From: mtgradwell@btinternet.com   
      
   David  wrote in message   
   news:bmosfv$dj5$1@mulga.cs.mu.OZ.AU...   
   ..   
   > > > Consider the situations where we have made (at least) 2 tosses but   
   > > > the game has not yet terminated.  With equal probability (1/4)   
   > > > the last two tosses must be one of the sequences TT, TH, HT or HH.   
   >   
   [Masrtin Gradwell]   
   > > The error here is in the phrase "with equal probability". If we   
   > > had no knowledge about the recent state of the game, then these   
   > > four possibilities would indeed be equiprobable. But we are   
   > > considering the situation where the game has not yet terminated.   
   > > Since the game only terminates when there has been a preponderance   
   > > of heads, the fact that it hasn't terminated suggests the likelihood   
   > > that there has been a preponderance of tails.   
   >   
   > You're right that the these 4 cases aren't equiprobable in general,   
   > but they are equiprobable after throwing 2 tosses.  I miss-stated   
   > that but I don't believe it affects the analysis below.   
      
   Yes the "(at least)" modifier was incorrect, the 4 cases are equiprobable   
   after the first 2 tosses, but not thereafter (if we are given that the game   
   has   
   not yet terminated).   
      
   >   
   > > There are 8 possible sequences of 3 coin tosses to start the game off:   
   > > ttt,tth,tht,thh,htt,hth,hht,hhh, and of these four (thh,hth,hht,hhh)   
   > > end the game straight away.   
   >   
   > No, HHH does not terminate the game, because the requirement is   
   > "exactly two heads up" of the last 3, not at least 2.   
      
   Yes. I somehow managed to misread the description of the game.   
   Sorry about that. This means my analysis was wrong in detail. I ended   
   up trying to solve some other problem instead of the one that was set.   
   I should have checked more carefully before posting.   
      
   The four cases are still not equiprobable after three or more coin   
   tosses, but they are equiprobable after the first two, and that is what   
   matters. So the expectation for the number of tosses required really   
   is 2 plus the expectations deriving from the 4 possible cases that arise   
   after two tosses, all summed and divided by 4.   
      
   ..   
      
   >   
   > OK, please run the simulation, and let us know what you get.   
      
   The problem I was trying to solve, due to my misreading, probably   
   would require a simulation, or a lot more maths than I'd like to tackle   
   by hand. But I think your approach was the correct one for tackling   
   the problem as stated, and was marred only by the incorrect use of   
   the parenthetical modifiier "at least". I hope that clears things up.   
      
   Martin   
      
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