# Statistics puzzle i05244c

Eddy Sean sre at al.cam.ac.uk
Wed Jun 15 07:17:55 EST 1994

```  [ Answering a bridge probability puzzler, I say 2/3 from Bayes'
rule, to a small chorus of nays.]

>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.

Heh. Not according to my reasoning. *Before* the first card is played,
there are four models: west holds both red aces; west holds either one
of the red aces; or east holds both red aces. The priors on these are
1/4 each. The likelihood of randomly choosing (spying) a red ace from
west's hand in these models is 2/13, 1/13, 1/13, 0/13. The posterior
probability that west holds the ace of hearts before we see the first
card is :
2/13*1/4 + 1/13*1/4
--------------------                  = 3/4
2/13*1/4 + 1/13*1/4 + 1/13*1/4 + 0/13*1/4

When we see west lead the ace of diamonds, the probability that he
holds the ace of hearts drops to 2/3, because the likelihood of our
data given the model that he holds only the ace of hearts then becomes
0.

The 50/50 answers (or 48/52 answers) are intuitively unreasonable; we
*do* have information that west has at least one red ace, which must
increase the probability that he holds the ace of hearts. The model in
which east got both red aces is eliminated by the data.

Anyone wanna bet? :) :)

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

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