has the long clubs). If the division were 2-2, then you can drop the queen without finessing. However, the uneven distribution is more likely, so figuring out who is likely to have the longer club suit is crucial to the hand. Given a known 5-3 spade division (which you may be able to deduce based on the the fact that one normally leads the 4th highest of one's longest and strongest suit against a No Trump contract), I myself would be tempted to guess the long clubs were with East - but I see the author's point. Similar thinking got me the top score on a dupicate hand once, but there the spades were 7-1.

Wikipedia probably has more information - maybe even some on the "Theory of Restricted Choice", which as I have mentioned before, seems reminiscent of the Monty Hall problem also.

Well, let's try this tack.

View the hands as four jars into which the cards are dealt. In dealing hands normally, one essentially distributes each card into a randomly chosen jar. But consider an alternative but equivalent (trust me for now) dealing technique:

- pick a suit (let's assume spades)

- randomly distribute the 13 cards among the four jars

- put the Qc into one of the four jars

- randomly distribute the remaining cards among the jars until each is filled

After the spades have been distributed, there are 39 cards remaining. A given hand (let's assume E) will have n spades and therefore can accept 13-n additional cards. So, the conditional probability that the Qc is put in E given that E has n spades is (13-n)/39.

Now let's address the "trust me". This isn't a proof, but it should give some confidence. We know that after a normal deal, the probability of a given card - in particular, the Qc - being in E is 1/4. Using the new dealing technique, it should be the same. The probability that E gets n spades is:

p(n) = B(13,n)(1/4)**n (3/4)**(13-n)

where B(a,b) is the binomial coefficient. Then the probability that Qc is in E is the sum from 0 to 13 of p(n)(13-n)/39, which in fact does work out to be 1/4.

Now to the bridge problem. After W's initial lead, N's lay down, and E's follow of a spade and lead of a spade, S knows that there are eight spades outstanding, that W has at least four (otherwise, one of the other suits would have at least four cards and would have been lead), and that E had at least two spades. The possible spade splits and remaining non-spade capacity are then:

W-E

Spades Not spades

------------------

4-9 ....... 4-9

5-8 ....... 3-10

6-7 ....... 2-11

Since the conditional probability of Qc being in a hand is an increasing function of the number of non-spades, the probability that it is in E

given our assumptions about W's strategy for the initial leadis always greater than or equal to the probability it is in W. Hence, the finesse should target E.Or maybe not. Critiques welcome.

- Charles

Wow, very neat blog. Would be awesome if you could post some simplified bridge traps, as I am definetely no expert but love the game!

I just noticed that ironically, there is a mistake in the easy part of my last comment: my description of a normal deal. Obviously, you deal the cards into the jars sequentially, not randomly.

Doesn't bode well for the challenging part!!

- Charles

This is the first time I came across the blog. Will come back later today. Let me give another explanation of 'finesse' because that is the whole point of this deal.

If you are South in another hand and have A Q in a suit, and one of the opponents has the King, then your strategy is start from the North hand, and lead a card. If East has the King he has to play first -- if he plays the King then you win with the Ace and the Queen is now good. If he plays low,you play the Queen, and hope that it will 'ride'. However, if West actually had the King, he will take the Queen (usually, but in bridge he can play a lower card and not win if he chooses to)

In this bridge hand, the hidden card is the Queen and S has A 10 x x and North has K J x x x -- so if you want to finesse for the Queen you can do it either way (lead toward the Ace or lead toward the King). If you were going to finesse, typically you would first cash the high card from the other hand, in case the Queen is a singleton. So the Monty Hall thing is tied to deciding which way to finesse.

Plus, there is another subtlety here worth mentioning. The S&N hands have 9 cards between them, so there are 4 outstanding. You also have the option of playing A-K and hoping the 4 cards are divided 2-2 and not 3-1 or 4-0. The probability that they are 3-1 is slightly more likely, but any additional information is important.

Jason:

I'm certainly no expert at bridge, but I did find THIS site, which seems to be describing the apparent paradox of which you speak. Or not, it's a bit of a head scratcher for me.

Jason,

Sorry I'm so long getting back to this.

You're quite right in evaluating the six sure tricks and the problem with the club suit.

JimV explained the "ruff" sufficiently for this example, since there's no trump suit. Generally speaking, in contracts with a trump suit, a player leads a losing card in a non-trump suit toward a known void in his partner's hand, planning for that hand to take the trick by trumping it.

I think that John Abbott's explanation of finessing is a bit clearer than JimV's, but both suffice.

I also agree with OneBrow's assessment of how another table could arrive at a 3NT contract with North playing the hand.

One other overall consideration comes to mind. N-S only have one stopper in each of the red suits, so it's very important for South to establish the clubs while he still has a stopper in each other suit.

Having thought some more about the problem, I have somewhat greater confidence that my conclusion is correct and Mr. Martin is not. Having reread his argument, I see where he may have gone astray.

His essential assertion seems to be that since you know

a priorithat whoever leads the first trick will go with their longest suit, the fact that W does so conveys no new information. By itself, that's true, but after N's lay down you have additional information, namely the possible spade splits. Note that my "proof" depends on knowing that E can have no more spades than W since we know W must have at least four and there are only eight outstanding. If there were more than eight spades outstanding, my simple argument would no longer work. The conclusion might be the same, but you'd have actually compute the probabilities in order to find out.Mr. Martin asks (in essence), "if N plays and E leads hearts, is [N] supposed to finesse against W for the Qc?", with the subtle implication that the response should be "of course not, that's a foolish idea". However, if I am right, then the answer is in fact "yes" - for the same reason. In that case the finesse fails, but that only confirms the obvious fact that playing the odds doesn't guarantee success (and that if you're going to be snarky, make sure you're right!). And it's still possible that Mr. Martin

isright - my confidence in my conclusion is only modestly high. But if he is, I don't think it's for the reason implicit in his query, ie, that the play shouldn't depend on the suit. Since after the initial lead and the lay down you have information about the possible splits for each suit, they are no longer equivalent unless they have equal numbers of outstanding cards.Mr. Martin's comment about "work[ing] it out the hard way" is disturbing since defining the sample space carefully and counting (or computing) is always "the gold standard" approach, but I'm somewhat skeptical.

- Charles

My "simple argument" is incomplete (why is left as an exercise for the reader, if any), but the conclusion is still right and can be gleaned without calculating. However, calculating the actual probabilities reveals that it's about twice as likely that the Qc is in E.

Sound familiar? (It's just coincidence, of course.)

- Charles

Although everyone has worked out the bridge part quite well, the Monte Hall part is still quite incorrect.

1) Once the bogus prize has been revealed, there are only two possible choices left for the prize, leaving you with a nominal 50-50 chance at $100 grand, meaning that you SHOULD NOT take the $40,000.

2) In fact, using the theory of restricted choice (which was mis-stated in the Sunday magazine article), the odds that the remaining door holds the grand prize are even greater than 50%.

In the Sunday magazine article, it was never stated that Monty Hall knew what was behind the door ahead of time, or that he deliberately chose a door with a booby prize. Those people, like me, who never wasted their time watching that dumb show, were confused as to why the odds would go above 50-50.(But not as confused as Jason)