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- Xref: sparky sci.math:17014 rec.puzzles:7964
- Newsgroups: sci.math,rec.puzzles
- Path: sparky!uunet!pipex!bnr.co.uk!bnrgate!corpgate!crchh327!crchh410!bcash
- From: bcash@crchh410.BNR.CA (Brian Cash)
- Subject: Re: Marilyn Vos Savant's error?
- Message-ID: <1992Dec16.072922.21082@bnr.ca>
- Sender: news@bnr.ca (News on crchh327)
- Nntp-Posting-Host: crchh410
- Reply-To: bcash@crchh410.BNR.CA (Brian Cash)
- Organization: Bell-Northern Research Ltd.
- References: <1992Dec15.012404.24027@galois.mit.edu> <1992Dec15.052211.24395@CSD-NewsHost.Stanford.EDU> <1992Dec15.063213.11742@bnr.ca> <1992Dec15.091823.10100@nmt.edu>
- Date: Wed, 16 Dec 1992 07:29:22 GMT
- Lines: 79
-
- In article <1992Dec15.091823.10100@nmt.edu>, jduncan@nmt.edu (John
- Duncan) writes:
- |>In article <1992Dec15.063213.11742@bnr.ca> bcash@crchh410.BNR.CA
- (Brian Cash) writes:
- |>>In article <1992Dec15.052211.24395@CSD-NewsHost.Stanford.EDU>,
- |>>snewman@Xenon.Stanford.EDU (Steven Newman) writes:
- |>>|>
- |>>|> 1) You meet a man on the street and ask him how many children he has.
- |>>|> He replies "two, and one is a boy." What is the probability that
- |>>|> his other child is also a boy?
- |>>|>
- |>>|> 2) You meet a man on the street and ask him how many children he has.
- |>>|> He replies "two, and the older one is a boy." What is the
- |>>|> probability that his other child is also a boy?
- |>>|>
- |>>|>The answer to problem 1 is 1/3, while the answer to problem 2 is 1/2.
- |>>|>(One must be careful in the interpretation of the statement in problem
- |>>|>1; it means "I have two children, and it is not the case that both of
- |>>|>them are girls.")
- |>>|>
- |>>|> - Steve Newman (snewman@cs.stanford.edu)
- |>>
- |>>Sorry! You made the same mistake as the first poster!
- |>
- |>actually, no. it is you that repeated the misteak :)
- |>
- |>>The odds of the other child being a boy is 1/2 in both cases.
- |>>Observe:
- |>>
- |>>Case 1) Child = boy Other = ?
- |>>Yes -> a) Child = boy , Other = boy
- |>>No -> b) Child = boy , Other = girl
- |>> c) Child = girl, Other = boy
- |>> d) Child = girl, Other = girl
- |>> Obviously, c and d are invalid.
- |>
- |>WRONG !!! c is not invalid
- |>one is still a boy so the chances are, in fact, 1/3 that the other is a boy.
- |>
- |>Regards
- |> --John P. Duncan (jduncan@prism.nmt.edu)
-
- Uh, I'm sorry, but I don't follow the logic in that.
- The man said "one is a boy". The "Child" in case 1 is
- the child the man refered to when he said "one".
- c states that the child he is talking about is a girl,
- and the one he doesn't mention is a boy. This is
- invalid.
-
- ex.
- "I have two coins that add to $.35 and one is not a quarter, how is this so?"
-
- One = dime Other = quarter
- NOT One = quarter Other = dime
-
- What am I doing wrong?
-
- I think you might be dividing the children in another way:
-
- a) child A = boy child B = boy
- b) child A = boy child B = girl
- c) child A = girl child B = boy
- d) child A = girl child B = girl
-
- Given this and "one is a boy" only d is invalid, but it also
- makes b and c the *same* (one boy, one girl...order is irrelevant).
- So now you have two events: a and b/c. Odds = 1/2
- I flip a coin 100 times. 99 of those times it came up heads,
- what are the odds that the other flip came up heads also?
- 1/2. Which coin is the "mystery flip" is irrelevant to the problem.
-
- The only cases where you can say {A = x, B = y} and {A = y, B = x}
- are two separate probabilities is when A and B are not interchangable
- (ie. Younger and Older are not interchangable but child A and child B
- are).
-
- Does anyone follow what I am saying?
-
- Brian /-|-\
-