There are exactly 8192 (213) possible suit holdings, from a void to AKQJT98765432. Surely you’ve held the majority of them, and probably all of the shorter lengths. For the extreme lengths, frequency drops fast, though most players have held a 9-card suit, some have held 10, and a lucky few 11. I am unaware of a 12-card suit, let alone 13, ever occurring in an official event, but they abound in fantasy problems and puzzles (for which I plead the Fifth). For more on the topic see Long Suit Story, but don’t let that interfere with your accounting job here.
Are you a suitable accountant? Test yourself or make your best guesses, but have no fear. You won’t be held accountable! Consider the following six categories, and which suit holdings fit in each. Assume every suit holding is in card-rank order (using T for 10) as in this Data File. For example, K83 would fit in every category except #4.
How many of the 8192 suit holdings fit in each category?
1. Odd number of cards 610 800 1024 1378 2380 4096
2. More odd ranks (KJ9753) than even 610 800 1024 1378 2380 4096
3. Exactly 3 high-card points 610 800 1024 1378 2380 4096
5. At least one straight flush 610 800 1024 1378 2380 4096
5. No adjacent card ranks 610 800 1024 1378 2380 4096
6. Digit string a prime number 610 800 1024 1378 2380 4096
*Solvers had to produce answers on their own. Multiple choice was only added for this writeup.
The object was to determine how many of the 8192 suit holdings (which include a void) fit in each of the six categories. Scoring is based on the cumulative difference (Error) between submitted answers and the correct answers, so the lower the better. Listed below are the four wizards and four others who were correct on at least four of the six categories. The majority scored well over 1000, and two had no answer correct. Ties are broken by date and time of entry.
An easy one to start, which almost everyone got right. The answer must be exactly half of the 8192 holdings, or 4096. One way to verify this is to consider that 13c0 = 13c13, 13c1 = 13c12, …, 13c6 = 13c7. Hence combinations of even lengths must produce the same total as odd lengths.
David Wu: For each odd-length holding, there exists an even-length holding with all the missing cards. Hence just cut 8192 in half.
Tina Denlee: The function “cards you do not have” is a bijective mapping between odd-length and even-length holdings.
Dan Baker: It is easily proved that the number of odd-size and even-size subsets of any non-empty set must be equal.
Richard Stein: This must be equal to its negative space (cards not in holding).
Charles Blair: Suspiciously easy.
Tina Denlee: For each available choice of the numbers of odd and even ranks, multiply. Then add.
Dan Baker: 6×1 + 15×(1+7) + 20×(1+7+21) + 15×(1+7+21+35) + 6×(1+7+21+35+35) + 1×(1+7+21+35+35+21) = 2380.
Richard Stein: Choose from odd ranks, and sum over choices from even ranks that total less.
The above synopses are correct but cryptic. This table by suit length is more understandable:
There are two ways to hold 3 HCP: King plus any combination of T98765432 (including none), or queen-jack plus the same. In either case there are 29 = 512 combinations, so the answer is 512 + 512 = 1024.
Venk Natarajan: Two ways: K or QJ with any combination of 2-10. Each one can happen 512 ways, so 1024.
Dan Baker: King or queen-jack plus any combination of the nine spots, so 2 × 512.
Charles Blair: 29 + 29 = 1024.
Looking at it from a different angle, this solver found a simpler solution:
Tina Denlee: There are 16 ways to hold high cards, two of which total 3 HCP, so the answer is 2/16 of the 8192 holdings.
I took the easy route, and perhaps others remembered too. Seven years ago I wrote Hits and Misses about the occurrence of various anomalies in a bridge hand, one of which was straight flushes. By computer examination, I found exactly 1378 suit holdings with at least one straight flush.
Dan Baker: There are 10 possible straight flushes, but have to avoid double-counting suits with multiple. My approach was to count holdings with AKQJT, then KQJT9 without A, QJT98 without K, etc., then remove those with two separated straight flushes like AKQJT87654 and AKQJT5432.
Tina Denlee: Don’t forget 5432A!
Evidently quite a few did, as the answer 1262 stems from five-high straight flushes being ignored.
Richard Stein: I used a simple inclusion-exclusion count by [iteration]. I hope it’s right.
We Richards count alike.
Here again I went the easy way: In my March 2016 puzzle The Seven Percent Solution, I generated a List by computer of the 610 suit holdings without touching cards (which includes a void). A mathematical solution eluded me, so I was duly impressed by:
Dan Baker: Let f(N, K) = number of such holdings of N cards chosen from K. Each holding either includes N of the lowest K-1, or the highest and N-1 of the lowest K-2. So f(N, K) = f(N, K-1) + f(N-1, K-2) with base cases f(0, K) = 1, f(1, K) = K, and f(N, 0) = 0. Sum f(N, 13) to get 610.
Neat, and I almost understand it.
Richard Stein: My intuition tells me this should be a Fibonacci number, so I believe my code is good, finding 610.
Tina Denlee: [Unlike the straight-flush category] ace and two are not adjacent. We are looking for Fibonacci(15).
Fibonacci it is. Tina recalls my What’s the Score? puzzle, where I asked which duplicate bridge score is a Fibonacci number. Coincidentally, that answer was also 610!
Besides single-digit primes 2, 3, 5, 7 and the lone 97, the lowest card must be a three to have any chance of being prime. Routine primality tests revealed 43, 53, 73, 83, 643, 653, 743, 853, 863, 953, 983, 7643, 8543, 8753, 9643, 9743, 76543, 87643, 98543 and 986543. Thus there are 25 primes, and each can be preceded by any combination of AKQJT (including none) which is 25; hence 25 × 32 = 800 suit holdings.
Dan Baker: Possible primes are [all 25 listed]. Add any combination of AKQJT to get 25 × 32.
Tina Denlee: Two sequences end with a 7; 21 with a 3; and don’t forget 2 and 5. Total is 25, and there are 32 ways to combine the honor cards.
Richard Stein: Assuming that a ten is T and not 10, primes are [all 25 listed].
Charles Blair: Thanks to Maxima for the ‘primep’ function.
Out of curiosity, and to keep up with the times, I posed these six problems to ChatGPT and Google AI (or whatever they call themselves lately). Not surprisingly both got the three easiest ones (1-3) right, and each got one other, but my expectations were upheld by their gross misses of the other two. For the record, by my scoring method ChatGPT had a respectable 441, easily beating Google with a dismal 1801. Artificial intelligence certainly has improved since this debacle a few years ago, but it’s still missing some neurons.
So what else is new. Real intelligence can have neurons missing too:
Charles Blair: Thanks for the ESH data file, but its name puzzles me.
A detective you are not. Sherlock Holmes could deduce “Every Suit Holding” in his sleep.
Cyrus Hettle: I assume suits are written high-to-low. In a world where the majority of people are left-handed,or card indices are on the top right and bottom left, would we write suits low-to-high?
Good question. The queen at right would probably know. If not, ask the nutty one below.
Tina Denlee: This reminded me of a math championship a few years ago. One question was: “Pick cards from a deckuntil you pick five spades. On average, how many cards have you picked?” This failed half the candidates.
I think the key to the solution is the linearity of expected non-spades before each spade, but my neurons are telling me it’s bedtime.Thanks for “half the candidates” as a segue to my August puzzle.
© 2026 Richard Pavlicek
by Richard Pavlicek by Richard Pavlicek