NetNews Usenet Archive 1992 #31/NN_1992_31.iso/spool/sci/math/17372

Internet Message Format | 1992-12-23 | 3.9 KB

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From: [e-mail] (Steven White)
Subject: Re: Marilyn Vos Savant's error?
Nntp-Posting-Host: horus.ps.uci.edu
Message-ID: <[e-mail]>
Newsgroups: sci.math,rec.puzzles
Reply-To: [e-mail] (Steven White)
Organization: University of California, Irvine
Keywords: savant
Lines: 66
References: <[e-mail]> <[e-mail]> <[e-mail]> <[e-mail]>
Date: 23 Dec 92 18:33:32 GMT

In article <[e-mail]>, [e-mail] (John C. Baez) writes:
|> >|>
|> >|> > 1)  You draw 4 cards from a well-shuffled standard deck.  You turn
|> >|> >     one over and it's an ace.  What's the probability that they are
|> >|> >     all aces?
|> >|>
|> >|> There are 4! C(52,4) = 52!/48! = 6497400 sets of 4 cards you could
|> >|> draw, counting different orders as different.  In one case in 13 you'll
|> >|> turn up an ace, for a total of 499800, while in only 24 cases are they
|> >|> all aces.  The answer to this question is therefore 24/499800, which is
|> >|> 1 in 20825, or, to 20 places from dc, .00004801920768307322.
|> >|>
|> >|> > 2)   You draw 4 cards from a well-shuffled standard deck.  You turn
|> >|> >      one over and it's the ace of hearts.  What's the probability
|> >|> >      that they are all aces?
|> >|>
|> >|> Of the 6497400 sets of 4 cards, in one case in 52 will you see the ace
|> >|> of hearts, for a total of 124950.  Again, 24 are all aces.  The answer
|> >|> is thus 24/124950, or 12/62475, about .00019207683073229291.  This is
|> >|> roughly four times as large as the previous answer.
|> >|>
|> >|>                     der Mouse
|> >
|> >No way for 2). 24 possibilites are all aces, but only 6 of those had the
|> >ace of hearts as the first card. Thus 6/124950=1/20825, same as 1.
|> >Its clear if you think about the questions a little more that 1 and
|> >2 have to have the same answer.
|>
|> Steve's right.  Let me do, out loud, the thinking Steve suggests.
|> Suppose you draw four cards, turn one over, and its an ace.  There are
|> four cases - hearts, diamonds, clubs, spades.  Let p_i be the
|> probability that one has drawn 4 aces, *given* that the ith case has
|> occured (i = hearts, diamonds, clubs, spades).  By symmetry all p_i are
|> equal, say p.  Now turn to problem 1: one has drawn four cards, turned
|> one over, and its an ace.  The chance that the ith case has occured is
|> 1/4, by symmetry.  So the answer to problem 1 is p/4 + p/4 + p/4 + p/4 =
|> p.
|>
|> Of course, its reasoning along these lines, but ERRONEOUS reasoning,
|> that might make one suspect the answers to my "improved" pair of
|> questions were also equal.

Here is the reworded 2:

> 2)   You draw 4 cards from a well-shuffled standard deck.  Given that
>      one is the ace of hearts, what's the probability that they are
>      all aces?

Actually, if you reword your reworded 2 slightly it can give the same
answer as 1:

 2)   You draw 4 cards from a well-shuffled standard deck.  You look at
      the cards, and tell me (truthfully) that one is the ace of hearts.
      What's (my assessment of) the probability that they are all aces?

This question is ambiguous. It depends on what you would have done if you
had not found the ace of hearts. If you had found the ace of clubs (but no
ace of hearts), would you tell me it has the ace of clubs? If you simply
reshuffle until you get the ace of hearts, then it's equivalent to your
reworded 2. But if you have other options, then it is different and more
complicated. Suppose you would have picked any of the aces that are in
the hand at random and told me you had that ace if you have any aces, but
if no aces you reshuffle. Then the answer is the same as 1.

                    Steve White
would tell me