🚀 go-pugleaf

In article <1993Jan27.112827.26276@cbfsb.cb.att.com>  
cooper@cbnewsg.cb.att.com (Ralph 'Hairy' Moonen) writes:
> 2) (The one that is subject to misinterpretation) A coin is tossed  
twice.
>    You are told that *at least* one of the tosses produced a heads. What
>    is the probablity of the other also being heads. The problem here is 
>    that people seem to interpret this question as if the sequences TH  
and
>    HT are different. But that is not how the question is worded.  
Therefore
>    the answer to this question is also 1/2. 

The reason TH and HT are treated as different is simply that they *are*  
different. There are four equally likely possibilities, one of which is  
ruled out by the statement: HH, HT and TH are all possible, and TT is not.  
Of the three equally likely cases, one results in both tosses being heads.  
Therefore, the probability is 1/3, not 1/2.

>The wording needed to come to
>    an answer of 1/3 is this: (and is not the original question)
> 
>    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. 

I don't see how anyone could possibly mistake one of these for the other.  
If THE SECOND (and *not* the first!) toss produced heads, then you are  
reduced to the single case TH. The probability of having two heads in this  
case is exactly zero.

--
Dave Seaman
ags@seaman.cc.purdue.edu