I sent an email on this subject to the folks at thestraightdope.com and they said this was already discussed. But I cannot seem to find it.
Anyway in the following old "Monty Hall thread: www.straightdope.com/classics/a3_189.html
there is a problem posed about a card in a hat. Cecil’s analysis was wrong, the answer was “1 in 2”. Did anyone ever catch him on it?
Regards,
-Ken
The column says that the answer is “2 in 3.” Or, do you mean that you disagree with that answer? It seems that he justified his answer pretty well, and I agree with it.
Here are some forum theads about the column http://www.straightdope.com/ubb/Forum1/HTML/000089.html
lagae and Larsy make a “1 in 2” argument, The Dude answers. Larsy makes a strange return argument. http://www.straightdope.com/ubb/Forum1/HTML/000090.html
But that one doesn’t discuss the cards in hat, I think. http://www.straightdope.com/ubb/Forum1/HTML/000465.html
The last one is entitled “Cecil’s Mistake In ‘Card In Hat’ Problem.” O, wait, that’s us!
.
Cecil’s answer looks fine to me; could you elaborate on where you think the mistake is?
No, his response was right. To state the problem again, “Suppose our task is to pick the ace of spades from a deck of cards. We select one card. The chance we got the right one is 1 in 52. Now the dealer takes the remaining 51 cards, looks at them, and turns over 50, none of which is the ace of spades. One card remains. Should you pick it? Of course. Why? Because (1) the chances were 51 in 52 that the ace was in the dealer’s stack, and (2) the dealer then systematically eliminated all (or most) of the wrong choices. The chances are overwhelming–51 out of 52, in fact–that the single remaining card is the ace of spades.”
If the dealer did NOT look at them, but just flipped 50 over at random (none of which were the Ace of Spades), then the chance would be 1 in 2 that you are holding the Ace of Spades. But since he looked at them and systematically removed 50, chances are 51 out of 52 that he was holding the Ace of Spades and flipped over every card but that one, so the chance is 51 out of 52 that the one card he did not flip over was the Ace of Spades, with a 1 out of 52 chance that you happened to pick the Ace of Spades when you drew that one card.
David
Card tricks I don’t know from. But this does remind me of a “logic” problem a former boss of mine once posed. The question: A basketball player has two free throws. He makes the first one. What are the odds he makes the second? I said one in two; either he makes it or he doesn’t. He said three in four, and drew diagrams to “prove” it. Now, in his favor he did have a bachelor’s degree in math, but counting against him was his frequent use of mind-altering chemicals. Seems to me the guy either makes the shot or he doesn’t. Making the first shot doesn’t affect the outcome of the second.
AskNott
Otto: The problem makes no sense. The odds of a shooter making a free throw at any given time is entirely determined by his skill level. If he has historically made 75% of his free throws, then that’s the chance that he’ll make the next one, if the conditions are the same. The fact that he just made one or a hundred is irrelevant.
Well, I don’t know diddly about basketball stats, but I wouldn’t be surprised if there was a measurable difference in the odds of a players second shot, depending on the outcome of the first shot. Shooting free throws aren’t independent events the way coin tosses are.
Anyway, it has nothing to do with the OP. The correct answer is indeed 2 in 3. Get a friend and try it out if you don’t believe it (thats how I convinced a skeptical friend).
Here is the email that I sent to Cecil a couple of weeks ago:
Dear Cecil,
I was reading the archive on “The Straight Dope” website and happened across a series of columns relating to the “Let’s Make A Deal” a.k.a. “Monty Hall” problem. (http://www.straightdope.com/classics/a3_189.html)
Of course the Monty Hall stuff is correct, however, Jordan Drachman sent in two similar problems and you set out to explain the correct answers to us. Well, I contend that the answer to the card-in-the-hat problem is 1/2, not 2/3 as you suggest. It can be proven using the example that you give for the boy-girl problem in your very own column. See below.
Puzzles from Jordan Drachman of Stanford, California:
>There is a card in a hat. It is either the ace of
>spades or the king of spades, with equal probability.
>You take another identical ace of spades and throw it
>into the hat. You then choose a card at random from
>the hat. You see it is an ace. What are the odds the
>original card in the hat was an ace? (Answer: 2/3.)
>
>There is a family with two children. You have been
>told this family has a daughter. What are the odds
>they also have a son, assuming the biological odds of
>having a male or female child are equal? (Answer: 2/3.)
Cecil writes:
>The answers given to Jordan Drachman’s questions–2
>in 3 in both cases–were correct. The odds of the
>original card being an ace were 1 in 2 before it was
>placed in the hat. We are now trying to determine what
>card was actually chosen based on subsequent events.
>Here are the possibilities:
>
>(1) The original card in the hat was an ace. You threw
>in an ace and then picked the original ace.
>
>(2) The original card in the hat was an ace. You threw
>in an ace and then picked the second ace.
>
>(3) The original card was a king; you threw in an ace.
>You then picked the ace.
>
>In 2 of 3 cases, the original card was an ace. QED.
me: The flaw here is that you have not shown that these
three cases are equally likely.
>The second question is much the same. The possible gender
>combinations for two children are:
>
>(1) Child A is female and Child B is male.
>
>(2) Child A is female and Child B is female.
>
>(3) Child A is male and Child B is female.
>
>(4) Child A is male and Child B is male.
>
>We know one child is female, eliminating choice #4. In 2
>of the remaining 3 cases, the female child’s sibling is
>male. QED.
me: Yes, I agree completely.
>Granted the question is subtle. Consider: we are to be
>visited by the two kids just described, at least one of
>which is a girl. It’s a matter of chance who arrives
>first. The first child enters–a girl. The second knocks.
>What are the odds it’s a boy? Answer: 1 in 2. Paradoxical
>but true. (Thanks to Len Ragozin of New York City.)
me: I also agree with this solution. It is this second
puzzle that is similar to the card-in-hat puzzle. Why?
Because “picking an ace” is the same as “first child
enters – girl”.
Let me rewrite the puzzle this way using some of your words:
Granted the question is subtle. Consider: we are to be
visited by the two cards just described, at least one of
which is an ace. (we know at least one is an ace 'cause
we threw an ace into the hat) It’s a matter of chance who
arrives first. The first card enters–an ace. The second
card knocks. What are the odds it’s a king? Answer: 1 in 2.
Paradoxical but true.
Best Regards,
-Ken Fuchs www.koblackjack.com
“Is that a theoretical shooter? Is is O’Neal? Is it Miller? A little more information could mean a lot.”
The shooter is entirely theoretical.
But they are equally likely. Others have discussed this pretty well.
<font color=#DCDCDC>rocks</font>
Just because there are three “possible” cases it does not necessarily mean that the three cases are equally likely. Take a look at the two variants of the sibling problem. In the first problem all three cases are equally likely, so the answer is 2 of 3. In the second problem the three cases are NOT equal, and the answer turns out to be 1 of 2. Both are correct answers to different problems. It also turns out that the card/hat problem is EXACTLY like the second sibling problem. Please read it again. If you agree with Cecil’s analysis of the card/hat problem, then by default you must disagree with his analysis of the second sibling problem.
Regards,
-Ken Fuchs www.koblackjack.com
I think Ken is confusing the odds at the start of the situation (1 in 2 chance that the card is an ace) with the odds at the end of the situation.
Ken is trying to argue that:
(a) Chance of original card is Ace: 1/2
(b) Throw in a new ace
© Draw out an ace
(d) Chance of drawn card being the “old ace” is 50% (it’s either the new ace or the old ace)
(e) Therefore, chance of this outcome is 1/4 (= 1/2 x 1/2). Ken notes that this is what Cecil calls outcome (1) and that Cecil assigns the odds of 1/3 to this outcome.
I side with Cecil. Hey, I always side with Cecil, he’s my boss’s boss.
The logic flaw in Ken’s approach can be seen if you apply the same analogy as for the Monty Hall problem. Instead of throwing in just one additional ace and then pulling it out, suppose you throw in 1000 aces and pull out 1000 cards, all of them aces. Ken, do you still think the chance of the remaining card being a king is 50%?
The logic is subtle, but the point is that you have changed the odds by throwing in and drawing out that extra ace.
Look at it turn-about. Suppose you have a single card in the hat, it is either an ace or a king, odds are 1/2. You pull a card and find it to be an ace. Your act of pulling the card has now CHANGED the original odds; it is no longer 50% possible that the card was a king. Pretty obvious, yes?
In the same way, you change the odds by tossing in a card and drawing out a card.
Cecil was right, I was wrong.
Sorry to waste everyone’s time.
-Ken Fuchs
Not a time waste at all, Ken. It’s a subtle point, and it’s easy to miss. I blew it the first time I heard it, too.
Now, that would make a fine T-shirt.
And the rest wasn’t a waste of time, either. It took me a bit of time to solve your parallel formulations of the two puzzles. Get this: the probability changes after both “visits”, I think. After the girl visits, the probability is one-half, before the visit–two-thirds. After the ace “visits”, the probability is two-thirds–before the visit, it is one-half!
I think
<font color=#FCFCFC>therefore
rocks</font>
Well…since this thread is up, here is another similar puzzle:
Freedonia, a tiny country of 3 million people in Europe, is plagued by a
horrible disease which renders its citizens blind and eventually kills them.
Doctors at the national medical university have determined that about 30,000
people will come down with this disease. Working under government orders, the
doctors have come up with a test for the disease and a treatment. The test is
95% accurate - this means that 95% of the time the test will accurately identify
a person with the disease as having it (5% false positive), and 95% of the time it will accurately
identify a person without the disease as not having it (5% false negative). Early treatment for the
disease is very effective for those who have the disease – the patient is
completely cured, and blindness is prevented. Unfortunately, administering the
treatment to a patient without the disease kills the patient.
You take the test. The test identifies you as having the disease. Should you undergo treatment, refuse treatment, or doesn’t it matter?
(note: if you take the test again you will get the same result).
-Ken Fuchs www.koblackjack.com
If nobody takes the treatment, 30,000 people die. If those who are identified as having the disease all take the treatment, 150,000 people die. So no, I shouldn’t take the treatment.
It is too clear, and so it is hard to see.
This is more or less the real-life analog of smallpox – more people die from the vaccine than from disease, and so widespread vaccination has been discontinued. Hence, ZenBeam’s comment.
If everyone taking the test takes the test twice, however, as per the footnote, the story is slightly different. (I am assuming that the results of test 1 do not affect the results of test 2; if that assumption is incorrect, then we’re back at ZenBeam’s comment.)
If you give the treatment ONLY to those for whom both tests indicate that they have the disease (if even one of the tests indicate you do not have the disease, then you do not take the test), then a total of 9,850 die compared to the initial 30,000.
What if you wait until you go blind before taking the cure?