A slightly less-than-honest bridge player (south) caught a glimpse of a card dealt to her opponent on the...

SOLUTION:  I believe the glimpse of "a red ace" is superfluous
information, once it is known that west has the ace of diamonds.  A
simulation helps illuminate matters:

1)  Shuffle a deck and deal out four hands.  Examine all four hands.

2)  Establish whether the conditions of the problem -- west has the ace
of diamonds, neither north nor south has the ace of hearts -- are met.
(If not, skip 3 and 4.)

3)  If so, record a tally in the "valid hands" scorecard.

4)  Does west have the ace of hearts?  If so, record a tally in the
"success" scorecard.

5)  Repeat steps 1-4 many times, and divide the number of successes by
the number of trials.

Here is a computer program (in RESAMPLING STATS):

MAXSIZE hands 10000 success 10000
URN 1#1 1#2 50#0 deck      'Ace of hearts (1), ace of diamonds (2)
                           'and 50 other cards (0's)
REPEAT 10000
  SHUFFLE deck deck
  TAKE deck 1,13 south      'deal 4 hands
  TAKE deck 14,26 west
  TAKE deck 27,39 north
  TAKE deck 40,52 east
  SUM south s1             'sum the values in north and south
                           'hands.  If sum >0, one of them has
                           'a red ace and the hand does not
                           'meet the conditions of the problem.
  SUM north s2
  ADD s1 s2 s3
  IF s3 =0                 'If neither south nor north has a red
                           'ace, this is a valid hand, so play
                           'on
    COUNT west =2 b        'determine whether west has the ace
                           'of diamonds, as the problem requires.
    IF b =1                'if so
      SCORE 1 hands        'score a "1" to the valid hands vector
      COUNT west =1 c      'determine whether west has the ace
                           'of hearts
      IF c =1
        SCORE 1 success    'Score "1" to the "success" vector
      END
    END
  END
END
COUNT hands =1 khands          'Count how many hands were valid
COUNT success =1 ksuccess      'Count how many were successes
DIVIDE ksuccess khands result
PRINT result

RESULT   =    0.48417

This appears to confirm the solution of one reader: 12/25 (48%).
(Number of cards west has divided by the total number of cards
outstanding that might be the ace of hearts.)

Discussion of the advantages and disadvantages of analytical
approaches versus resampling simulation can be found in a work in
progress by Julian Simon that provides resampling solutions to
Mosteller's "50 Challenging Problems in Probability" (as well as to
many other puzzles), available in the resampling section of the
University of Maryland's "InforM" Gopher system.

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

----

This isn't really correct.  We can see a flaw in step 2.

There's a problem here: to satisfy the conditions of the problem, it is
not sufficient that West simply has the diamond ace.  West must also
lead the diamond ace, and it is far from certain that West will always
lead the diamond ace whenever he has it.

A more accurate simulation would [Editor: a verb is missing here in
the original] the conditions of the problem as follows:  West has a red
ace, neither North nor South has the heart ace, and West leads the
diamond ace.  Unfortunately, we don't have any criteria given in the
problem for determining when West leads the diamond ace from his hand.
We either need to write a bridge-playing program, or make some
simplifying assumptions (e.g. West always leads an ace when he can, and
he chooses that ace randomly.  Notice that this assumption is different
than the one that David Karr used in his analysis.  [I think.  I
haven't read it in a while.])

There's nothing wrong with using simulations, but you must be sure that
you are simulating the right thing.  In this case, your simulation
corroborates your supposition that the glimpse of a red ace is
superfluous information simply because your simulation ignores that
information.

-- Randall M. Gee <rmgee@soda.berkeley.edu>

----

As I have pointed out, the problem is not well posed.  There are,
however, ways of making it precise for which this method will give a
wrong answer.

For one thing, how does West behave?  If she always leads the Ace of
hearts if she has it, the chances that she has it are zero.  If she
chooses between the Aces indifferently when she has both, the chances
are nonzero, and if she always leads the Ace of diamonds the chance of
her also having the heart is higher still.  In other words, from her
play you know more than that she has the Ace of diamonds:  you know
that she has not led the Ace of hearts, which changes the probability
that she has it.  Such inferences are very important in real bridge
problems, of which I gave an example.

For another, the glimpse is not necessarily irrelevant, if you consider
it also to be a random event.  That is, suppose you see the rank and
color of one card at random.  Then you will see a red Ace twice as
often on hands containing two red Aces as on hands with one.  So simply
eliminating hands with no red Aces does not do the trick; you also have
to overweight hands with two red Aces.

In one sense, there is nothing wrong with solving the problem by
simulation; you have just simulated the problem incorrectly.  On the
other hand, we might reasonably ask whether the errors of logic to
which a simulation approach is subject are the same or different from
an analytical approach.

-- Tom Permutt <permuttt@clark.net>