Statistics puzzle i05244c

Warren Gallin wgallin at gpu.srv.ualberta.ca
Tue Jun 14 13:28:55 EST 1994

In Article <SRE.94Jun14185054 at al.cam.ac.uk>, sre at al.cam.ac.uk (Eddy Sean) wrote:

>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  

Doesn't that only apply BEFORE the first card is played?  Once the ace of
diamonds is down, south has no information on what might be in either hand
(since all that was known before was the hand contained a red ace), so Bayes
is unnecessary. its is eithe 12/25 if the trick is not completely played
(thanks to Bradley Sherman for pointing that out to me) or 12/24 if the
trick is completely played.

Warren Gallin,
Department of Zoology, University of Alberta
wgallin at gpu.srv.ualberta.ca

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