From: dnk@OMIT.cs.mu.oz.au   
      
   In  JXStern  writes:   
      
   > On Fri, 17 Oct 2003 03:42:29 GMT, dnk@OMIT.cs.mu.oz.au (David) wrote:   
   > On Fri, 17 Oct 2003 03:42:29 GMT, in comp.ai you wrote:   
      
   > >In  JXStern    
   writes:   
   > >   
   > >>On Thu, 16 Oct 2003 06:12:07 GMT, dnk@OMIT.cs.mu.oz.au (David) wrote:   
   > >>>So the overall expectation = 2 + (14/3 + 8/3 + 10/3 + 6/3)/4 = 31/6   
   > >   
   > >>In other words, four tosses, where three was the min possible.   
   > >   
   > >5 and a bit, actually.   
      
   > Um, yes, on what 31/6 equals, I must have slipped a cog.   
      
   > However, even with a little actual thought, I have a problem   
   > following your logic on the others, and my guestimation logic   
   > still says four.   
      
   I suspect that's because you are answering a different question, see below.   
      
   > Let's simply enumerate all   
   > sixteen possibilities.  We want a series of three that   
   > had exactly two H's:   
      
   > HHHH -   
   > HHHT -       Y4   
   > HHTH -       Y3   
   > HHTT - yes   Y3   
   > HTHH -       Y3   
   > HTHT - yes   Y3   
   > HTTH - yes   
   > HTTT - yes   
   > THHH -       Y3   
   > THHT -       Y3   
   > THTH - yes   Y4   
   > THTT - yes   
   > TTHH - yes   Y4   
   > TTHT - yes   
   > TTTH - yes   
   > TTTT -   
      
   I've edited your list above to add Y3 or Y4 to the terminating ones.   
   While we differ on the details :), we both get 9 as the answer.   
      
   > Out of 16, we get 9.  The answer to the original question is four, as   
   > your 1% sequence logic also shows -- 7/16 odds of non-completion at   
   > four tosses.  Yes?   
      
   If the question is   
      
    "How many tosses to have at least 50% chance of termination?"   
      
   the answer is indeed 4.  But the expected number of tosses is the   
   average number over all possible games, and it's bigger than 4   
   because some games may take a long time, e.g. there's a better than   
   5% chance that the game will take at least 10 tosses to terminate.   
      
   Imagine some other game that has a 50% chance of terminating after   
   exactly 1 or 9 turns, the expected tosses to termination will be 5,   
   even though you can't actually terminate after exactly 5 tosses.   
      
   > Joshua Stern   
      
   David   
      
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