Main Puzzle 8M67 by Richard Pavlicek
I know, my dear Watson, that 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 Moriartys 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 problems, though this one may be a three-piper. Evidently the percent chances of the three bridge hands, rounded to decimal hundredths to obtain three digits each, identify the nine numbers in the safes combination. Further, the total of the percentages is seven, 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.
Clarifications: Each of the three problems is separate and presumes one bridge hand (13 cards) is randomly dealt from a full deck. Controls are counted: Ace = 2, King = 1. Touching means adjacent in rank, e.g., A-K are touching but A-Q are not. Poker hands consist of five cards and rank: straight flush, four-of-a-kind, full house, flush, straight, three-of-a-kind, two pair (nothing worse is possible).
Can you find the percent chance (to decimal hundredths) for each of the three bridge hands? And you can save yourself some effort with my assurance that all three add to 7.00 percent.
In 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 Id better hold off on the vodka: He was a member of the Slovakian youngsters team in the 2014 World Youth Championships (Istanbul).
|13||Gary Leung||Hong Kong||0.00|
|Puzzle in a Puzzle: Why must the error sum always end in an even digit? (solution at end)|
This 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 doesnt work that way. If you state that you have one seven and actually have two, it would be a lie by grammatical standards. As an analogy, suppose partner uses Blackwood and you show one ace. Would he be pleased when you put down the dummy with two?|
Problem 2 was more difficult, in that no formula exists, other than a humongous summation suitable only for computer. 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. 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 below, multiplied by the number of permutations, and summed all to find 16,264,952,268 qualifying hands. Dividing by the number of bridge hands gave ~0.02561355, or ~2.56 percent.
|Void1 4-card suit210|
|Singleton13 5-card suit126|
|Doubleton66 6-card suit28|
|Tripleton165 7-card suit1|
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: Couldnt 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. Then 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.
|Void1 4-card suit715 8-card suit916|
|Singleton13 5-card suit1277 9-card suit331|
|Doubleton78 6-card suit1645 10-card suit52|
|Tripleton286 7-card suit1499 11-card suit1|
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 didnt 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 Ill trust you for Problem 3 to get the 7.00 percent sum, although my calculations total 7.03 percent. I dont know where the excess 191 million hands come from.
No problema! Whats 191 million hands among friends?
|Puzzle in 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, so the absolute sum must again be even.|
Charles Blair: Since Harkers 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 youd ever throw a puzzle like this our way. Really original and a great change of pace!
Thanks, although the word original might have Mr. Doyle rolling over in his grave.
Sherlock Holmes: So you see, Watson, the numbers in the safes 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 its almost a sure bet that 256-106-338 will open the safe.
Acknowledgments to Sir Arthur Conan Doyle (1859-1930)
© 2016 Richard Pavlicek