sarah@rdin.UUCP (sarah) (08/23/83)
c(FbiLhLhD*@iQhL
Here is the nasty gold/silver problem, exactly as it appeared:
Here goes: you are in a room with three cabinets, each of which has
two drawers. One cabinet has a gold coin in each drawer. Another
has a silver coin in each drawer. The third has a gold coin in one
drawer and a silver coin in the other.
You pick a cabinet at random and open a random drawer. It contains
a gold coin. What is the probability that the other drawer of that
same cabinet contains a gold coin?
OK, folks, I've read your solutions claiming to prove that the probability
is 2/3. However, this answer seems to rest on one or both of the following
assumptions:
1) Any of the three remaining drawers (having tossed out the SS cabinet)
can be chosen.
2) The gold coin you know about was *not* necessarily the first one
you chose.
When as a matter of fact, Larry Kolodney states:
So, there are three gold coins. Since we have an equal chance of
choosing any one of them, if we do choose one it is equally likely
that it is the one in the silver gold cabinet, or that it is one of
the two in the gold gold cabinet. {He then continues on to support
the 2/3 answer.}
The problem with the 2/3 answer is that once we have that gold coin in our
grubby little paws, we have reduced the problem to two cabinets (not three
remaining drawers). If the problem had asked about the probability before
we knew we had one gold coin, or if we were free to choose from the three
remaining drawers, the answer would be different. However, by choosing a
gold coin, we know:
1) The silver cabinet is out of the question.
2) The one remaining drawer we must open to determine which cabinet
we have chosen must contain either a gold coin or a silver coin.
The probability associated with picking a gold coin in the first place
is *not* part of the original question as stated. It is a *given* that
the gold coin is chosen.
Can you please explain yourself clearly once more for the benefit of all
us confused people out here?
Sarah Groves
New York
philabs!rdin!sarahlarry@grkermit.UUCP (Larry Kolodney) (08/24/83)
OK. ONE MORE TIME AND THEN I QUIT.
From sarah@rdin:
The problem with the 2/3 answer is that once we have that gold coin in our
grubby little paws, we have reduced the problem to two cabinets (not three
remaining drawers). If the problem had asked about the probability before
we knew we had one gold coin, or if we were free to choose from the three
remaining drawers, the answer would be different. However, by choosing a
gold coin, we know:
1) The silver cabinet is out of the question.
2) The one remaining drawer we must open to determine which cabinet
we have chosen must contain either a gold coin or a silver coin.
The probability associated with picking a gold coin in the first place
is *not* part of the original question as stated. It is a *given* that
the gold coin is chosen.
~~~~~~~~`
Just refer to the Article about the shredded National Enquirer.
Or better, here's actually an intuitive way to think about it.
Lets say we pick a coin at random from one of the six drawers, before
we do anything else. What is the prob. that IT is gold. I hope you
said 1/2, because there are 3 of each type of coin.
After we pick the gold coin, we have to pick another coin at random.
Wouldn't you assume that since we've already eliminated two silver and
only one gold coin the odds should be better for the gold coin?
If yo agree with this than 1/2 is obviously wrong! Now, why 2/3. Well
you have 2wice as many gold coins as silver left, so 2/3 is it!
The crux of the argument is to convince yourself that choosing the
drawer after the first coin toss is just as random as before. Since
either DRESSER is just as likely, they are interchangable, and picking
the drawer in the same dresser is just like picking any of the drawers
from one of the dressers with a gold coin in it.
Okay?
--
Larry Kolodney (The Devil's Advocate)
{linus decvax}!genrad!grkermit!larry
(ARPA) rms.g.lkk@mit-aijim@ism780.UUCP (Jim Balter) (08/24/83)
You are not reading the solutions carefully.
this answer seems to rest on one or both of the following
assumptions:
1) Any of the three remaining drawers (having tossed out the SS cabinet)
can be chosen.
No such assumption was ever made, although it happens to turn out that
the results are the same if you do choose any of the other drawers not
in the S/S cabinet, instead of just the one in the same cabinet, because all
of the permutations of those drawers satisfy the stipulations in the problem,
so picking a drawer other than the one in the same cabinet is just equivalent
to picking the one in the same cabinet given a different random arrangement
of the drawers. If you find my language too convoluted, please just ignore
it instead of using it as a further grounds for claims of errors in the
analysis.
2) The gold coin you know about was *not* necessarily the first one
you chose.
No such assumption has ever been made, and I don't understand how you could
have gotten this impression. Certainly the coin known about is the coin
chosen.
The problem with the 2/3 answer is that once we have that gold coin in our
grubby little paws, we have reduced the problem to two cabinets (not three
remaining drawers).
You only say this because you have a mental image in which cabinets play
a bigger role than drawers. You don't know which cabinet (among those
containing G and G or G and S) you chose; nor do you know which coin
(among the *three* (G, G, S) with a G in the other drawer of the same
cabinet (the given)) you are going to find in the second drawer. Imagine the
coins are attached by long strings, instead of sharing cabinets. I put them
all in another room, and bring out a gold one. You know that the S/S
combo is out. You have a G in your hand. Which of the remaining three coins
is tied to the string? Suppose that, just before you pull on the string to
retrieve the coin, my accomplice in the other room randomly reattaches it to
one of the three. Does that affect the probabilities?
If the problem had asked about the probability before
we knew we had one gold coin, or if we were free to choose from the three
remaining drawers, the answer would be different.
The probability of what? Of finding a gold coin in the first drawer?
That is 1/2, but surely you don't mean that. Of finding a gold coin in
the second drawer if you do in fact find a gold coin in the first drawer?
That is the problem as stated. If you would try to formalize what you
mean here, you might get a hint of the imprecise thinking which is misleading
you.
Many people seem to be having a problem with subjunctive situations.
You are to determine the probability after reading the problem; the problem
describes a situation, in which there is a drawer open containing a gold coin.
If you are the subject of the problem, or you are just reading about it,
the situation is the same.
Also, you talk about the three remaining drawers. Imagine
that you are in the room with the cabinets. Which of the drawers are those
three? Are there only two cabinets? The notion of throwing out the cabinet
with 2 silver coins merely indicates that the odds of *having selected it
given that one of its drawers contains a gold coin* is (2-2)/3,
just as the odds of having selected the cabinet with 1 silver coin is (2-1)/3,
and the odds of having selected the cabinet with 0 silver coins is (2-0)/3
(the answer!).
However, by choosing a gold coin, we know:
1) The silver cabinet is out of the question.
Hmmm, yes, 0/3, yes, that sounds right. But, this non-quantitative language
suggests you might be about to make a mistake.
2) The one remaining drawer we must open to determine which cabinet
we have chosen must contain either a gold coin or a silver coin.
Hmmm, yes, but are those two possibilities, a gold coin or a silver coin,
equally likely? Why do you insist on ignoring this question when it has
been brought up repeatedly? We are looking for a coin which shares a cabinet
with a gold coin. There are three such coins; two are gold.
The probability associated with picking a gold coin in the first place
is *not* part of the original question as stated. It is a *given* that
the gold coin is chosen.
You are certainly right. Couldn't agree with you more. Never said otherwise.
I am amazed that so many people, who never would have questioned the answer
given in the back of the textbook, will so thoroughly question the same
answer when given on the net. I think that is a move in the right direction,
but perhaps a trip back to the textbook might help in this case.
Someone wrote a program, ran it, and posted the results, which were
much closer to 2/3 than 1/2. If you can come up with a program consistent
with the problem that gives a different result, please post it.
Otherwise, please just contemplate quietly.
Jim Balter (decvax!yale-co!ima!jim), Interactive Systems Corp
--------mitchell@psuvax.UUCP (08/24/83)
We have three cabnets, each with three drawers. We pick a drawer (or equivalently a coin) at random and find it is gold. There are three possibilities: either of the drawers of the gold-gold cabinet and the gold drawer of the gold-silver cabinet. In two of the cases (the drawers of the gold-gold cabinet) the other drawer is gold; in one case (the silver-gold cabinet) the other drawer is silver. Therefor the probibility is 2/3 that the other drawer is gold. This is trickier to think about if you think of first choosing the cabinet and then the drawer. One way to do this is think of making 1 of 3 equally likely choices, and after that 1 of 2, making 6 equally likely choices and leaving us with the first analysis. Incidentally, the arogance of some of the articals on this issue has been so obnoxious as to leave it irrelevant whether or not they are correct. One of the most obnoxious I reread carefully and still couldn't decide what answer he was claiming, far less his reasoning for it.
laura@utcsstat.UUCP (Laura Creighton) (08/27/83)
"contemplating quietly" may do no good at all. I understand *now* where my error was. I wrote a program to simulate the process immediately after sending my incorrect answer to rabbit!ark, to see if I were right. I knew then that I wasn't but i could *not* see what the bugs in my thinking were. When the explanations came in I could not understand how they contradicted my thinking either, at first. Perhaps all of this discussion should have gone on in mail. The best explanation I received was from metheus!howard, at least in terms of identifying what my problem was, but from the amount of mail I received that went "I don't get it either, but I don't have the guts to admit it, so if you find out will you let me know" I think that there is more interest in this question out there. Do you solve all your thinking problems in isolation? Have you never asked someone else to help you discover why a program of yours was no longer working? Why is this problem any different? because you are sick of it, perhaps? why don't you then use your 'n' key? Laura Creighton utzoo!utcsstat!laura