# Statistics puzzle i05244c

Andrew Cockburn afc at gnv.ifas.ufl.edu
Wed Jun 15 08:55:29 EST 1994

```In article <wgallin.1122006175I at NEWS.SRV.UALBERTA.CA>, wgallin at gpu.srv.ualberta.ca (Warren Gallin) writes:
> 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

This solution ignores the problem of whether the red ace that was seen was
D or H.  If the red ace was D, then this is correct.  However, if the red
ace could have been H, then the probability of W having both red aces is
higher.

The approach that I would take is

1) consider all possible hands,
2) eliminate all hands in which W does not have the AD (exactly half),
3) figure out how many of these have the AH in W (12/25, since one card
4) multiply the probability of each remaining hand times the probability of
exposing a red ace from that hand (effectively the 12 two red aces in W
hands have twice the probability of the 13 AD hands),
5) normalize.

So P(AH in W)=(12*2)/(12*2+13)=24/37 or a little less than 2/3
P(AHin E)=13/(12*2+13)=13/37 or a little more than 1/3

Andrew Cockburn

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