Monty Hall

THE MONTY HALL TRAP BY PHIL MARTIN

"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 sign of relief.

"I'll give you one last chance," says . "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 could 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 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.

*Easy to spot? Ha!

About a year after this article was first published (Bridge Today Magazine, 1990), the Monty Hall problem appeared in Marilyn vos Savant's column in Parade magazine.

Ms. Savant , stated, as I had, that the probability of your having chosen the grand prize remained one third. She received more than 1,000 letters scoffing at this conclusion and demanding that she print a "correction" and an apology. Many letters were from preeminent mathematicians and scientists.

Eventually, the debate found its way to the front page of the New York Times (July 21, 1991). In the end, the scoffers had to eat their words. But I forever lost my right to call this problem easy. - Phil Martin   Explain again

The Monty Hall trap in Bridge:

Noord

A 5

K 8 7

J 5 3

K J 7 4 2

Zuid

K 7 2

A 6 4 2

A 7

A 10 5 3

Problem 1: West has 5 spades and leads a small one, you duck and he continues with spades. How to play the clubs for 5 tricks?

Problem 2: West has 4 spades and leads a small one, you duck and he continues with spades. How to play the clubs for 5 tricks?

Problem 3: West has 4 hearts and leads a small one. How to play the clubs for 5 tricks?

Solution 1: If you think west has only 8 vacant places (13 minus 5 spades) and east has 10 vacant places (13 minus 3 spades) so the club queen must be found with east you'are wrong.
West leads his longest suit and that has consequences:
Suppose that at the other table your opponents are also in three notrump with north dealer.
East has 5 diamonds and leads a small one, now north has to find west with club queen?
That would mean your opponent has to play differently?
The relative distribution of the spade suit is Monty Hall's empty door.
You expect the spades to be 4.5 with west and 3.5 with east, the actual distribution gives west one more card in spades.
After playing a club to the king and a club towards the ace and east following with a small club, the situation is that west has as much vacant places as east.
West has club queen = 53% - East has club queen = 60% - Play for the drop = 60%

Solution 2 : Playing for the drop is wrong.
You expect 4,5 spades with west, actually he has four.
West leading spades means that his maximum diamond length is 4, his expected diamond length is less than 4.
Less spades and diamonds means more clubs.
West has club queen = 65% - East has club queen 48% - Play for the drop = 61%

Solution 3: After a heart lead also play west for the club queen.
Just like in problem 2 the lead of a four card heart makes a five card unlikely and other four cards less likely (restricted choice) so there is more room for club length.
West has club queen = 67% - East has club queen = 46% - Play for the drop = 60%
The percentages above are only valid if west leads his longest suit.

Conclusion what opponents tell you about their cards (bidding or leads) hardly effects how you should handle a suit.
What you discover yourself about opponents distribution indicates how to play a suit,
This does not apply if west leads from a six card or longer than the vacant places method should be used.

From the book:
For experts only
Pamela and Matthew Granovetter

Explaination: I shuffle 52 cards you pick one.
What are the odds you have the spade ace?
1/52
There are now two decks:
One of 1 card (your deck).
And one of 51 cards (my deck).
If you had to guess where is the spade ace?
Exactly in my deck odds: 51/52.
Does it make a difference if I show you 50 cards?
No. (You have learned nothing you already knew that there were 51 non spade aces in a deck)
Would you like to change decks?
Of course? The chance spade ace was in my deck was 51/52.


Play a Monty Hall simulation
Or play the game


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