I have in front of me an anthology of bridge (as in the card game) essays entitled For Experts Only, edited by Pamela and Matthew Granovetter. Essay number six was written by Phil Martin, and is entitled “The Monty Hall Trap.” Sounds interesting, but I am most definitely not an expert at bridge. In fact, I know nothing about the game beyond the basic rules. So I was hoping there was someone out there who could explain to me what Martin has in mind. Below the fold I have transcribed a lengthy excerpt from the article, starting at the beginning. All italics are in the original.
“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 Phil Martin. You and the audience breathe a sigh of relief.
“I’ll give you one last chance,”says Monty Hall. &kdquo;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 50-50 shot at the grand prize and would refuse the $40,000. But he didn’t. Showman that he is, he intentionally showed you a booby prize to heighten the suspense. Because 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:
Spades: A, 5
Hearts: 8, 7, 5
Diamonds: J, 5, 3
Clubs: K, J, 7, 4, 2
Spade: K, 7, 2
Heart: A, 6, 4, 2
Diamonds: A, 7
Clubs: A, 10, 5, 3
South: 1 NT North: 3 NT
Problem One. West leads a low spade. You duck in both hands; East continues spades. It appears from the carding that spades are 5-3. What is the percentage play to run the club suit?
Problem 2. What is the percentage play if it appears spades are 4-4?
Problem 3. Suppose, instead of a spade, West leads from a broken four card heart suit. Now what is the percentage play in clubs?
Solutuion 1. Some players would reason this way: East began with three spades to his partner’s five. That leaves East with 10 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 5-3, 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 distirbution 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 5-3 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 rthe 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 with positive attitudes.)
If you’re not too 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 (4-0 breaks are irrelevant) and no suit longer than five cards. If the pattern includes a second five-card suit, divide that frequency by two, because half the time West would lead the other five card suit. Then compute the odds for each club play.
Martin then goes on the provide a table listing the results of these calculations. There’s a bit more to the article beyond this, but I shall stop here for now.
I’m afraid I can’t make heads or tails of this once the bridge jargon starts. If anyone can offer some help I’d greatly appreciate it.
I should also mention that this essay was originally published before the Marilyn vos Savant controversy about the Monty Hall problem. In the anthologized version Martin has a footnote indicating that in light of the controversy he can no longer describe the problem as easy.