This column began with the notion that I’d introduce you to Jeff Rubens’ conception of “*short shorts,*” the bridge version, that is, as described and illustrated in his book *Expert Bridge Simplified: Arithmetic Shortcuts for Declarer*, 2009, The Bridge World Press. I’ll get there, but I will take two columns to do it.

*Simplified*, as in the book’s title, doesn’t mean simple. The book proposes to teach you enough about permutations, combinations, and the differences between these so that the book is self-contained. Don’t believe it. If you’re not familiar with Bayes’ Theorem and prior probabilities going in, you’re not going to master all that’s in the book. Not to say you won’t enjoy parts of the book and learn from it. You know, for example, that if you hold seven cards of the heart suit, the other six will split 3-3 about 35.5% of the time and that one of the opponents will hold 4 hearts about 48% of the time, that is, 24% for each. But suppose one opponent pre-empts and you “know” that opponent holds seven or eight cards of the bid suit. What now is the probability that his partner has exactly 4 hearts of the six outstanding? (40%) [Rubens sometimes taxes my ability to read his tables. If I were to ask what is the probability that the partner has 4 or 5 or 6 of the outstanding hearts, I think the answer is 71%, but I’m not sure I’m reading the table correctly.] Rubens has a table for that and similar information on page 113 you’ll be hard pressed to find elsewhere.

Here’s what I’m going to do. You’ve heard of “*restricted choice.*” If you understand how it works, and why, you can skip this column. In my experience, people can mouth the words, but can’t explain it. I’ve read handfuls of explanations, only one (Rubens’) of which I find compelling in the sense that someone who didn’t understand before reading understands afterwards. I’m going to take a crack at it. I think I have figured out that part of the problem results from the way writers use the word "choice". They make it sound that sometimes when you play a card to a trick you make a choice and sometimes you don’t. No wonder you are confused, because every time you play to a trick you make a choice. Even when you play a singleton, you make a choice—a choice not to revoke, for example. But I’m getting ahead of myself.

Before I get to restricted and unrestricted choices, I’ll visit another real world application of prior probabilities that people know but don’t understand: Monty Hall’s 3-door game from the TV show *Let’s Make a Deal*. To review: Monty has a contestant, here called Patsy, to whom he says: You face three doors, with labels A and B and C. Behind two of the doors are goats and behind the third is a car provided by our sponsor. You pick a door, whichever door you please. Once you settle on the door and we open it, no shenanigans, you can keep what is behind the door. Clear? Yes.

Let’s say Patsy chooses Door B. Monty assigns B to Patsy, while he assigns A and C to himself. Then he approaches one of his doors, say Door C, and flings it open to reveal a goat preening behind it. Then, he says to the Patsy, “Would you like to change your door choice from your current pick, Door B, to this other one, Door A?”

By now, after the original show from the mid-1970s and newspaper *Parade*’s intense coverage in the early 1990’s, many people know the problem and its solution, given the assumptions, which I’ll state clearly in a moment: Patsy should switch doors and double her chances of winning the car. Few can explain why that’s the right answer, which is why we are here.

Nobel prize winners have been known to get this wrong, claiming it’s 50:50: two doors, each equally likely to hide the car. Read the Wikipedia entry for the Monty Hall problem for the history of this puzzle and the conflicting arguments as to why the wrong answer, 50:50, appeals to so many smart people.

Now, to be formal.

Assume:

- Monty knows which of three Doors, A and B and C, hides the prize but wants to open one that that does not hide it, so to string out the suspense for both the contestant and the audience.
- The Patsy chooses her door at random and that the prize has 1/3 chance of being behind. The prize has 1/3 chance of being behind each of the doors, including the one she chooses.
- Monty opens a door, reveals the prize is not behind that door, and asks Patsy if she would like to switch her choice to the other door.

What should Patsy do? Members of the audience invariably offer conflicting advice. Your advice?

*Here’s a way to think about it*. We have two sets of doors, the Monty set of two doors and The Patsy set comprising a single door.

I hope we can agree that the prize has a 2/3 chance of being behind one of Monty’s two doors and a 1/3 chance of being behind Patsy’s singleton. Let’s write the number 2/3 on each of Monty’s doors and the number 1/3 on Patsy’s door, so that we don’t forget these so-called “prior facts”.

Monty opens one of his doors and there is no prize there. What is the probability that the Monty set has the prize? Still 2/3, yes? What is the probability that the unopened Monty door with the number 2/3 written on it has the prize? 2/3, right? [This is key.] What is the probability that Patsy’s door hides the prize? Still 1/3, yes? What does the number on the door say? 1/3, right? Which door do you prefer to open if you want the prize? The one that says 2/3. So do you switch your door from Patsy’s to Monty’s? Yes. Every time? Yes. Rubens uses the same intuition in explaining the 3-door’s solution to the skeptical reader on his page 180. He focuses on Patsy’s door’s probability starting at 1/3 and remaining 1/3, so the single other Monty door must have increased to 2/3. He says this in such an abbreviated way that you’ll likely overlook the profound thought given in his less than one-third of a sentence of explanation of the key thought. He assumes, but doesn’t state, that the sums of the probabilities must add to one. You’d never get to page 180 if you didn’t have that baked into your bones.

Monty’s doors hid the prize 2/3 of the time. The fact that he knew one of them did not hide the prize didn’t alter the fact that the other one did hide the prize 2/3 of the time. Here, I’m going to paraphrase the advice of the bridge player/writer, Terence Reese, on this subject: **A game contestant, such as Patsy, playing to optimize the chances of winning the car, should assume that Monty, in picking which door to open to reveal the goat, not the prize, had no discretion.** In bridge terms that we’ll get to later, Monty’s choice as to which door to open to reveal the goat was restricted because there was only one door he could choose to show a goat. If you make that assumption and act on it, you will be invoking the so-called principle of “restricted choice,” and you will switch doors. I do not, by the way, think that it’s obvious that following the boldface advice in this paragraph explains that Patsy should switch doors. I think the why’s of the restricted choice literature are opaque. I hope my, or Rubens’, explanation in the preceding paragraph touches your intuition. You knew before the rigmarole of Monty’s showing a door opening, that his two doors had a 2/3 chance of hiding the prize. That didn’t change because he, with perfect knowledge, showed you behind a door that didn’t hide the prize.

So here’s the thread between Monty, Terence Reese, and bridge. This is tough stuff. I needed to get your attention.

**Now the bridge version.**

You hold nine cards in a suit: AJ1098 in dummy opposite 7654 in hand, with KQ32 outstanding. You have easy entry between the hands. Play to win the maximum number of tricks. Assume East-West want to win their maximum and will false card as follows. If a defender has to play from a holding including both the 2 and the 3, he will play one of them at random. If he has to play from a holding including both the King and Queen, he will play one of them at random.

Declarer plans to consider taking two finesses. After taking the first, losing, finesse, he considers playing for the drop, for the suit to split 2-2.

I have read several writers’ attempts to explain why taking the second finesse, rather than playing for the drop has a 2/3 chance of succeeding. I have spent hours trying to find a compelling way to explain. In the end, I decided I can do no better than to paraphrase Rubens — paraphrase, rather than quote, because he treats a different situation with some different considerations.

Rubens, page 171, as much as writes:

At the outset, the four possible distributions of the honors are (--|KQ; Q|K; K|Q; KQ|--). Assume those four cases were equally likely. When the first finesse loses, the fourth case (KQ|--) is eliminated, but the other three remain equally likely. “A second finesse will win in two out of three so its chance of success is (approximately, because of our assumption) 2/3.” [I corrected an obvious typo in this quote, which some would mark with a snarky *sic*.] Later, on page 287, in the book, Rubens addresses the same situation, except that the suit layout is ♣32 opposite ♣AJ109876, functionally equivalent. He doesn’t mention restricted choice in discussing how to play this suit. Rather, he discusses the merits of playing two finesses vs. banging down the Ace. He derives the merits, the wisdom, of playing two finesses and shows the second finesse is twice as likely to succeed as is playing for the drop, but he writes as though he and the reader are ignorant of restricted choice. Maybe it’s better that way — to derive the optimal playing strategy without confusing readers with a “choice” label that they mightn’t understand. He’s had decades of experience as editor learning what readers understand and don’t.

Many debate who was the greatest bridge player no longer alive and separately, who was the greatest bridge writer, but I think most would agree who was the greatest player who wrote and conversely, who was the greatest writer who played: Terence Reese. Look him up on Wikipedia and save me the space of having to say more. About taking the second finesse in this layout he says, almost offhandedly, on page 36 of his 1958 *The Expert Game*, which the 2007 ACBL survey of experts rated the best bridge book of all time, “On the next round [declarer] should finesse again. Since East could have played either card indifferently from KQ, the fact that he has played one affords an indication that he does not hold the other.” Two pages later he states in **boldface **the how of putting restricted choice to work: **a defender should be assumed not to have had a choice rather than to have exercised a choice in a particular way. **This boldface statement I reworded earlier for advice to the Patsy in playing Monty’s game.

I’ve thought about Reese’s words and the term “restricted choice” often over the past 40 years and I get confused almost every time. My trouble comes from the word “choice” and its meanings. If I have to play to a trick and I have only one card in the suit, I might say I have no choice. If I have a doubleton or more, then I say I have a choice. But in the context of “restricted choice”, I now know that if I am following suit to a trick I want to win and my holding is, say, K2, then I have no choice, I must play the King. If I hold KQ, then I have a choice, as the King and Queen are equals at this point.

So when I’m East and declarer finesses into me and I play the Q, Reese would say I have no choice if I hold singleton Q or Q2, but do have a choice if I hold KQ. Alan Truscott, who first wrote about these matters a few years before Reese, called it “restricted choice.” For him, playing Q from Q or Q2 is a restricted choice, but playing Q from KQ is not restricted. Playing Q from singleton is restricted because it’s the only one you have and from Q2 is restricted because you want to win the trick and it’s the only way to do that. I can’t think that I’ve ever read that the constraints implied by the word “restricted” in “restricted choice” can come either from the rules of the game, such as you have to follow suit, or from your own preferences, such as I prefer to win this trick rather than not. I think some writers would say that playing Q from Q2 is not a restricted choice but they wouldn’t be focused on this issue. They’d be using the word in its everyday sense.

I’m tempted by the flexibility of the English language to use a different word to cleave the concepts at issue here. How about “discretion”? Reese could have said**: A defender, playing to optimize tricks taken in a deal, should be assumed not to have exercised discretion. **So, in the textbook examples, it’s twice as likely that an East observed to play the Jack holds a singleton Jack rather than that he holds a doubleton J10. Assume East had no *discretion *in playing that Jack.

This column concludes without my having said much about Rubens’ book, but the next column gives more. I end with a bridge deal over which I have discretion because both Reese and Rubens provide illustrations. I restricted my choice to ones in their two books. I’ll use the old master’s example from page 39, which he describes as follows: “In the 1955 World Championship match the American declarer would have saved an adverse swing had he drawn the right conclusion on the following hand:

From Terence Reese,

Schapiro and I succeeded in making Four Hearts after a spade lead and continuation. At the other table the American South played in Three No-Trumps. It looks as though, after a spade lead and return, he should make nine tricks by way of one spade, two hearts, two diamonds, and four clubs after the drop of the ♣Q under the Ace. In practice he went one down because he did not finesse ♣9 on the second round. He suspected that East’s Queen was a false card from Q10. It might have been, but the way to look at such problems is this: if East has held Q10 he might have dropped the Queen and he might have dropped the 10; with the Queen alone he could drop only the Queen; that makes the Queen singleton more likely, even after the greater frequency of a 4-2 division has been taken into account.”

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Great article. As they say: All roads lead to Rome. My view on the Monty Hall doors is similar, perhaps from a different angle.

Patsy has three strategies available after she is shown the empty door.

She can insist on her primary choice. The probability is the same as at the beginning as Monty can show one empty door from the other two (so no additional information was gained).

She can throw a coin in the air to stay with the primary choice or change to the other closed door. Therefore, the probability is ½. This mistake was made by many, even from some mathematicians. The reason I think is, that the conditional probability is somehow non-intuitional.

The third possibility (changing doors) offers the best chances, namely 2/3. Why? Let us say we use the following strategy. Patsy will pick the door, where no car is behind (two from three doors, so a probability of 2/3), then Monty will show her the other empty door and all she has to do is change the door and drive home (with 2:1 chances).

When I have to think carefully about such matters I prefer to use the word option to indicate one possibility. I use the word choice to indicate a decision between two or more options. In your example with the KQ32 missing, of the layouts where there's a difference the Q is the only option to win the trick in 2 layouts and, thus, the K is in the other hand. In one layout the Q is one of two options. So when the Q appears it is now twice as likely that the K is in the other hand and finessable.

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Great thinking, Roman!

Thanks !

Another great post, looking forward to the next one