From: JXSternChangeX2R@gte.net   
      
   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.  I   
   agree.  Seat of the pants reasoning:   
      
   Either the first three tosses give you the condition and you're done   
   right there, or the last two had a 50% odds of being either HT or TH,   
   and you have a 50-50 change of getting the third H on the next, so   
   you're over 50% odds on the whole deal at that point, plus other more   
   complicated scenarios.   
      
   Now, if you change the question to 99% certainty of the result, ...   
      
   J.   
      
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