From: JXSternChangeX2R@gte.net   
      
   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.   
      
   Let's simply enumerate all   
   sixteen possibilities.  We want a series of three that   
   had exactly two H's:   
      
   HHHH -   
   HHHT -   
   HHTH -   
   HHTT - yes   
   HTHH -   
   HTHT - yes   
   HTTH - yes   
   HTTT - yes   
   THHH -   
   THHT -   
   THTH - yes   
   THTT - yes   
   TTHH - yes   
   TTHT - yes   
   TTTH - yes   
   TTTT -   
      
   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?   
      
   Joshua Stern   
      
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