Statistics puzzle i05244c

Eddy Sean sre at al.cam.ac.uk
Tue Jun 14 18:50:54 EST 1994


   >Martin Gardner says that new mathematical puzzles are very
   >difficult to devise.  What do you think of this one?
   >
   >A slightly less-than-honest bridge player (south) caught a
   >glimpse of a card dealt to her opponent on the left (west) - it
   >was a red ace (she could not tell which suit).  This opponent --
   >west -- opens the game by playing the ace of diamonds.  South
   >sees that neither she nor the revealed cards of north have the
   >ace of hearts, which must be in either east's hand or west's
   >hand.  What is the probability now that west has the ace of
   >hearts?
   >For a solution using the resampling method, contact pcbruce at wam.umd.edu)

Er. What's wrong with good old Bayes' rule? Plug and chug; the answer
is 2/3. Intuition may fail, but I doubt that Bayes does.

 P(west | red ace) =  likelihood * prior / normalized over possible models;

                =    P(red ace | west) * P(west)
                      ------------------------
               P(red ace|west)*P(west) + P(red ace|east)*P(east)  

 The likelihood that a randomly espied card from west's hand 
 was a red ace is 2/13 if west holds both red aces, 1/13 if
 west only held the ace of diamonds. The priors are both 1/2.

                =  2/13 * 1/2          = 2/3
                   ----------          
                 2/13*1/2 + 1/13*1/2    

--
- Sean Eddy
- MRC Laboratory of Molecular Biology, Cambridge, England
- sre at mrc-lmb.cam.ac.uk



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