Vacant spaces is a useful technique for calculating odds quickly at the table. Unfortunately, many players use it in situations where it isn’t appropriate and, in fact, will give you a wrong answer. To see why, let’s consider a few problems.
Problem 1: The opponents are known to hold 5 spades and 7 cards in each of the other three suits. Spades are known to be 2-3 or 3-2. The remaining cards are distributed randomly. So far, I think we can agree that each player is equally likely to hold a third spade.
You turn to LHO and ask him, “How many hearts do you have?”
“Four,” he answers.
Assuming he has answered honestly, what is the probability that he holds the three-card spade suit?
One way to figure this out is to consider every possible distribution consisting of 2 spades and 4 hearts, calculate the number of cases for each distribution, and add them up. Then do the same for every possible distribution consisting of 3 spades and 4 hearts and compare the two totals. This is easy to do in a spreadsheet, but quite difficult to do in your head at the table. Fortunately, there is an easier way. Simply consider the vacant spaces. LHO is known to hold 2 spades and 4 hearts, leaving 7 vacant spaces. RHO is known to hold 2 spades and 3 hearts, leaving 8 vacant spaces. Therefore, the odds are 7 to 8 that LHO holds the long spade.
While this works, it is important to remember that this is not the correct way to calculate the odds. It is merely a shortcut. We can prove that, in this case—and, in fact, in many cases—the shortcut will give the same answer as the long method. But in some cases that isn’t true. It depends on the nature of the constraints we are given.
Problem 2: Again, the opponents are known to hold 5 spades, distributed either 3-2 or 2-3, and 7 cards in each of the other suits, distributed randomly. This time, we ask LHO a different question: “What is your longest suit? If you have two equally long longest suits, pick one at random.”
“Hearts,” he answers. So far, we have learned nothing useful. All we have done is divided our universe into three equally likely subgroups (not partitions, since there is overlap) and narrowed it down to one of these subgroups. By symmetry, in each subgroup each opponent is still equally likely to hold a third spade. So the probability that LHO holds the third spade is still one half.
“Just how long is this heart suit?” you ask.
“Four cards,” he answers.
“Aha!” you think. “LHO has 7 vacant spaces to RHO’s 8. Therefore, the odds are 7 to 8 (about 46.7%) that LHO holds the long spade.”
This time, however, vacant spaces has led us astray. Even without doing the long calculation, we can see this can’t possibly be true. In fact, LHO must be better than even money to hold the long spade. How do we know this? Before LHO answered that last question, we knew two things: (1) LHO held somewhere
between 4 and 7 hearts (More precisely, his average length was 4.6), and (2) the probability of his having the long spade was one half. When we find out his actual heart length is below average, the probability of his holding the long spade must go up.
In fact, if you do the calculation the long way, you will discover that the probability of LHO’s holding the long spade is 61.5%, significantly different from the 46.7% vacant spaces gave us. (I won’t go through the details of the calculation here. I may post a spreadsheet later if anyone is interested.)
So what went wrong? Why did vacant spaces not work? Because there is a difference between knowing LHO has four hearts and knowing LHO has four cards in his longest suit. In Problem 1, LHO can have 5, 6, or 7 cards in either minor. In Problem 2, that is not true. Vacant spaces fails to take this extra constraint into account, giving the same answer in either case. In short, vacant spaces fails when there are constraints placed on suits other than the one you have direct information about.
What if LHO had answered “Five cards”? Then the long calculation shows us he is 41.1% to hold the third spade. Vacant spaces would make the odds 6 to 9, or 40%. This is much closer, but it still slightly understates the odds.
Problem 3: Same setup. But this time we ask, “What is your longest suit? If you have two equally long suits, pick the higher ranking.”
“Hearts,” he answers.
“And just how long is this heart suit?” you ask.
“Four cards,” he says.
This is not the same as Problem 2. In Problem 2, if LHO has a 4-card minor, he might, in answering the first question, choose to tell us about that minor instead of hearts. This time, he had to answer hearts. So the probability of his having a 4-card minor has gone up. Accordingly, the probability of his having three spades should go down. And indeed it does. The probability of his having three spades is now 52.4%. Vacant spaces, however, still significantly understates the probability, giving the same 46.7% as before.
What if LHO had answered “Five cards”? Now vacant space is almost right. The probability of his having two spades is 40.1%; vacant spaces still predicts 40%. If LHO answers “six” or “seven,” vacant spaces gives precisely the right answer. This is because the additional constraint is no longer relevant. If LHO has six hearts and two spades, we already know he can’t have a six- or seven-card minor.
Note that the second half of Problem 3 is roughly the situation you would be in if LHO were to open one heart, playing five-card majors, and hearts subsequently proved to be 5-2. In general, when we learn about suit distribution from the bidding, the situation resembles Problem 3 more closely than Problem 1, since the information we get pertains specifically to our opponent’s longest suit rather than to a random suit. This means, unfortunately, vacant spaces is often an inadequate tool for analyzing these situations.
What can we conclude from this investigation? When I began it, I was hoping to find some way to tweak vacant spaces to make it useful in more situations. So far, I haven’t been able to do that. I haven’t found any shortcut for doing the long calculation when we have information about an opponent’s longest suit.
The best I can do is a general observation: When you know the length of an opponent’s longest suit, vacant spaces will underestimate the probability of that opponent’s holding a card of interest. This error diminishes rapidly as the length of this longest suit increases.
If anyone can do better, I would love to hear from you.
As an afterthought, I believe I should say a few words about the term “biased information,” which I used in “The Monty Hall Trap.” Some have objected to that term and I see their point, so I avoided it in this presentation, making the same arguments in a different way. Perhaps "biased" is the wrong word, but the basic idea I was trying to convey is this: Some information, because the person who presents it has an agenda, implies constraints not contained in the information itself. In the Monty Hall Problem, for example, we know Monty is not about to end the suspense by opening a door and exposing the grand prize. So you must take that fact into account in your analysis. The datum “There is a goat behind Door Number 2” is not the only thing that matters.
The same thing can be true in a bridge setting. When someone opens the bidding one heart, he has an agenda, namely, finding a trump suit. He is choosing to bid hearts precisely because he has heart length. He is behaving much more like our opponent in Problems 2 and 3 above, answering “What is your longest suit?” than like our opponent in Problem 1, answering “How many hearts do you have?” This does not mean you ignore this information. It simply means he is supplying additional information beyond just his heart length, and you must take this additional information into account when setting up your constraints. I used the term “biased information” to express that concept because it was easier to say than the more precise “information that implies constraints beyond the information itself.”