A bridge player glimpses a red ace in West's hand; West then leads the ace of diamonds.

probability/card/red.ace New

New entry, edited from material collected for the archive and not yet in a released version.

Problem

A bridge player glimpses a red ace in West's hand; West then leads the
ace of diamonds.  What is the probability that West also holds the ace
of hearts?

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?

-- Brad Cohen <bcohen@wam.umd.edu>

Show solution

In article <1994Jun9.144826.1@ube.ubalt.edu> eajqrda@ube.ubalt.edu
(Dick Adams) writes:
>stiemark@astro.ocis.temple.edu (Mark Stieffenhofer) writes:
>> Well, since west is holding 12 cards at this moment and east is
>> holding 13, I'd have to say 12/25.
>
>I'm going to stay with .5 for two reasons.

I disagree with both of you.

There are 25*26 different ways the two red aces could have landed
among the cards held by East and West.  So just knowing that neither
North nor South has a red ace gives us

   P(W has AH,AD) = P(W has neither) = 12*13/(25*26)
   P(W has just AH) = P(W has just AD) = (13^2)/(25*26)

But South knows that West has a red ace.  One way to model this is
that South gets to ask West, "Do you have a red ace?" and West must
answer yes or no truthfully.  A more realistic setup might be that
South has prearranged with a confederate standing behind West to
signal her if West has a red ace.

But a much more plausible mechanism was that South just happened to
see the rank and color of just one card that was dealt to West.  (If
South saw more of West's cards than that, it changes the probabilities
and should have been specified in the question.)  This eliminates the
case <W has neither ace>, but it also affects the probabilities of the
other cases, because the card that South saw is twice as likely to be
a red ace in the case <W has AH,AD> as in the other two cases.
Specifically, given that South and North have no red aces and South
saw the rank and color of just one of West's cards, we have

   P(W has AH,AD & S saw a red ace) = (2/13)*(12*13)/(25*26)
   P(W has AH & S saw a red ace) = P(W has AD & S saw a red ace)
                                 = (1/13)*(13^2)/(25*26)
   P(W has neither & S saw a red ace) = 0

Normalizing (multiplying all the probabilities by a factor so that
they sum to 1), we see that after the deal, based on South's
knowledge,

   P(W has AH,AD) = 12/25 = 24/50
   P(W has AH) = 13/50
   P(W has AD) = 13/50

Well, there's the 12/25 [Editor: "12/15" in the original] answer, but
West hasn't played yet.  The probabilities known to South *after* West
plays depend heavily on the likelihood that West would play the ace of
diamonds given each of the possible hands.  For example, if West would
always have played AH rather than AD if he had both cards, then P(W
had AH,AD) is *zero* after West plays AD.  (Notice I just changed
"has" to "had" in order to avoid the quibble about the fact that once
West plays AD, he no longer "has" it.)

Since I don't know enough to give a better estimate of these
probabilities, I'll assume that West picks one of his 13 cards at
random and plays it.  So given South's knowledge after the deal,

   P(W has AH,AD & W plays AD) = (1/13)(24/50)
   P(W has AH & W plays AD) = 0
   P(W has AD & W plays AD) = (1/13)(13/50)

Once again we normalize to 1, so that as far as South knows after W
plays the ace of diamonds,

   P(W had AH,AD) = 24/37
   P(W had only AD) = 13/37

Whew!

-- David A. Karr <karr@cs.cornell.edu>