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Brace yourself for another article about probabilities. Our starting point will once again be the Restricted Choice scenario. Suppose you have a 9-card fit in a key suit holding ace-king-ten in your hand. When you play one round from the top, LHO follows with the queen.

If we take for granted that LHO can only hold either the stiff queen or queen-jack tight, and that he will play either honor with equal probability in the latter case, it has been established over and over again that the odds for a successful second round finesse are about 2/3 (Vacant Spaces may distort the odds depending on how much about the rest of the hand is already known).

In a previous article I have demonstrated how Bayes' formula can be used to compute the odds based on other assumptions on LHO's strategy. Of course, such an assumption must come from elsewhere; one particular hand cannot be taken as proof for a general strategy.

For example, if you assume that LHO always follows up the line, the odds are 100%. On the other hand, if LHO is the kind of player who will play the queen whenever he has it, the odds are only about 50% because his play says nothing whatsoever about the location of the jack.

Whatever LHO's strategy really is, the odds are always somewhere between 50% and 100%. Restricted Choice basically says that taking the finesse is superior because with the stiff queen LHO had no choice at all, but with QJ he had a choice to make, so he might have played the jack at least some of the time.

Now, if the finesse is always better than the drop, it appears futile to invest any energy in an exact computation of probabilities. But that is only true if the entire problem is limited to this one suit. Global considerations can make a difference.

Consider the following construction:

North

♠

K4

♥

A763

♦

QJ10543

♣

6

South

♠

A1097532

♥

84

♦

AK

♣

AK

Against a 6♠ contract played by South, West leads some heart honor, immediately establishing a trick for the defense and removing a dummy entry. When declarer wins and cashes the ♠K, East plays the six, West follows with the queen.

What is the best way to make the contract? There are three reasonable lines:

* Plan A: Cash the ♠A. This wins if the trumps split 2-2 but loses otherwise. If the trumps are 3-1, there is no chance to recover.

* Plan B: Take the spade finesse. This plan works if East holds the ♠J but loses otherwise.

* Plan C: Cash two high diamonds. If they stand up, cross to dummy via a club ruff and lead a high diamond. If East follows, we can discard our heart loser; if he ruffs, we can safely overruff. One way or another, we will lose only one trick.

Plan C mainly works in case of a 3-2 break in diamonds, but a 4-1 break will do as well if the player with short diamonds also has no second trump. A player with eleven cards might have bid at some point; since I did not give the bidding, we cannot make this more precise, but let us assume the opponents had the chance to interfere (perhaps over 2♣-2♦) and chose not to do so.

The initial odds for a 3-2 break in diamonds are roughly 68%, and the first two tricks have changed very little about it. The combination of a diamond singleton, a trump singleton in the same hand and silence in the bidding will be very small in general, so let us say the odds for Plan C are about 70%.

According to our Restricted Choice results, we might deduce that the odds for the spade finesse are nearly the same (and that Plan A is inferior), so it appears to be a coin-flip between Plan B and Plan C.

The results from the previous page are wrong, or at least incompete. The main point is, we can no longer assume that LHO must hold either the stiff queen or queen-jack tight. If West holds QJ8, he might be tempted to falsecard on the first round of trumps, so that we waste our only other opportunity to play from dummy to take a finesse which he knows to be off.

Depending on the bidding, West might even know that he can even afford to play the queen from Q8. This case is different from the typical RC scenario because we have not seen East's play in the second round; we must decide first if we play another trump at all.

What we have are basically four possible layouts after the first round of trumps have been played:

* Layout L_1: LHO started with the stiff ♠Q.

* Layout L_2: LHO started with ♠QJ.

* Layout L_3: LHO started with ♠Q8.

* Layout L_4: LHO started with ♠QJ8.

From the dealing process, all four layouts are about equally likely (let us ignore rounding errors from Vacant Spaces). But after trick two, the relative odds have certainly changed, because the defenders had to make relevant choices.

In order to arrive at precise odds for the success of each plan, we need precise assumptions on the probabilities that West would play the queen from either of these four holdings. In fact, East also had to make a choice in every layout except the last one, and they should be included as well.

Let us denote the odds that the defenders would have played in the trump trick as they did by p_1, p_2, p_3 and p_4 (corresponding to the respective layout). Further, let us denote by X the state of the play that has been reached after trick two.

According to Bayes' formula, we have for each Layout L_i (using the standard notation for conditional probabilities I have introduced in earlier posts):

P(L_i|X) = P(X|L_i) * P(L_i) / P(X)

with P(X) = Sum (P(X|L_j) * P(L_j)),

where the sum is taken over j=1,2,3,4. If we make the simplifying assumption that all P(L_j) are equal, we arrive at

P(L_i|X) = p_i / Sum (p_j).

This is the formula all further computations are based on. Whenever we take an exact strategy by the opponents for granted, we can compute the consequent odds for each layout in the trump suit. (No inferences about the probabilities in the diamond suit can be drawn from this approach.)

Case 1: Suppose both opponents will play randomly from equals, but that neither of them will waste on honor unless he has to. In this case, we have p_3=p_4=0. Furthermore, we have p_1=1/2 (East might have played either of his spot cards), and p_2=1/4 (both opponents had to make a 50-50 choice).

These assumptions yield

Sum (p_j) = 1/2 + 1/4 + 0 + 0 = 3/4.

Not surprisingly, this brings us back to the original RC scenario. We obtain

P(L_1|X) = 2/3

and

P(P_2|X) = 1/3.

(By the way, the results are the same if we assume that East will play his spots up the line.) As you see, the 2:1 odds in favor of the spade finesse are as we initially said they would be, namely almost the same as the odds of a 3-2 break in diamonds. Regarding the decision between Plan B and Plan C, we could go either way in this case, although Plan C still has a slight advantage.

Case 2: Suppose West is a beginner who always play up the line, in particular from QJ tight. This means Layout L_1 is the only possible one, and we have P(L_1|X)=1. If we knew this was the case, clearly we would not put our contract on the line by playing on diamonds. Again, East's choices between his spot cards do not matter here.

Case 3: Let us try some numbers which include the possibility of falsecarding. We suppose that East would not drop the jack from an initial holding of J86, so we start with p_1=1/2 and p_2=1/4 once more.

It is hard to judge how often West would play the queen from Q8 or from QJ8. Intuitively, I would say that is slightly easier for him to falsecard in the catter case. Let's say he would drop an honor from this holding half the time (and the queen or the jack randomly), so we get p_4=1/4; East has no choice to make in this layout.

With West holding Q8 and East holding J6, I believe the odds that West would falsecard but East would not are somewhat lower. Let us assume p_3=1/8. This yields:

Sum (p_j) = 9/8,

P(L_1|X) = 4/9,

P(L_2|X) = P(L_4|X) = 2/9

and

P(L_3|X) = 1/9.

Using these figures, Plan A will work 1/3 or 33% of the time (Layouts L_2 and L_3), Plan B will work 5/9 or 56% of the time (Layouts L_1 and L_3). Neither of these percentages is close to the odds for Plan C, so declarer would be well-advised to forget about the trump suit and play on diamonds instead.

The numbers for the defenders' choices are only very basic approximations, and they might be off a lot. It appears, though, that Plan C is superior in any case.

Let us change the hand layout slightly:

North

♠

K4

♥

A63

♦

QJ107543

♣

6

South

♠

A1097532

♥

84

♦

AK

♣

AK

Dummy now has another diamond (the heart loser is still there, of course). In this new layout, Plan C requires a 2-2 break in diamonds (or the diamond shortness in the same hand as the trump shortness). The odds are a little above 40%, hence now the trump finesse is best.

If West is the kind of player who would falsecard a lot in this situation (perhaps because he can see two small diamonds in his hand and must fear that declarer will make the right decision if he follows small, or just for the fun of it), the odds will be less clear. The trump finesse appears to remain a favorite, though.

In another article I wrote that I prefer the Bayesian approach to the simpler method of just counting the relevant initial layouts. The Bayesian way is more accurate because it allows to take strategy into account.

On the other hand, the Bayesian way is hardly doable with greater precision in complex situations. If you think I am doing all the above computations in my head during play, don't kid yourselves. I use a lot of approximations (and a considerable part of the assumptions about the defenders' strategy is guesswork anyway).

When falsecarding becomes an issue, counting initial holdings is very dangerous, to say the least. One major aspect of the "shortcut" is that apparently irrelevant holdings are often ignored from the start. When layouts in two different suits must be considered, originally irrelevant holdings may become relevant and have a significant impact on the computations.

In the end, it is of course up to you which way you go. Like I said before, I still think it is useful to know the theory.

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