jim@ism780.UUCP (Jim Balter) (08/23/83)
I find it incredible that so many bright people can miss such a simple,
straightforward problem. Maybe it isn't so incredible; bright programmers
are excellent at coming up with clever rationalizations to protect their
egos from the pain of being in error.
I was always taught that past probabilities can not be used to determine
the probability of a future occurence.
I doubt that you were taught that, although you appear to have learned it,
most likely by using flipped coins as your only example.
The effect of past occurrences upon the probability of future occurrences
is the basis of induction, and enables us to think at all. The proper
concept is that the probability of the occurrence of one event is of no
bearing upon the probability of occurrence of an *independent* event. As an
illustration, I play Russian Roulette and get a blank chamber.
I spin the barrel and fire again; what are my odds? What would my odds
be if I didn't spin the barrel? In the first case the two events are
independent, but not in the second.
If the drawer contains a gold piece we know
that the cabinet does not contain two silver pieces. At this stage we
have two possible cabinets. One that contains a gold piece and a silver
piece, and another that contains two gold pieces. One of the drawers
containing a gold piece is open so the remaining drawers contain a gold
piece, a gold piece and a silver piece. Thus the probability for finding
a gold piece when we open another drawer is 2/3.
Wrong. At this point you claim to be ignoring the constraint that you had to
open the drawer in the same cabinet (see below). So, there are three
cabinets, and you don't know the contents of any drawer other than the open
one. The odds of opening a random drawer and finding a gold coin in this
situation is of course 2/5.
But we have another constraint. Since
we are trying to determine the probability that the
OTHER DRAWER OF THE SAME CABINET contains a gold piece
Yes that is the constraint. Since we don't know which coins are in which
drawers other than the g/g s/s g/s distribution given, being constrained
to select the drawer in the same cabinet gives the same results
(probability-wise) as being constrained to not pick a drawer in the s/s
cabinet (because all permutations with each constraint satisfy the other
constraint), which as you showed above is 2/3.
we have to change or [sic] model so that the drawer
that we opened on the selected cabinet is open in BOTH POSSIBLE CABINETS.
This is nonsensical, but it does seem to satisfy the need to arrive at the
wrong answer. Please, quit trying to imagine how many teeth there are in the
horse's mouth, and try the experiment (but be sure you follow the original
statement of the problem).
As for intuition, why would anyone intuitively expect a woman who has
already borne one boy to have the same chance for two boys after her
second child as a woman who has as yet borne no children?
And for the latest permutation of the problem with selecting the bottom
drawer first instead of a random drawer, why would that possibly matter,
any more than the stipulation that you pick the drawer which was manufactured
first? The problem made no claim at all about the effect of top
or bottom drawers upon the distribution of the coins in the drawers.
For all you know, the damn drawers are side by side. Sigh.
Jim Balter (decvax!yale-co!ima!jim), Interactive Systems Corp
--------jim@ism780.UUCP (Jim Balter) (09/11/83)
re: I am going to give a quiz, subject to two conditions: 1) It will be given during class one day this term. 2) At no time prior to the quiz will it be possible to infer from these conditions that the quiz will be given on a certain day. I say that the paradox is now genuine. That is, it is impossible for the teacher to fulfill these conditions. --- Sorry, but no. If the teacher gives the test on the first day, (1) is certainly true, and on what basis could anyone have inferred that the test would have been given on that day? If the test is given on the last day, you can infer that, so the test cannot be given on the last day *if* the above conditions are satisfied. But if the test has not been given prior to the next to last day, you can infer that *either* the test will be given on the next to last day *or* the conditions will be violated. You cannot infer either alone. So, the test can be given on the next to last day (or any previous) without violating the conditions. The problem is that the student is treating the inviolatability of the conditions as an axiom in his induction, which isn't valid because such an axiom can only be part of the metalanguage, and so is not available to the student's logical calculus. --------
jim@ism780.UUCP (Jim Balter) (09/19/83)
re: Jim Balter's solution to the circle problem is wrong. I have run the simulation and the result for n=4 agrees with my result, 1/2. Moreover I have an inductive proof that the probability for arbitrary n is equal to n/(2^(n-1)). --- Either my message got garbled in transmission, or you didn't read it properly. My result agreed with yours (although it was arrived at in a significantly different way). To repeat (with the proper term "arc" substituted for the improper "chord" in my original sending): For each of the n points, imagine an arc extending clockwise from that point through the other n-1 points. Since the n different arcs include all the ways that the points might be formed into a semicircle, and since no two of the arcs can both fall within a semicircle (with the exception of singular cases), the odds that all the points fall within a semicircle is the sum of the odds that any of the arcs falls within a semicircle; since the points are equivalent, p = n * p_one_arc. Since, in order for one of the arcs to fall within a semicircle, all of the last n-1 points must be within 1/2 the circle from the first point, p_one_arc is clearly (1/2) ** (n-1). So, p = n/(2**(n-1)). Jim Balter (decvax!yale-co!ima!jim), Interactive Systems Corp --------
levy@princeton.UUCP (Silvio Levy) (09/21/83)
My apologies to Jim Balter. I read his P(n) as n/(2*(n-1)) (which gives
the same result for n=3). I don't fully understand his proof, but it seems
to be sound and considerably simpler than mine.
-- Silvio Levydavis@hplabs.UUCP (Jim Davis) (09/21/83)
Steve Summit writes: > The proof that a = b was pretty obvious. Here's one I discovered > by accident and still can't quite figure out: > > 1 = 1 reflexive property of equality > > 2/2 = 2/2 another name for 1 > > 2/2 2/2 > -1 = -1 raise -1 to both sides > > 1 2 b/c th b > -1 = sqrt( ( -1 ) ) a == c root of a > > 2 > -1 = sqrt( 1 ) reduce ( -1 ) > > -1 = 1 reduce sqrt( 1 ) > > The mistake is probably in the fourth step, but no book I've seen > places restrictions on a, b, or c in that identity. Yes, the error in the argument is in step 4. If (x = y) then it is the case that (sqrt(x) = +or- sqrt(y)). If you examine any reasonable text of mathematics you will find 1/n 1/n that (x = y) implies ( x = omega(k) y ) where each of the n omegas th is chosen from the n roots of unity. -- Jim Davis (James W Davis) ...!ucbvax!hplabs!davis davis.HP-Labs@UDel-Relay ----------------------------------------------------------------
jfp@emory.UUCP (09/21/83)
Soon after posting the problem I found, by coincidence, a problem
in The Mathematical Intelligencer, Volume 5, No. 3, asking for the
probability of n points chosen at random being within a small circle
of radius r. The editor noted that the N points lying within a
hemisphere on the surface of an n-sphere had been solved by
J.G. Wendel, the answer being
n-1
1 ----
----- \ ( N - 1 )
N-1 / ( k )
2 ----
k = 0
Unfortunately there was no reference to the proof.
If anyone knows of the reference, or has their own proof,
I'd be very interested to hear. I like Jim Bailey's proof
in the case n = 2, and I also don't follow Silvio Levy's complaint.
---------
John Pedersen. {sb1,akgua}!emory!jfp