Given that a bridge hand contains an ace, or given that it contains the ace of spades, in which case are...

probability/card/more.aces Added material

Problem

Given that a bridge hand contains an ace, or given that it contains the
ace of spades, in which case are you more likely to have two or more aces?

Show solution

Consider all the different one-ace hands you could have been dealt, and
all the more-than-one-ace hands you could have been dealt.  The odds of
two or more aces, given at least one ace, is the size of the second set
divided by the size of the first set.

Now restrict the hands to those that contain the ace of spades.
This eliminates all but one fourth of the one-ace hands.  But more than
half of the two-or-more-ace hands have an ace of spades in them, so we
keep most of the second set.  So whatever the odds were in the case where
any ace would do, they are better now (in fact they've more than doubled).

Additional material

rhoads@sceloporus.rutgers.edu (Glenn Rhoads) writes:
>The second probability seems unintuitively high to me.  This probability
>says that of all the hands that have the ace of spades, 56% of them have
>another ace!  A very surprising result.

It's not all *that* unintuitive.  Suppose you were betting on there
being another ace.  Then you'd be betting that of all the places the
Ace of Hearts could have gone, it happened to go into the same hand.
Well, chances are it didn't, but you can also win if the Ace of
Diamonds went in the same hand, or if the Ace of Clubs did.  The
probability that any one of these cards will win the bet for you is
not great (it's slightly less than 1/4), but you have three chances to
win.  Given enough chances to win, even at poor odds, and chances are
one of them will pay off.

Here's a much easier way to figure the exact odds than counting
combinations.  Imagine that instead of dealing the deck normally, what
you do is take the deck in order---first the Ace of Spades (SA), then
the other three aces (let's say HA,DA,CA in that order), then the rest
of the cards---and taking each card in turn, randomly place it into
one of 52 spaces on the table in front of you, choosing with equal
probability among any open spaces.  The 13 leftmost spaces become your
"hand" at the end.

This procedure gives the same probability of any hand as a normal
deal, but it's a lot easier to compute.  First of all, you're not
supposed to consider any hand that doesn't contain the SA, so it's
given that the SA gets dealt to one of the 13 spaces on the left.  Now
the only way *not* to get two aces is for the next three aces all to
get dealt on the right.  The HA has 51 places to go, of which 39 are
on the right, so it has 39/51 probability to go on the right.  But if
the HA goes on the right, it uses up a slot there, and the DA has only
38/50 probability to go on the right.  And if the HA and the DA do go
on the right, the CA has only 37/49 probability to do so.  So the
probability that all three of these cards go on the right, one after
the other, is (39/51)*(38/50)*(37/49) = 0.4388 (approx.)---and if this
*doesn't* happen at least one of them will wind up in your hand, so
the probability of that is 1 - 0.4388 = 0.5612.

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