Vacant Places

In the game of bridge the law (or principle) of Vacant Places can be used both to aid in a decision at the table and to derive the entire suit division probability table. The principle of vacant places can be summarized as follows:

When the distribution of one or more suits is known exactly, then the probability of finding a particular card in a particular hand is proportional to the number of unknown cards (vacant places) in that hand.

The principle of vacant places follows from Conditional Probability theory, which is based on Bayes Theorem. For a good background to bridge probabilities, and vacant places in particular, see . See also

How Vacant Places calculations work at the table

K962

AJ873

Suppose that we are the declarer and our trump suit (let's say hearts) is Kxxx in the dummy and AJxxx in hand. There are four cards missing, Qxxx (in this context, the T can be considered an x). We play small to the K and lead small back. The last of the small cards appears on our right. There is one card to be concerned with, namely the Q. Because nobody would play the Queen of the suit when they have a small card to play, we can consider the three small cards separately from the Q. We can therefore perform a vacant places calculation. Assume for now that we know nothing as yet about the other suits. Our right-hand opponent has two known cards (the xx of our suit that he's already played) and therefore 11 vacant places in which the Q may reside. Our left-hand opponent has only admitted to one card (the x he has already played) and so has 12 vacant places. Therefore, in 12 of the 23 possible cases, LHO has the Q and we must play our Ace. In the other 11 cases, RHO has the Q and we must play the J. The odds in favor of the Ace play are therefore 12/23 = 52.174% which is exactly the probability that appears in the standard tables.

Now let's suppose that LHO dealt and opened a (weak) 2S vulnerable. Between our hand and dummy, let's say we have four spades. We now arrive at the same situation as before. In addition to the one heart we know he has, we also know that he has six spades in his hand (he might even have a seventh spade but we don't know that for sure). Therefore, there are now only 6 vacant places in which the HQ may be hiding in LHO's hand. Assuming for now that RHO definitely has two spades (he might have a third but we don't know that for sure), and the two hearts we know about, he now has 9 vacant places for the HQ. There is therefore a 60% chance (9/15) that RHO has the missing Queen and that the finesse of the J will succeed.

How Vacant Places calculations work away from the table

Let us imagine that we are required to build up a set of probability tables to help show how a suit might be splitting, for example, the Probability of suit distributions in two hidden hands on the page Bridge probabilities. Let's suppose that we are missing three cards in the suit and we know nothing about the distribution of other suits (i.e. we are looking for the a priori probabilities). When we "deal" out the first card of the three, we can put it in either hand. Each hand, by definition, has 13 vacant places, so it is a toss up which hand it goes into (13/26 = 50% for either hand). Now let us suppose that we want to know the probability of the suit being divided 3–0. The first card is already in, let's say, the East hand. Now he has only 12 vacant places so the probability of that hand getting the second of the three cards is 12/(12 + 13). This must be multiplied by the initial 1/2 probability to find the probability of East having both of the first two cards. Now let's deal out the third (and last) of the missing cards. By this time, East has only 11 vacant places, while West still has 13. The probability of East getting all three of the missing cards is 1/2 × 12/25 × 11/24 which is exactly 0.11, which is the value that we see in the fourth row of the table (3 - 0 : 0.22 : 2 : 0.11).

Now, let's calculate the individual probability of a 2-2 split when missing four cards (the following row in the table). This time, proceeding similarly to before, the calculation is:

13/26 × 12/25 × 13/24 × 12/23 = (3 × 13) / (23 × 25) = 0.067826.

The probabilities of other suit divisions can be calculated similarly.