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North

♠

64

♥

AQ74

♦

AKQJ

♣

AJ10

South

♠

AK5

♥

863

♦

10874

♣

752

Many bridge players, including me, have been confused by the widely accepted explanation of equals, called Restricted Choice. Confusion arises because cards that bridge players need to consider to be equals, such as the missing ♣K and ♣Q in the deal above, are marked differently, making it difficult to consider them as equals. Any explanation of equals in bridge has to deal with the oxymoron of *distinguishable equals*.

The math applicable to a deal where declarer is missing the ♣K and ♣Q, is the same math applicable to 2 reds balls and 24 white balls randomly and evenly divided between two urns. The contents of the urns correspond to East's and West's hands and drawing a ball from an urn corresponds to a defender playing a card. There is no question that the two red balls are equals, but when a defender plays the ♣K or ♣Q, doubt arises: "Are they really equals when a defender, holding both, always chooses to play the ♣Q?" My goal is to make the cards-model exactly match the balls-and-urns model, with its utterly simple view of equals.

I find the clearest explanation of Restricted Choice to be the one provided by Hugh Kelsey and Michael Glauert in their book *Bridge Odds for Practical Players*, pages 92 thru 94. I will use their North and South hands and pattern my explanation after theirs, but with important differences in emphasis.

North

♠

64

♥

AQ74

♦

AKQJ

♣

AJ10

South

♠

AK5

♥

863

♦

10874

♣

752

You are the South declarer in a 3NT contract. West leads the ♠T and you win the first trick with the ♠K. You try the club finesse which loses to an honor in East's hand. East returns a spade to knock out your last spade stopper. For your ninth trick, you must choose between the heart finesse (clearly a 50% chance) or a second club finesse (better or worse?). Let's look in detail at this question.

**Simplest Conditional Probability Problem in Bridge**

There are four ways the club honors can be distributed between the East and West hands:

** Distributions of ♣K and ♣Q**

**Distribution West East Probability**

** [1] ♣K,♣Q --- 24%**

** [2] ♣K ♣Q 26%**

** [3] ♣Q ♣K 26%**

** [4] --- ♣K,♣Q 24%**

The distribution probabilities in the table show how frequently declarer can expect, over the long run, each distribution to occur when the shuffles and deals are fair and honest. It is important to realize that the calculation that we are about to perform is based on nothing but these distribution probabilities. This point will come up repeatedly.

Here is a way of describing *all* probability calculations, but will show the nature of conditional probability clearly:

**probability = successful relevant distributions / all relevant distributions**

*Before* the first club finesse all four distributions **[1]+[2]+[3]+[4]** are relevant but only **[1]** is successful.

24% / (24% + 26% + 26% + 24%) = 24%

*After* East won the first club finesse with an honor, **[1]** is no longer relevant because the information from East's first card informs us that East started with at least one club honor. That leaves **[2]+[3]+[4]** as relevant distributions. Thus, we see the nature of conditional probability: *As cards are played, the** information they provide reduces the number of **relevant distributions. * Of the distributions that remain relevant, the second club finesse will succeed when East started with only a single club honor, as in distributions **[2]+[3]**.

So, the conditional probability that the second club finesse will succeed is the successful relevant distributions, **[2]+[3]**, divided by all relevant distributions, **[2]+[3]+[4]**:

(26% + 26%) / (26% + 26% + 24%) = 68%

which is much better than the 50% expected from the heart finesse.

You may have noticed that I have been referring to the ♣K and ♣Q as club honors but never by their individual ranks. That was deliberate because that makes the cards-model exactly match the balls-and-urns-model, where the two club honors correspond to the two red balls. Other gimmicks have been used over the years to make equals indistinguishable, such as painting equal cards with red paint; or saying that the light is so dim that declarer can see that the card defender played is a face card, but not which face card; or in the case of queen and jack equals, calling them "quacks". The time has come to dispense with all these gimmicks and take the bull by the horns and **insist that because the ♣K and ♣Q are equals, when either card is played, that card must be thought of, not as the ♣K or ♣Q, but as one of the ♣KQ equals**.

Think of ♣KQ as a single symbol that can have a value of 0, 1 or 2. Before a defender plays, think ♣KQ=0. After a defender plays the first ♣K or ♣Q, think ♣KQ=1. The only important attribute equal cards have as a group is "How many?". Ignore their individual ranks.

** Distributions of the ♣KQ Equals**

**Distribution West East Probability**

** [1] ♣KQ=2 ♣KQ=0 24%**

** [2] ♣KQ=1 ♣KQ=1 52%**

** [3] ♣KQ=0 ♣KQ=2 24%**

Thus, the conditional probability that the second club finesse will succeed is now the successful relevant distributions **[2]** divided by all relevant distributions **[2]+[3]**:

52% / (52% + 24%) = 68%

In this calculation, the ♣KQ equals are indistinguishable, making the ♣KQ symbol equivalent to the number of red balls pulled from an urn in the balls-and-urns-model. This is the proper way that declarer should view equals.

**Repulsion of Equals**

There is a law of repulsion between played equals and unplayed equals.

For the case of two equals that we have been considering, the play of the first equal by a defender "repels", with nominal 2-to-1 odds, the unplayed equal into the other defender's hand. *Repel *is simply a descriptive term for the pattern found in probability calculations involving equals. *Nominal* simply means the usual value when there are no other large impacts on the probability calculation, such as a defender opening the bidding with a weak two or showing a void early in the play.

When there are three equals, the play of the first equal "repels" the other two with a phenomenal 85% probability that the other defender holds at least one of the unplayed equals. However, after a defender plays two equals, the unplayed equal is "repelled" with almost 3-to-1 odds into the other defenders hand. However, after both defenders have played an equal, the third unplayed equal is repelled equally by each played equal, so it's a toss up as to which defender holds the unplayed equal.

This way of thinking, that played equals "repel" unplayed equals, provides an easily remembered and straightforward way to help you plan the play when equals are involved. I propose calling this pattern *Repulsion of Equals*.

**Restricted Choice**

Let us now switch horses in midstream and take a look at just how complicated an explanation can become when it considers the ♣K and ♣Q to be *distinguishable equals*.

** Distributions of ♣K and ♣Q **(repeated for convenience)

**Distribution West East Probability**

** [1] ♣K,♣Q --- 24%**

** [2] ♣K ♣Q 26%**

** [3] ♣Q ♣K 26%**

** [4] --- ♣K,♣Q 24%**

Previously we said that East won the first finesse with a club honor, then we changed from "club honor" to "one of the ♣KQ equals". Now we consider the effect of saying that East won specifically with the ♣K.

Distributions **[3]** and **[4]** are the only ones where East could have started with the ♣K so they are the only ones that remain relevant to the conditional probability calculation.

The second finesse will succeed when East started with only the ♣K, distribution **[3]**. At first glance, the probability seems like it should be **[3]** divided by **[3]+[4]**, or:

26% / (26% + 24%) = 52%

but this answer is not correct. To explain why, we need to *require that East, holding both the ♣K and ♣Q, randomly pick **between the two*, playing ♣K half the time and ♣Q the other half of the time. This has the effect of splitting distribution **[4]** into two parts: **[4K]** where East plays the ♣K first; and **[4Q]** where East plays the ♣Q first.

** Restricted Choice View of ♣K and ♣Q Distributions **

**Distribution West East Probability**

** [1] ♣K,♣Q --- 24%**** **

** [2] ♣K ♣Q 26%**

** [3] ♣Q ♣K 26%**

** [4K] --- ♣K,♣Q 12%**** **

** [4Q] --- ♣K,♣Q 12%**

After East played the ♣K, the relevant distributions are **[3]** and **[4K]**, the only distributions where the ♣K would be played first. The conditional probability that West started with only the ♣K is **[3]** divided by **[3]+[4K]**, or:

26% / (26% + 12%) = 68%,

which agrees with the value we calculated earlier.

Even though Restricted Choice calculates the correct answer using a conditional probability calculation, the explanation is built on the *requirement* that an opponent, when holding equals, must pick randomly between them, in effect, telling the opponents how to play their cards to make the explanation work. This questionable requirement opens Pandora's box and lets loose the following troubles:

- The explanation fails for three or more equals. Declarer missing, say, the ♣KQJ identifies them as equals but when West holds the ♣Q East and holds the ♣K and ♣J, the explanation cannot
*require*that East pick randomly between the ♣K and ♣J when East has no reason to consider those two cards to be equals. - The explanation is needlessly complicated, giving rise to game theory questions. Can a defender gain an advantage by playing one of the equals more frequently? Can declarer take advantage of a defender who plays this way? Game theory has definitive answers to both of these questions: Defender's optimal strategy is to pick randomly from equal cards, and declarer's optimal strategy is to ignore the ranks of equal cards that defenders play. These answers are conclusive but bridge forums continue to beat this dead horse.
- The explanation is misleading in two ways. Bridge players are misled into thinking that all conditional probability calculations involving equals
*require*that a defender pick randomly from equals. Bridge players are also misled into thinking that Bayes Theorem applies because Restricted Choice considers defenders choice, or lack of choice,*before*playing an equal. The explanation presented earlier in the section titled "The Simplest Conditional Probability Calculation in Bridge", is not subject to these misconceptions because it is based on nothing but the distribution probabilities resulting from a fair and honest deal. The defender's choice, or lack of choice, is*ignored completely*. The focus is clearly on the time*after*defender played one of the ♣KQ equals. Equals in bridge can be explained without being so misleading.

**Comparison of the Two Explanations**

Repulsion of Equals takes a simpler view of the initial ♣K and ♣Q distribution and Restricted Choice takes a more complicated view.

All conditional probability calculations involving the distinguishable equals of the cards-model require that equals be viewed, one way or another, as indistinguishable. *Repulsion of Equals correctly places responsibility on declarer* to view equals as indistinguishable by insisting that declarer only count the played equals and pay no attention to ranks. * Restricted Choice incorrectly places responsibility **on the defenders* to view equals as indistinguishable by requiring them to play equals randomly.

This transfer of responsibility from declarer to defenders is at the root of all the Restricted Choice troubles listed earlier, but even worse, it makes a match of the cards-model with the balls-and-urns-model impossible. The Repulsion of Equals explanation, on the other hand, provides an exact match between the two models, with none of the troubles inherent in the Restricted Choice explanation.

A final point. Even the name *Restricted Choice* sows confusion. Kelsey and Glauert, in their book *Bridge Odds for Practical Players, *find the name to be "unfortunate", with "overtones of complexity". They chose to rename it *Freedom of Choice*. Unfortunately, both the old and new name incorrectly place the emphasis on defender's choice, or lack of choice, *before* defender plays an equal. The name *Repulsion of Equals* is actually a brief description of the law of repulsion between played and unplayed equals.

North

♠

64

♥

AQ74

♦

AKQJ

♣

AJ10

South

♠

AK5

♥

863

♦

10874

♣

752

**I call on bridge teachers everywhere to teach Repulsion of Equals to lift the burden bridge players have been saddled with by the poor name and the inadequate, needlessly complicated, and misleading explanation called Restricted Choice.**

If you are a bridge teacher, here's how you might introduce Repulsion of Equals to a student:

Teacher: You are declarer in the deal shown above. Do you agree that the missing ♣K and ♣Q are equals?

Student: Yes.

Teacher: Should you regard equals differently from non-equals during the play?

Student: Probably, but I'm not sure how.

Teacher: Well, the way you treat equals differently is by ignoring their actual ranks and think, instead, of counting them. So when a defender first plays the ♣K or ♣Q, think of the card as "the first of the ♣KQ equals". Does this seem right to you?

Student: I don't see anything wrong with it, but what's wrong with just thinking of the ♣K as the ♣K?

Teacher: Didn't you just agree that the ♣K and ♣Q were equals and that equals need to be treated differently?

Student: Yeah, O.K. So how does it help to just count the equals?

Teacher: There are two benefits. When you ignore the actual ranks of the equals you automatically use your optimal strategy against the defenders, whose optimal strategy is to randomly pick from equals. The main benefit, though, is that you can now use an easy way to think about defenders play of an equal, called Repulsion of Equals. It says that played equals always "repel" unplayed equals, no matter how many equals we are talking about. For our case of two equals, the play of the first ♣KQ equal by a defender "repels", with 2-to-1 odds, the unplayed equal into the other defender's hand.

Student: What do you know about this thing called Restricted Choice?

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