🚀 go-pugleaf

In article <C1J1GE.5wy@news.rich.bnr.ca>
bcash@crchh410.BNR.CA (Brian Cash) writes:

> >From: cooper@cbnewsg.cb.att.com (Ralph 'Hairy' Moonen)
> 
> Everything stated is corect, until this:
> >   A coin is tossed twice. You are told that *at least* one of the tosses
> >   produced heads. What is the probabilty that THE SECOND (and *not* the
> >   first!) toss produced heads to make both tosses heads. In this case,
> >   the permutations of HT and TH are distinhuished and different. the
> >   answer now is 1/3. People read the original question and see the above
> >   one, because it's easy to make this mistake. 
> 
> What is the prob. that the 2nd toss is heads and not the first, making
> both tosses
> heads?  Zero!  If the first toss is not heads, then they can't be both heads!
> I think you want to re-state this.
> I think this is what you mean:
> 
> A coin is tossed twice. You know that *at least* one of the tosses is heads.
> What is the prob. that the second toss is heads?  Ans = 2/3.
> What is the prob. that they are both heads? Ans = 1/3.
> 
> 
> 
> The main point with the sibling/coin problem is whether or not the the
> probabilities are independent or not.  If you say "One is x, what is the
> prob. that the other is also x?", then the probabilities are independent and
> the fact that "One is x" doesn't change anything.  If you say "At least
> one is x 
> (out of two), what is the prob. that number 2 is x?" links the prob.s because
> if the first is not-x, then the second _must_ be x, so the answer is different.
> 
> Does this sound reasonable?   (I got nailed for missing this on the
> first pass :)
> 
> Brian /-|-\

But what would the answer be if Monty Hall asked the question?

:-)


--Ted Schuerzinger
email: .zed@Dartmouth.EDU
"I should have realized it would be bad vodka when all the label said
was 'Russian Vodka'."