The Bridge Hands With an Ace
December 14th, 2013 at 12:45:51 PM
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With apologies to kubikulann and to avoid confusing others, I shall state the problem exactly
as it was put to Marilyn Vos Savant by Martin Gardner, as reported in her book The
Power of Logical Thinking, which can be read on line courtesy of Google Books.
The Greens and Blacks are playing bridge. After a deal, Mr. Brown, an onlooker, asks Mrs.
Black: "Do you have an ace in your hand?" She nods. There is a certain probability that her
hand holds at least one other ace.
After the next deal, he asks her: "Do you have the ace of spades?" She nods. Again, there is
a certain probability that her hand holds at least one other ace. Which probability is
greater? Or are they both the same?
Martin Gardner
Hendersonville, North Carolina
Unlike some of her much simpler puzzles, her reply drew no indignant responses from Ph.D's
saying that she was wrong; however I think her analysis has an important error and that she
got the right answer merely by luck. If so, here is a case where both Marilyn and her army
of Ph.D. critics were wrong.
In what follows, nCr will stand for the number of ways of choosing r objects from a
collection of n objects. It is sometimes called "n choose r" and is a built-in function on
scientific hand calculators.
Method I (Statman):
If the ace of spades is dealt, there are 48C12 ways of dealing the remaining cards not
containing another ace and 51C12 ways of dealing them total with a probability of
48C12/51C12 = 0.43884, so there is a Pr of 0.56115 that the hand will contain at least one
other ace.
The Pr that if a hand contains at least one ace of whatever suit it will not contain another
ace is the number of ways of dealing 13 cards containing only one ace divided by the number
of ways of dealing hands containing at least one ace. The latter is the number of ways of
dealing all hands less the number of ways of dealing hands without aces. This can be
expressed as 4*48C12/(52C13 - 48C13) = 0.630362 so the Pr that if a hand contains an ace the
Pr it will contain at least one other ace is 0.36963.
Method II (Marilyn Vos Savant)
"There are fewer opportunities to get a certain ace than any ace at all but each of these
groups gives an equal number of ways to get more aces, so the smaller group has the greater
success ratio." Although the conclusion is correct, the reason is wrong. Both groups do not have
an equal number of ways of getting more aces. Reducing this to numbers, if the ace of spades
is dealt it has 52C12 - 48C12 ways of getting at least one more ace. This is the number of possible
completions less those that contain no aces, so each one that remains contains at least one ace
and the Pr of getting another ace is (52C12 - 48C12)/51C12 = 0.56115. So far so good!
If it is known that the hand that has been dealt contains at least one ace, it could be any
of 52C13 - 48C13 hands, of which 4*48C12 contain exactly one ace; therefore this hand has
52C13 - 48C13 - 4*48C12 ways of getting more than one ace. This evaluates to 1.6341E11
whereas 52C12 - 48C12 = 1.36711E11. The hand with an ace of unspecified suit has more ways
to get additional aces but because of the size of this group it loses out in the ratio.
Marilyn reached the correct conclusion for he wrong reason.
Method III
Oza. Rather complicated, but his results agree with Statman's.
as it was put to Marilyn Vos Savant by Martin Gardner, as reported in her book The
Power of Logical Thinking, which can be read on line courtesy of Google Books.
The Greens and Blacks are playing bridge. After a deal, Mr. Brown, an onlooker, asks Mrs.
Black: "Do you have an ace in your hand?" She nods. There is a certain probability that her
hand holds at least one other ace.
After the next deal, he asks her: "Do you have the ace of spades?" She nods. Again, there is
a certain probability that her hand holds at least one other ace. Which probability is
greater? Or are they both the same?
Martin Gardner
Hendersonville, North Carolina
Unlike some of her much simpler puzzles, her reply drew no indignant responses from Ph.D's
saying that she was wrong; however I think her analysis has an important error and that she
got the right answer merely by luck. If so, here is a case where both Marilyn and her army
of Ph.D. critics were wrong.
In what follows, nCr will stand for the number of ways of choosing r objects from a
collection of n objects. It is sometimes called "n choose r" and is a built-in function on
scientific hand calculators.
Method I (Statman):
If the ace of spades is dealt, there are 48C12 ways of dealing the remaining cards not
containing another ace and 51C12 ways of dealing them total with a probability of
48C12/51C12 = 0.43884, so there is a Pr of 0.56115 that the hand will contain at least one
other ace.
The Pr that if a hand contains at least one ace of whatever suit it will not contain another
ace is the number of ways of dealing 13 cards containing only one ace divided by the number
of ways of dealing hands containing at least one ace. The latter is the number of ways of
dealing all hands less the number of ways of dealing hands without aces. This can be
expressed as 4*48C12/(52C13 - 48C13) = 0.630362 so the Pr that if a hand contains an ace the
Pr it will contain at least one other ace is 0.36963.
Method II (Marilyn Vos Savant)
"There are fewer opportunities to get a certain ace than any ace at all but each of these
groups gives an equal number of ways to get more aces, so the smaller group has the greater
success ratio." Although the conclusion is correct, the reason is wrong. Both groups do not have
an equal number of ways of getting more aces. Reducing this to numbers, if the ace of spades
is dealt it has 52C12 - 48C12 ways of getting at least one more ace. This is the number of possible
completions less those that contain no aces, so each one that remains contains at least one ace
and the Pr of getting another ace is (52C12 - 48C12)/51C12 = 0.56115. So far so good!
If it is known that the hand that has been dealt contains at least one ace, it could be any
of 52C13 - 48C13 hands, of which 4*48C12 contain exactly one ace; therefore this hand has
52C13 - 48C13 - 4*48C12 ways of getting more than one ace. This evaluates to 1.6341E11
whereas 52C12 - 48C12 = 1.36711E11. The hand with an ace of unspecified suit has more ways
to get additional aces but because of the size of this group it loses out in the ratio.
Marilyn reached the correct conclusion for he wrong reason.
Method III
Oza. Rather complicated, but his results agree with Statman's.
December 14th, 2013 at 12:49:41 PM
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i'd say this one is pretty intuitive. i don't have a hard time at all grasping why the hand with the ace of spades is more likely to hold a second ace than a hand where you just notice "an ace."
December 14th, 2013 at 1:27:14 PM
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Sodawater: Does this also mean that you are more likely to have a good hand if you look at your cards than if you don't look at them? I can't answer this with mathematics.
December 14th, 2013 at 2:33:22 PM
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Quote: puzzlenutSodawater: Does this also mean that you are more likely to have a good hand if you look at your cards than if you don't look at them? I can't answer this with mathematics.
That depends on how good a bridge player you are.
Wisdom is the quality that keeps you out of situations where you would otherwise need it
December 14th, 2013 at 3:53:10 PM
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3 No Trump!Quote: dwheatleyThat depends on how good a bridge player you are.
Someday, joor goin' to see the name of Googie Gomez in lights and joor goin' to say to joorself, "Was that her?" and then joor goin' to answer to joorself, "That was her!" But you know somethin' mister? I was always her yuss nobody knows it! - Googie Gomez
December 14th, 2013 at 6:36:26 PM
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Quote: dwheatleyThat depends on how good a bridge player you are.
As a 30+ year bridge player, I don't see the ace-suit as a meaningful distinction in a bridge question. Maybe the only reason they picked that is that you get 13 cards in bridge, so they used that game to illustrate the math? Aces can be the trickiest cards in bridge unless considered within the entire "gestalt" of distribution and bidding, often over-valued. I'm just going to assume I'm overthinking it.
If the House lost every hand, they wouldn't deal the game.
December 15th, 2013 at 12:56:51 AM
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Quote: s2dbaker3 No Trump!
Don't look at your hand, just bid. Looking will only discourage you.
I'm not getting this puzzle at all but it's late on a Saturday, I just got home, and I've had a few. It will probably make sense tomorrow.
December 15th, 2013 at 5:29:03 AM
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Beachbumbabs: You are right that this is just a mathematical puzzle and has little to do with the game of bridge, however it always has amused me that bridge partners are always trying to convey to their partners what they hold in their hand. The only proper way to do it is with "bidding conventions" but some money players use other means.
December 18th, 2013 at 3:35:22 AM
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SOLVING THE PARADOX
Here is a classical paradox, in that both answers seem correct in terms of reasoning, but lead to different values. When this happens, you have to look at the subtle difference in the wording: the reasonings are answering different questions.
Never forget that probabilities and statistics (except in quantum physics) are about information. When two situations give a different probability, that means there is some more information in one of the situations. So you should look at what info is given in the question.
The expression "you notice that one of them is an ace" is the culprit. It is not detailed enough to put it into probabilities. Practically, how can you "notice an ace" without knowing its suit?
SAME VALUE
One possibility is that some other player happened to see one card, and told us it was an ace.
Then, proper Bayesian analysis makes you subdivide the possibilities in four cases (the four suits), all equiprobable (one fourth). Each case is analysed according to the "spade" example, leading to 0.56116 . The overall proba is .56116(1/4)+ .56116(1/4) + .56116 (1/4)+ .56116(1/4) = .56116, of course.
DIFFERENT VALUES
Another possibility is that, for some reason, you know that other player is folding when he has no ace. He saw his cards, and did not fold. So your information is that "there is at least an ace". (Notice the difference: he didn't see ONE card, he saw ALL thirteen of them.)
Then the second formula is correct: P(at least two aces)/P(at least one ace) = .36963
EXPLAINING THE DIFFERENCE
Suppose there is "at least one ace" in the hand. If there are more than one, and the player happens to see one card randomly, then the probability of seeing an ace is larger than if there is only one ace in the hand. By Bayesian (or inverse) reasoning, that means that if you chance to see an ace, it increases the probability of several aces compared with when you simply knew "there are aces(s)".
The error in the paradox is to treat the cases (1 ace), (2aces) etc. as if identical in terms of probability of letting you see one ace.
To help understand, think of the situation where, in addition to the ace of spade, you had seen, say, a two of hearts. New info! All the calculations are affected, aren't they? Well, the fact that a card was seen AND that it was an ace is new information, too.
Other possibilities exist. The other player may have seen several cards and told you he saw an ace. Or even, he may have decide to tell you about the ace only in specific cases.
What if, having seen his hand, he decides to show you the ace of spades? Did he draw a card to show at random? Did he choose? Had he any reason to prefer spades if he had several aces? Etc etc etc All different values.
Always remember the Monty Hall problem. Always ask: what is the mechanism giving additional information, and in what respect is it affected by the content? If it is, then there is information to be used. Monty Hall knows where the car is and opens a door knowingly. Hence you can use the info he is betraying.
Here is a classical paradox, in that both answers seem correct in terms of reasoning, but lead to different values. When this happens, you have to look at the subtle difference in the wording: the reasonings are answering different questions.
Never forget that probabilities and statistics (except in quantum physics) are about information. When two situations give a different probability, that means there is some more information in one of the situations. So you should look at what info is given in the question.
The expression "you notice that one of them is an ace" is the culprit. It is not detailed enough to put it into probabilities. Practically, how can you "notice an ace" without knowing its suit?
SAME VALUE
One possibility is that some other player happened to see one card, and told us it was an ace.
Then, proper Bayesian analysis makes you subdivide the possibilities in four cases (the four suits), all equiprobable (one fourth). Each case is analysed according to the "spade" example, leading to 0.56116 . The overall proba is .56116(1/4)+ .56116(1/4) + .56116 (1/4)+ .56116(1/4) = .56116, of course.
DIFFERENT VALUES
Another possibility is that, for some reason, you know that other player is folding when he has no ace. He saw his cards, and did not fold. So your information is that "there is at least an ace". (Notice the difference: he didn't see ONE card, he saw ALL thirteen of them.)
Then the second formula is correct: P(at least two aces)/P(at least one ace) = .36963
EXPLAINING THE DIFFERENCE
Suppose there is "at least one ace" in the hand. If there are more than one, and the player happens to see one card randomly, then the probability of seeing an ace is larger than if there is only one ace in the hand. By Bayesian (or inverse) reasoning, that means that if you chance to see an ace, it increases the probability of several aces compared with when you simply knew "there are aces(s)".
The error in the paradox is to treat the cases (1 ace), (2aces) etc. as if identical in terms of probability of letting you see one ace.
To help understand, think of the situation where, in addition to the ace of spade, you had seen, say, a two of hearts. New info! All the calculations are affected, aren't they? Well, the fact that a card was seen AND that it was an ace is new information, too.
Other possibilities exist. The other player may have seen several cards and told you he saw an ace. Or even, he may have decide to tell you about the ace only in specific cases.
What if, having seen his hand, he decides to show you the ace of spades? Did he draw a card to show at random? Did he choose? Had he any reason to prefer spades if he had several aces? Etc etc etc All different values.
Always remember the Monty Hall problem. Always ask: what is the mechanism giving additional information, and in what respect is it affected by the content? If it is, then there is information to be used. Monty Hall knows where the car is and opens a door knowingly. Hence you can use the info he is betraying.
Reperiet qui quaesiverit
