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The Monty Hall Trap

The Monty Hall Trap

by Phillip Martin, New York City

"Behind one of these three doors," shouts Monty Hall, "is the grand prize, worth one hundred thousand dollars! It's all yours--if you pick the right door."

"I'll take door number one," you say.

"Let's see what's behind door number--No! Wait a minute!" says Monty Hall. "Before we look, I'll offer you twenty thousand dollars, sight unseen, for whatever's behind door number one."

"No! No!" shouts the audience.

"Of course not," you say. "Even assuming the booby prizes are worth nothing, the expected value of my choice is thirty-three and a third thousand dollars. Why should I take twenty thousand?"

"All right," says Monty Hall. "But before we see what you've won, let's take a look behind door number two!"

Door number two opens to reveal one of the booby prizes: a date in the National Open Pairs with Phillip Martin. You and the audience breathe a sigh of relief.

"I'll give you one last chance," says Monty Hall. "You can have forty thousand dollars for what's behind door number one."

"No! No!" shouts the audience.

"Sure," you say.

If Monty Hall had chosen a random door to open, you would calculate that you now had a fifty-fifty shot at the grand prize and would refuse the forty thousand dollars. But he didn't. Showman that he is, he intentionally showed you a booby prize to heighten the suspense. Since you already knew that at least one of the other two doors held a booby prize, you have learned nothing. You still have the same one chance in three that you started with.

This scenario exemplifies a common probability trap: treating biased information as random. Whenever the information itself has a direct bearing on whether or not you receive it, you must be careful to take that into account.

Here, the trap is easy to spot. But the same trap can crop up more subtly in a bridge setting. Let's make up a deal:

S A 5
H 8 7 5
D J 5 3
C K J 7 4 2
S K 7 2
H A 6 4 2
D A 7
C A 10 5 3
1 NT 3 NT

Problem 1

West leads a low spade. You duck in both hands; East continues spades. It appears from the carding that spades are five-three. What is the percentage play to run the club suit?

Problem 2

What is the percentage play if it appears spades are four-four?

Problem 3

Suppose, instead of a spade, West leads from a broken four-card heart suit. Now what is the percentage play in clubs?

Solution 1

Some players would reason this way: East began with three spades to his partner's five. That leaves East with ten unknown cards; West, with eight. So East is five to four to hold the club queen.

This is falling for the Monty Hall trap. If West had led a random suit and that suit had happened to split five-three, the reasoning would be valid. But that's not what happened. West, with malice aforethought, chose to lead his longest suit. Is it any surprise that he has more spades than his partner?

Suppose your opponents at the other table somehow reach three notrump from the North hand. East leads a red suit and (surprise!) he has more cards in that suit than West. Is your opponent supposed to finesse against West for the club queen while you finesse against East?

No, like the booby prize behind door number two, the relative distribution of the spade suit is biased information. You knew ahead of time that West was apt to be longer in whatever suit he led. "Discovering" what you already knew cannot change the odds.

How, then, do you determine the percentage play? Given that spades is West's longest suit, the expected spade break is roughly four and a half-three and a half. West rates to have one more spade than East. So the actual five-three break is only one card away from expectation. It is equivalent to a random suit's breaking four-three.

That means you can counter the bias by pretending that East has only one extra unknown card instead of two. You cash the club king and lead toward the ace. East follows. Now he has zero extra unknown cards. So it's a toss-up. The finesse and the drop are equally likely to fail (or to succeed, for those of you with positive attitudes).

If you're not convinced, you can work it out the hard way. Calculate the frequency of all of West's possible patterns assuming he has one, two, or three clubs (four-zero breaks are irrelevant) and no suit longer than five cards. If the pattern includes a second five-card suit, divide that frequency by two, since half the time West would lead the other five-card suit. Then compute the odds for each club play. Your calculations should bear some relation to those you see here:

West's Pattern A Priori Frequency Adjusted Frequency Total
5-5-2-1 672 336  
5-4-3-1 3360 3360  
5-3-4-1 5600 5600  
5-2-5-1 3360 1680  
Singleton club     31%
5-5-1-2 288 144  
5-4-2-2 2520 2520  
5-3-3-2 6720 6720  
5-2-4-2 6300 6300  
5-1-5-2 2016 1008  
Doubleton club     47%
5-5-0-3 24 12  
5-4-1-3 480 480  
5-3-2-3 2240 2240  
5-2-3-3 3360 3360  
5-1-4-3 1680 1680  
5-0-5-3 224 112  
Trebleton club     22%
Declarer's Play Frequency of Success
Finesse West 53%
Finesse East 60%
Play for drop 60%

Solution 2

With spades four-four, most players would play for the drop, reasoning that an even split in spades would not change the a priori odds. Actually, an even split is unexpected. We expect spades to split four and a half-three and a half. So West's spades are shorter than average. Furthermore, his diamonds are shorter than average. A priori, West's expected diamond length is four. But, after a spade lead, his maximum diamond length is four. His expectation must be something less than four. So West is short in two suits. His expected club length increases accordingly and it becomes right to finesse him for the queen.

If you work it out the long way, you find that finessing against East works about forty-eight percent of the time; playing for the drop, about sixty-one percent; and finessing against West, about sixty-five percent.

Solution 3

After a heart lead, one might finesse against East, since he has only two hearts to West's four. But this is by far the worst of the three plays. As in problem two, the heart lead reduces West's expected spade and diamond lengths. In theory, he can't have five of either suit and, by restricted choice, he is less likely than normal to have four. So his expected pointed-suit length goes down and his expected club length goes up. Despite the fact West has more specifically known cards than East, it is right to finesse West for the club queen. Finessing against East works forty-six percent of the time; the drop, sixty percent; finessing against West, sixty-seven percent. These arguments assume that West can be relied upon to have led his longest suit. If the auction makes certain leads unattractive, or if West has led from a sequence, or if West is simply known to be perverse, then he might have a longer suit and none of this applies.

Sometimes, biased information can come from the auction.

S K 10 7 6
H A 7 4 3
D A 5 2
C J 6
S A J 5 3 2
H 8 2
D Q 9 3
C A 5 4
1 C
1 S Pass 2 C Pass
2 D Pass 3 S Pass
4 S Pass Pass Pass

West leads the deuce of clubs (third and fifth)—low, nine, ace. You play another club. East wins with the queen and leads the king of clubs to tap dummy. West follows to all three clubs. What is the percentage play in spades?

East has five clubs to his partner's three. But, since his club length and his decision to bid clubs were intimately linked, this is biased information.

If East had not opened, you would still know that clubs were three-five (He would not have played the nine at trick one from king-queen-nine.) and it would be clear to finesse West for the spade queen. But he did open, so you know more. You know that he cannot have a five-card red suit. This decreases his likelihood of holding a stiff spade.

If you work it out, you will find that, as in problem one, it's a toss-up between the two plays. Both the finesse against West and the drop are about sixty percent.

Paradoxically, when you discover the club break yourself, you have a clear percentage play. But when East tells you about it, you have no clue what to do. This is comforting news for those of us who like to open the bidding.

Reprinted by permission of Bridge Today.

© 1989 by Bridge Today


I was introduced to the Monty Hall problem in the 70’s by a friend who had run across it in his college probability course. Years later, when I began writing this article, I decided it would make a good introduction, serving as an easy-to-understand example of the principle I was trying to explain. Although the problem is well-known now, at that time I had never encountered it in print.

About a year later, the problem appeared in Marilyn vos Savant’s column in Parade magazine. Ms. Savant gave the right solution, then received thousands of irate letters, some of them from college math professors, complaining that her solution was wrong. Eventually, the controversy found its way to the front page of the New York Times.

Actually, those who complained weren’t entirely wrong. The correct answer depends on your assumptions about Monty Hall’s actions. As I state in my article, if you assume he is simply opening a door at random, the probability you are holding the grand prize is indeed 50-50. If you assume he will always show you a booby prize, then his action is irrelevant; the probability remains one third.

Mathematicians are trained not to assume facts that aren’t explicitly stated, which is the right approach in a mathematical proof. Game players make assumptions about their opponents’ behavior all the time, which they need to do if they hope to win. The truth is, despite the brouhaha in the media, I have never known a bridge player to have the slightest difficulty understanding this problem. So, for my intended audience, I stand by my claim that this problem is easy.

A similar philosophical difference can be found in the “July 4th Problem,” popularized on the radio show Car Talk. Tom says, “Did you know that, of the first five U.S. Presidents, three died on July 4?” Ray answers, “No. In fact, I knew nothing about the first five Presidents except that they are, in order, Washington, Adams, Jefferson, Madison, and Monroe. But, based on your statement, I can now identify one of those five as having died on July 4.” Which President did Ray identify, and how did he know?

The answer is “Monroe.” And Ray identified him by reasoning that, if Monroe wasn’t one of the three, Tom would have said, “Did you know that, of the first four U.S. Presidents, three died on July 4?”

Some people consider this problem flawed. I don’t think so. It simply depends on the context. If you are taking a final exam in a logic course, Ray’s assumption is unwarranted. But, if you assume Tom was genuinely trying to impress Ray with an unlikely fact, his assumption is entirely valid.