Puzzle 8M67 Main
The Seven Percent Solution
by Richard PavlicekPuzzle 8M67 Main |
| by Richard Pavlicek |
“My dear Watson, I know you share my love of all that is bizarre and outside the conventions of everyday bridge.”
“I say, Holmes… I thought you had given up bridge.”
“I had, until last night, when I discovered a wall safe in the basement of Moriarty’s flat behind some loose bricks. I knew the model, a rotary dial requiring nine single-digit numbers, but the only clue to its combination was a conundrum attached to the same brick wall. Apparently the evil mastermind has been calculating bridge probabilities — and bizarre ones at that.”
Bridge hands three, percent to be:
1. Seven controls and one seven
2. No touching cards in any suit
3. Best poker hand four-of-a-kind
For the 7% solution, sum all three
“Mumbo jumbo. Makes no sense.”
“You know my methods, Watson. I thrive on statistical challenges, though this one may be a three-piper. Evidently the percent chances of the three bridge hands, rounded to decimal hundredths to produce three digits each, identify the nine numbers in the safe’s combination. Further, the total is seven percent, which reduces the task forthwith, since solving any two and subtracting from 7.00 will reveal the third.”
“Absolutely brilliant, Holmes. You astound me.”
“Elementary school arithmetic, my dear Watson.”
Find the percent chance (to decimal hundredths) for each of the three bridge hands.
Clarifications: Each of the three problems is separate and presumes a bridge hand (13 cards) is randomly dealt. “Controls” are counted: Ace = 2, King = 1. “Touching” means adjacent in rank, e.g., 
A-K are touching while A-Q are not. “Poker hands” consist of five cards (chosen from 13) that produce the highest ranking hand by poker rules. All three answers add to 7.00 percent.
Martin Vodicka WinsIn March 2016 this puzzle was presented as a challenge, inviting anyone who wished to submit a solution. Participation was healthy, with 48 persons (second most in this series) giving it a try, and the 21 best solvers are listed below. My arbitrary cutoff at 0.10 percent error sum was chosen because of a wide gap to the next.
Anyone for a vodka martini? Close enough! Congratulations to Martin Vodicka, Slovakia, who was the first of 14 to submit perfect solutions, right on the mark for all three problems. Martin is a new participant, and a quick research shows I’d better hold off on the vodka: He was a member of the Slovakian “youngsters” team in the 2014 World Youth Championships (Istanbul).
| Rank | Name | Location | Error Sum |
|---|---|---|---|
| 1 | Martin Vodicka | Slovakia | 0.00 |
| 2 | Tina Denlee | Quebec | 0.00 |
| 3 | Jamie Pearson | Ontario | 0.00 |
| 4 | Dan Baker | Texas | 0.00 |
| 5 | Charles Blair | Illinois | 0.00 |
| 6 | Gareth Birdsall | England | 0.00 |
| 7 | Jon Greiman | Illinois | 0.00 |
| 8 | Tim Broeken | Netherlands | 0.00 |
| 9 | Grant Peacock | Maryland | 0.00 |
| 10 | Hendrik Nigul | Estonia | 0.00 |
| 11 | Tom Slater | England | 0.00 |
| 12 | Bruce Zastera | Illinois | 0.00 |
| 13 | Gary Leung | Hong Kong | 0.00 |
| 14 | Jean-Christophe Clement | France | 0.00 |
| 15 | Jean-Baptiste Courtois | Belgium | 0.02 |
| 16 | Eric Lanier | Wisconsin | 0.02 |
| 17 | David Brooks | Australia | 0.02 |
| 18 | Jonathan Mestel | England | 0.04 |
| 19 | Stan Zhang | California | 0.06 |
| 20 | Michael Selby | California | 0.10 |
| 21 | Baptiste Couet | France | 0.10 |
Puzzle within a Puzzle: Why must the error sum always end in an even digit? (solution at end)
| Puzzle 8M67 Main | | Top The Seven Percent Solution |
SolutionThis puzzle was a change of pace, more mathematical than bridge oriented, and not even containing a bridge diagram. Based on participation and comments, it seemed well received, and the plethora of perfect solutions was impressive. Keen statisticians out there! I may do something similar in the future.
Several respondents asked whether computer-aided solutions were acceptable and/or considered ethical. Absolutely! The intent of my puzzles is to provide a mental challenge (hopefully enjoyment too) and how people pursue this is unimportant. Computer programming is certainly a mental challenge, and many, like me, find it an enjoyable pastime. The only unethical practice would be not to determine the solution yourself.
Problem 1 is a direct calculation. “Seven controls and one seven” can occur in two forms: (1) three aces, one king and one seven, or (2) two aces, three kings and one seven. For Form 1 there are four ways to select three aces, four ways to select a king, four ways to select a seven, and 40c8* ways to select eight other cards (non-aces, non-kings, non-sevens) so it yields 4 × 4 × 4 × 40c8 = 4,921,899,840 hands.
*combinatorial notation for “40 choose 8” (number of ways to select 8 items from 40 items)
For Form 2 there are six ways to select two aces, four ways to select three kings, four ways to select a seven, and 40c7 ways to select seven other cards, so it yields 6 × 4 × 4 × 40c7 = 1,789,781,760 hands. Add Forms 1 and 2 (6,711,681,600) then divide by the number of bridge hands (635,013,599,600) to get ~0.01056935, or ~1.06 percent.
Dan Baker: Three aces (4 ways), 1 king (4 ways), 1 seven (4 ways), and 8 other cards (40c8); or 2 aces (6 ways), 3 kings (4 ways), 1 seven (4 ways) and 7 other cards (40c7).
Tom Slater: Hopefully straightforward as 4c3 × 4c1 × 4c1 × 40c8 + 4c2 × 4c3 × 4c1 × 40c7.
A few respondents apparently misinterpreted my hand descriptions as Boolean statements. For example, a hand with two sevens would test true for having “one seven.” Unfortunately the English language doesn’t work that way. If you state that you have one seven and actually have two, you would be lying by grammatical standards. As a bridge analogy, suppose partner used Blackwood and you showed one ace. Would he be pleased if you put down the dummy with two?
Problem 2 was more difficult, in that no formula exists, other than some humongous summation that only a computer could solve. My approach was to start with the 8192 (213) possible suit holdings and remove all those with touching cards, leaving the 610 listed here and summarized below by suit length.
| Void | 1 | 4-card suit | 210 |
| Singleton | 13 | 5-card suit | 126 |
| Doubleton | 66 | 6-card suit | 28 |
| Tripleton | 165 | 7-card suit | 1 |
Then for each of the 23 generic hand patterns with at most a 7-card suit, I found the number of combinations per the table, multiplied by the number of permutations, and summed them all to find 16,264,952,268 qualifying hands. Dividing by the number of bridge hands gave ~0.02561355, or ~2.56 percent.
Dan Baker: This can be found by counting the possibilities for each suit length 1-7 (via a recursive function) then combining them for each possible hand pattern, giving a total of 16,264,952,268 hands.
Tom Slater: Couldn’t see a quick way to do this, so I crunched the sequence 1-13-66-165-210-126-28-1 for individual suit combinations and applied to all distributions up to 7-card suits.
Jamie Pearson: Problem 2 was my favorite, basically boiling down to: How many ways can you pick X cards in a suit with none touching for all X? I think the easiest approach is to partition 13-X into the number of gaps required across three cases (X-1, X, or X+1 gaps depending on whether 0, 1, or 2 of the ace and deuce are selected).
Problem 3 was the hardest. Counting the number of hands with four-of-a-kind is difficult enough, but it was also necessary to exclude hands with a straight flush in any suit. I took a similar approach as with Problem 2, starting with the 8192 suit holdings and removing those with a straight flush, leaving 6814 listed here and summarized below by suit length.
| Void | 1 | 4-card suit | 715 | 8-card suit | 916 |
| Singleton | 13 | 5-card suit | 1277 | 9-card suit | 331 |
| Doubleton | 78 | 6-card suit | 1645 | 10-card suit | 52 |
| Tripleton | 286 | 7-card suit | 1499 | 11-card suit | 1 |
Next I computer-checked each of the 18 generic hand patterns without a void for all possible combinations and permutations to find 21,490,969,152 hands with four-of-a-kind. Dividing by the number of bridge hands gave ~0.03384332, or ~3.38 percent.
A number of respondents were slightly off on this problem because they failed to exclude the low-ball straight flush (5-4-3-2-A), no doubt influenced by the ace always being high in bridge. Some followed up with corrections, like our previous winner:
Grant Peacock: As you may have noticed, I wrote an entire C program to solve this that fits in your comment field, but it had a slight bug, in that it didn’t exclude hands with a 5-4-3-2-A straight flush. So I think 21,490,969,152 hands (3.38 percent) is my final answer.
Tina Denlee: I found the exact number of suitable hands for Problems 1 and 2, and I’ll trust you for Problem 3 to get the 7.00 percent sum, although my calculations total 7.03 percent. I don’t know where the excess 191 million hands come from.
No worries! What’s 191 million hands among friends?
Puzzle within a Puzzle (solution). With three answers forced to total 7.00 it is impossible to get exactly two right. With one right the two errors must be equal but opposite signed, and the total of two equal values must be even. With none right the signed sum of the three errors must be zero, which requires either three even numbers or one even and two odd numbers, hence the absolute sum must again be even.
Double, Double, Doyle and TroubleCharles Blair: “Since Harker’s bust was one in three, the chances were exactly as I told you.” The Adventure of the Six Napoleons -Arthur Conan Doyle
Jamie Pearson: Given all the probability analysis that appears on your site, I was wondering if you’d ever throw a puzzle like this our way. Really original and a great change of pace!
Thanks, although “original” might be an overbid… [Mr. Doyle rolls over in his grave]
Sherlock Holmes: So you see, Watson, the numbers in the safe’s combination are 106, 256 and 338, though I suspect not in that order. The digits in each group total 7, 13 and 14, so Moriarty would arrange them: 13 (prime and Fibonacci), 7 (prime only) and 14, so it’s almost a sure bet that 256-106-338 will open the safe.
| Puzzle 8M67 Main | | Top The Seven Percent Solution |
Acknowledgments to Sir Arthur Conan Doyle (1859-1930)
© 2016 Richard Pavlicek