From: msb@sq.sq.com (Mark Brader)
Subject: Re: Aces and Queens (P.S. to: The paradox of the second ace)
Followup-To: rec.puzzles
Organization: SoftQuad Inc., Toronto, Canada
Date: Thu, 17 Dec 92 09:51:46 GMT

SPOILER for the aces and queens question, at the end.
Followups are directed, somewhat arbitrarily, to rec.puzzles.

> This paradox is discussed in one of Martin Gardner's collections.

Can someone identify which one, or provide information about the
original appearance of this column?

> probability (a), below, is less than probability (b):
> (a) P(2nd ace | any ace), that is, the probability that a randomly chosen
>     bridge hand has a second ace, given that it has at least one ace
> (b) P(2nd ace | ace of spades), that is, the probability that a randomly
>     chosen bridge hand has a second ace, given that it has the ace of
>     spades

Here's a way to take some of the surprise out of this result.

From a truly randomly permuted deck, you may as well "deal" the bridge
hand by taking the first 13 cards off the top of the deck.  (Do not
try this form of dealing at a bridge club, kids!)  Then the various
probabilities involved are probabilities as to HOW HIGH IN THE DECK
the various aces were located before the "deal".  Number the positions
from 1 (top of the deck) to 52 -- then positions 1 to 13 make the hand.

Okay, consider probability (a).  Advance through the deck looking at
successive cards: position 1, 2, etc.  If you haven't hit an ace by
position 13, you throw out the "deal" and reshuffle.  If you do hit
one, then probability (a) is the chance that you will hit *another*
ace by the time you get to position 13.

Now consider (b).  Start the same way: look through the deck card by
card.  If you haven't hit the ace OF SPADES by position 13, you throw
out the "deal" and reshuffle.  If you do hit it, then probability (b)
is the chance that EITHER you will hit another ace by the time you
get to position 13, OR you hit one BEFORE you got to the ace of spades,
or both.

See?  Now it's not at all surprising that (b) is larger.  The key thing
is that in finding out that there is an ace of spades in the hand, it
doesn't have to be the FIRST ace you come to.

> (I wonder, is this something any decent bridge player would know by
> instinct?)

Well, I'm a decent (not expert) bridge player, and I found it 100%
counterintuitive, until I thought of the above line of reasoning.
Probability questions arise quite often in actual bridge play, but
that is not one that would realistically do so.

A simple example of a question that does commonly arise (from a hand I
played a few hours ago), is "given that East and West have between
them the King, Queen, and two other diamonds, what is the probability
that West either has exactly two, or has the King or Queen and no
others?"  (About 53% -- fortunately, she did.)
 
> Here is a variant I haven't seen elsewhere.  Consider:
> (c) P(ace | queen), that is, the probability that a randomly chosen
>     bridge hand has an ace, given that it has a queen
> (d) P(ace | queen of spades), that is, the probability that a randomly chosen
>     bridge hand has an ace, given that it has the queen of spades
> Despite what you might guess, (c) and (d) are *not the same*.

Just as with aces, a hand with the queen of spades is more likely to
have a second queen than just any hand with a queen is.  A hand with more
queens is less likely to have more aces -- there are fewer "slots" left
to fill.  Therefore, (c) is the higher probability.  (That was fun.)
-- 
Mark Brader	"Many 'business-oriented' packagings of these dialects ...
utzoo!sq!msb	 omit the games section.  Those responsible will doubtless be
msb@sq.com	 reincarnated as worker insects of some sort." -- "J. E. Lapin"

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