>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
>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