There is much disparity in the verity of parity, so to increase familiarity I will offer some clarity.
But first some hilarity from a bidding vulgarity:
North’s launch into Blackwood borders on insanity. Alas, the same has been said about me, but let me assure you, I haven’t spoken to any saucer people in over a week. After a trump lead (East pitching a heart) won in hand, how can 6 be made? Surely you would start with a diamond to the king, but then what? Even looking at all four hands it’s not obvious.
Declarer must next play three rounds of clubs to pitch the diamond, forcing East to win the defenders’ one trick. This prevents a second trump lead, then declarer can establish a diamond trick by ruffing finesse and crossruff with impunity. Now that you see how to make 6 , take another look. Yes, the slam can be defeated if East pitches a club on the opening lead.
Enough bridge! Back to parity, for which the above deal is a rarity. Almost everything is even. Each hand has an even number of: high cards (AKQJ), high-card points, honor cards (AKQJT), control cards (AK), controls (A=2 K=1), touching* cards, and every card rank except deuces (an even number of all card ranks is impossible).
*Consecutive cards within the same suit. A three-card sequence like East’s Q-J-10 counts as two touching cards (Q-J touch, J-10 touch); a four-card sequence is three touching cards, etc.
Further, each hand has an even number of cards in three suits, the most possible.* Note that zero is an even number.
*Every bridge hand must have exactly three even suits and one odd suit, or vice versa. There is no mystery in this but just plain arithmetic; suit lengths must add to 13, and the sum of four numbers can be odd only when one or three of them are even (or odd).
Test your parity dexterity by entering O (odd), E (even) or S (same chance).
In a random bridge hand, are these more likely to be odd or even:
1. Number of clubs?
2. Number of minor-suit cards?
3. Number of aces?
4. Number of kings and queens?
5. Number of high-card points?
6. Number of touching cards?
7. Number of honors (AKQJT)?
8. Number of even-ranked cards?
Quit
Congratulations to Noa Kant (Netherlands) who now joins Peter Randall, winner of Two Eight-Baggers, as the only persons who have won every contest they entered at this web site. (Peter has not entered since, so he may be sitting on his 1/1 for posterity.) Time will tell if Noa stands pat, too.
Most answers can be derived by straightforward combinatorics. In each case I chose to count hands with an odd number, hence if that total is below 50 percent, the answer is even. Obviously the same result would be reached by counting hands with an even number.
Richard Stein: A script of several R commands did the trick for all of these with straight combinatorial calculations — except #6.
An odd number of clubs can be held in seven ways: one club and 12 non-clubs, three clubs and 10 non-clubs, etc., through 13 clubs and no other cards. The following table counts the hands:
Since the odd percent is below 50, an even number of clubs is more likely, albeit not by much.
Here I use the same combinatorial format, but there are 26 minor-suit cards, and 26 others (major-suit cards).
Note that the above total is exactly half of the 635,013,559,600 possible bridge hands.
Paul Gilbert: This is easily provable due to major-minor symmetry.
Right, and true of course for any two suits. If the number of minor-suit cards is odd, the number of major-suit cards must be even, and vice versa. Hence every bridge hand must have an odd number of one or the other, so the relationship must be equal.
This is easy, as there are only two ways to hold an odd number of aces: one ace and 12 non-aces, or three aces and 10 non-aces.
Therefore, an even number of aces (0, 2 or 4) is more likely. Of the eight problems, this is the only one with a significant difference.
There are four ways to hold an odd number of kings and queens, and the hands are counted below.
Evidently bridge hands are tough on marriages, as more often than not there’s an odd spouse.
Puzzle within a puzzle: What is “Never odd or even?” (answer at end)
Rather than implement an unwieldy hand count, this one has a neat solution. The parity of HCP depends solely on the number of kings and jacks. Either contributes an odd number of HCP (king = 3, jack = 1) so the HCP total will be odd only when the number of kings and jacks is odd. Obviously, this must be identical to the previous problem (kings and queens) so the odd percent is 50.02370976.
Nicholas Greer: Number of HCP should have the same parity as the number of kings and jacks, which must be the same as kings and queens. Six more shortcuts like this, and I might be able to solve this accurately.
This was the stickler of the contest, with no obvious algorithm, besides counting individual hands.
Richard Stein: [Solving this] involved enumeration of the 39 hand patterns, and the parity of touching cards in every possible suit length. Just don’t ask me why I bothered to do it!
We Richards think alike, although I had to do it, because I posed this stupid problem. The first step was to determine the apportionment of each suit length for odd-even touching cards. For example, a three-card suit comprises 286 holdings, of which I found that 110 had an odd number of touching cards, and 176 had an ever number. Note that zero is even. The following table shows the breakdown:
This table alone suggests the answer is even, as there are more cases of even touching (4160) than odd (4032), and only five suit lengths (tinted gold) favor odd — yet hardly valid proof for a full hand. The next step was to check each generic hand pattern to find the ways to obtain an odd total. To illustrate, consider the 4-4-3-2 pattern. Of the 16 possible odd-even suit sequences, eight of them produce an odd total as summarized below.
Applying the above to each of the 39 generic hand patterns produced the following table:
Adding the three Odd Hands columns produces a total slightly under 50 percent, hence even is more likely.
Charles Blair: If my program can be trusted (ha-ha), even touching cards outnumbers odd by a little under 10 million.
On the mark. The exact difference is 9,765,680 hands.
Charles Blair: Fun fact: For suits headed by any odd-ranked card (K-J-9-7-5-3) and counting all possible holdings thereafter, the number with odd touching cards is exactly equal to the number with even touching cards.
Interesting. Checking suits headed by a five shows four odd (5432 542 532 54) and four even (543 53 52 5). For suits headed by a seven, if the five is included, the equality remains because the six changes nothing (e.g., 765 and 75 are both even); and without the five leaves eight odd (76432 7642 764 763 762 743 732 76) and eight even (7643 7632 7432 742 74 73 72 7). Suits headed by the king, jack or nine keep the same relationship. Cute! I know many of my projects are useless, but the Blair Witch Projects bring me hope.
Back to easy solutions! There are 20 honor cards, and 32 non-honor cards. Hands with an odd number are counted below:
Excluding Problem 2 (same chance) this was the closest of all. The number of hands with an even number of honors is 317506794880, only 30160 more than above.
Noa Kant: Fun challenge! Some cases are only microscopically different.
Julien Reichert: Some non-zero differences are astonishingly small!
There are 28 even-ranked cards (A-Q-10-8-6-4-2) and 24 others. Hands with an odd number of even-ranked cards are counted below:
Noa Kant: [This assumes] even-ranked cards are defined as A-Q-10-8-6-4-2. If instead only 10-8-6-4-2, the answer would be the same as for honors.
If anyone is still interested, the favored parity of odd-ranked cards (K-J-9-7-5-3) is even. Indeed, this has to be so with the same number of hands, because every hand with an odd number of even-ranked cards must have an even number of odd-ranked cards, and vice versa. If this doesn’t confuse you enough try “She sells sea shells by the seashore.”
Puzzle within a puzzle (answer): It’s a palindrome!
Julien Reichert: Hoping to feel lucky about touching cards.
Sorry, no luck there, but consider yourself lucky that I’m not Harry Callahan.
Grant Peacock: Just like your Two Eight-Baggers puzzle, I wrote software to solve this, but its output might not mean anything.
Two of Grant’s answers confirm this. Apparently he quit NBC to work for the government.
© 2021 Richard Pavlicek
by Richard Pavlicek by Richard Pavlicek