If I asked you to guess a number from 1 to 100, you’d have a 1-percent chance of getting it right, or perhaps a fraction better presuming the number wouldn’t be 1 or 100. Alas, a dismal chance with or without the fraction.
You open the bidding 1 , and partner forces all the way to slam! Upon seeing dummy, you are fuming to be in about a 1-percent contract. “Relax,” consoles partner, trying to suppress your anger, “with you at the helm it might even approach 2 percent!”
And so it goes with the following holdings in a bridge hand. At first you might guess about a 1-percent chance, which is the correct whole portion, but they’re all a fraction better. Before reading the solutions, test your combinatorial skills, or just make your best guesses of the percents shown. (Each percent is correct for one of the six questions.)
What is the percent chance that a random bridge hand has:
1. at least one five as its lowest card rank? 1.06 1.29 1.40 1.53 1.75 1.96
2. the beer card, devil’s bedposts and little cassino? 1.06 1.29 1.40 1.53 1.75 1.96
3. three suits (4+ cards each) none of them spades? 1.06 1.29 1.40 1.53 1.75 1.96
4. exactly 4 HCP and no suit longer than four cards? 1.06 1.29 1.40 1.53 1.75 1.96
5. exactly five spades, four hearts and no void suit? 1.06 1.29 1.40 1.53 1.75 1.96
6. a pinochle and exactly one marriage (K-Q same suit)? 1.06 1.29 1.40 1.53 1.75 1.96
*Contest participants did not have the benefit of multiple choice, which was added for this writeup.
This puzzle contest is the 82nd in a series that began 15 years ago. See my Puzzle Hall of Fame for an overall ranking of top solvers, top locations, and a compendium of all 82 leaderboards. How many times does your name appear?
Everyone who did well is ranked below by error sum: cumulative absolute difference in hundredths from the correct percents. Ties are broken by date and time of submission (earliest wins). My cutoff to the top eight was clear, as the next best error sum was 53, and it grew to a worst of 214. Geez, if you guessed 1.50 for every answer you’d only be off by 149.
The easiest way to calculate this is to count hands without twos, threes or fours; then subtract from this hands without twos, threes, fours or fives. Logically, the difference will be hands without twos, threes and fours that contain at least one five. To find the percent chance, simply divide by the number of bridge hands (635,013,559,600) and multiply by 100 — or to save the last step, just divide by 6,350,135,596. Percents are shown to 12 decimal places but rounded to 2 places for the puzzle.
At least three solvers were on to this method. Dan Baker and David Wu echoed the expression of:
Richard Stein: (40c13-36c13) / 52c13
A few respondents complained they had never heard of the last two, but with the resources available today, even a blind squirrel could look them up… well, maybe not. The devil’s bedposts is the four of clubs (picture the spots to understand why). The little cassino is the two of spades, from the card game cassino. Of course, everyone knows the beer card, which might say something about the priorities of our society.
This was the easiest problem of the set, as the answer would be the same for any three specific cards. There are 49 remaining cards from which 10 must be chosen to fill the hand.
Richard Stein: 49c10 / 52c13
The stated condition comprises four specific hand patterns, so the straightforward solution is to count the combinations of each:
Richard Stein: Combinations of 1=4=4=4 + 0=5=4=4 + 0=4=5=4 + 0=4=4=5
Cyrus Hettle: Three suits, none of them spades? I’d rather be dead.
This was the most difficult problem, because HCP calculations cannot be quantified as plain numbers. Instead, each specific high-card arrangement that adds up to 4 HCP must be calculated separately for each qualifying hand pattern, then a humongous summation provides the number of hands. Fortunately, I did the math many years ago for every HCP total with every hand pattern, or rather my computer did from the coding I wrote, so I’ll defer to those findings. The answer is available in my study of High Card Expectancy and summarized below:
The answer can also be found with my Shape-HCP Probability Calculator by entering constraints of 1-4 for each suit length and 4 HCP.
Needing only the nearest hundredth, this solver found it empirically:
Richard Stein: I found this by simulation. Random number generator, please don’t let me down!
These solvers were right on the money without any shortcuts:
Dan Baker: I broke this down into specific sets of high cards, then 9cN to fill the low cards in each suit. Total ends up at 8,882,797,392.
David Wu: Hardest problem by a mile. I had to make a spreadsheet of all the possibilities, then write a program to calculate them all out.
Charles Blair: This involved a lot of programming!
As with Problem 3, this is a straightforward case of determining the possible patterns, calculating the number of hands for each, and adding them up, as shown in the table below:
Cyrus Hettle: Take 5=4 majors with assorted holdings [3=1 2=2 1=3] in the minors.
Richard Stein: Combinations of 5=4=3=1 + 5=4=2=2 + 5=4=1=3
Dan Baker: 13c5 × 13c4 × (26c4 - 2(13c4))
Cryptic but absolutely correct… what else from our math guru!
Some solvers were bewildered by this, but it was easy enough to Google an explanation. Pinochle is a melding and trick-taking card game, the most unusual feature being its deck: A-10-K-Q-J-9 twice in each suit, and yes, the 10 is the second-ranked card.
Pinochle not only names the card game but also a specific meld: Q J. A marriage can be any of K-Q, K-Q, K-Q or K-Q, which complicates the problem. If the marriage is in spades, only three cards are needed ( K-Q J) but otherwise four (e.g., Q J K-Q); and a further difficulty is to eliminate hands with two or more marriages. My approach was to count the hands in two groups, depending on whether the K is held.
Note that finding an exact number of marriages requires deducting the counts of a greater number. For example, in Line C there are 3 ways to choose 2 marriages, but each also includes hands with 3 marriages, so Line B is deducted 3 times.
Charles Blair: By method of inclusion and exclusion, I think there are 7,298,761,821 hands that include the K, and 3,818,557,509 that do not.
Not sure how his method works, but methinks he thinks pretty well.
Our winner found the answer a different way, which I must admit is more elegant than mine:
Dan Baker: We always need the J and Q. Of the remaining 7 kings and queens, there is 1 way to select 1 [ K]; 9 ways to select 2 [ K-Q, K-Q, K-Q, K + any]; 24 ways to select 3 [ K-Q + NS*, K-Q + NS, K-Q + NS, K + 2 NM]; and 20 ways to select 4 [ K-Q + 2 NS NM, K-Q + 2 NS NM, K-Q + 2 NS NM, K + 3 NM]. Therefore, 43c10 + 9 × 43c9 + 24 × 43c8 + 20 × 43c7 = 11,117,319,330.
*NS = not K, NM = no marriage
[Addendum] This solver could have won save an arithmetic error in his original calculation:
David Wu: [corrected] 49c10 + 3(48c9) - 6(47c8) - 6(46c7) + 9(45c6) + 3(44c5) - 4(43c4) = 11,117,319,330
This checks, but being less intuitive would tend to be error prone.
When I was in the U.S. Army in my late teens, pinochle was a popular card game, second only to poker. We played the partnership version using a double deck without nines, hence 80 cards (A-10-K-Q-J in each suit four times). My commanding officer, an astute player, recognized my card sense and took me under his wing to learn his bidding methods and become his partner. We soon played matches against other good pairs for substantial stakes, winning more often than not. When I later took a liking to bridge, Lt. Colonel Harris was disappointed, but we still played occasionally.
And finally, from pinochle being nine-less, to words from the mindless:
Charles Blair: I read that Helen Sobel, partnering Oswald Jacoby, bid a grand slam after his 5NT bid, and afterwards said, “Were you asking for a marriage?”
Cyrus Hettle: [Solving these] made a rough evening and a worse night. I’ll need a “coconut and lime” in the morning.
The Donald: What’s trump? Oh, never mind… it’s me.
© 2025 Richard Pavlicek
by Richard Pavlicek by Richard Pavlicek