Absolutely Winning Bridge Hands

post by jefftk (jkaufman) · 2019-12-25T02:50:14.711Z · LW · GW · 4 comments

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In bridge the best you can do is commit to take all the cards and then follow through. Even better if you do this without the help of a trump suit. Playing today I was curious: what fraction of hands are "absolutely winning" for 7NT, in that they'll let you make this contract no matter what your partner or opponents happen to hold?

Here's one example hand:

• Clubs: A, K, Q
• Diamonds: A, K, Q
• Hearts: A, K, Q
• Spades: A, K, Q, J

And here's a very different example:

• Clubs: A
• Diamonds: A
• Hearts: A
• Spades: A, K, Q, J, 10, 9, 8, 7, 6, 5

Any absolutely winning 7NT hand needs to have all the aces, so that whatever your opponent leads you can take the first trick. Then all the other cards need to be "good", in that no other player can have a higher card in their suit.

Since a card being "good" requires that you also have every higher card in the same suit, all that matters for each suit is its length. If I tell you that an absolutely winning hand has four spades, you know they are A, K, Q, J. What we're doing, then, is assigning positive lengths to suits, where all lengths must sum to 13. But it's simpler if we ignore the aces, since you have to have all four, and assign non-negative lengths that add to 9. How many ways can we do this?

Since I'm more of a programmer than a mathematician, here's a way to solve this with code:

print(len([
  (c, d, h, s)
  for c in range(10)
  for d in range(10 - c)
  for h in range(10 - c - d)
  for s in range(10 - c - d - h)]))

Which gives 715. [1] There are 52-choose-13 bridge hands (about 635B) so your chances of getting an absolutely winning 7NT hand is 715 in 635B or just a bit better than one in a billion.

While very unlikely, this is in the "it could happen, maybe" range and not all the way to "no way, unless the shuffle was rigged".

EDIT: the above is wrong in two ways. First, the number of spades is completely determined by the number of the other three suits, and so for s in range(10 - c - d - h) should not have been included. And second, if you have seven or more of a suit then which lowest cards you have doesn't matter, because by the time you get to them no one else will have any of the suit. Here's a revised solution that fixes these two:

from math import factorial

def choose(n, r):
  return factorial(n) // factorial(r) // factorial(n-r)

count = 0

# Start by choosing a length for each suit.
# Most lengths will lead to one valid
# combination, though a few will lead to
# several.

# Since every suit needs to have an ace,
# there are nine remaining cards in the hand
# to be divided among the four suits.
hand_size_ignoring_aces = 9

for c in range(hand_size_ignoring_aces + 1):
  for d in range(hand_size_ignoring_aces + 1 - c):
    for h in range(hand_size_ignoring_aces + 1 - c - d):
      s = hand_size_ignoring_aces - c - d - h

      # If a suit has >= 7 cards there are
      # extra possibilities: 
      #   A K Q J followed by any 5 others.
      # If your suit is length N, you need
      # the top 13-N cards in order, and
      # then any remaining cards in that
      # suit are also good because no one
      # else has that suit

      # At most one suit can have >= 7 since
      # you have 13 total, so find out if
      # we're in this case.
      longest = max(c,d,h,s)

      # We excluded the aces, but it's
      # clearer here with them included:
      longest += 1

      if longest < 7:
        # Situation doesn't apply, we must
        # just have the top cards in the
        # suit in order
        count += 1
      else:
        others_have = 13 - longest
        in_order = others_have
        out_of_order = longest - in_order
        count += choose(
          others_have + out_of_order,
          out_of_order)
print(count)

This gives 3,756 hands, or about six in a billion.

[1] This is 13-choose-4, and while there's probably an obvious-in-retrospect reason why, I'm not seeing it.

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