Main     Puzzle 8M97 by Richard Pavlicek    

Doubled in Spades

“Professor! You missed an incredible game at Jourdan’s last night. In the first two rounds, Grover and I were doubled in spades on every board! The way Grover bids, that’s usually our opponents’ best strategy, but this time it backfired. Amazingly, we made every one! Plus over four thousand on six boards, 4360 to be exact. Have you ever seen anything like that?”

“Can’t say that I have, Timothy. I’ve certainly been plus more than that, but six consecutive contracts in the same suit is remarkable. Did Jourdan’s have a big crowd?”

“No, just nine tables but a perfect Mitchell. Even more bizarre is that we played every spade contract from one to six, and each score was higher than the previous one — like a progressive blitzkrieg.”

“That is extraordinary; but even so, I’m betting you and Grover didn’t win.”

“[Sigh] Good bet. After the opening streak, every spade contract went sour, and we barely broke average.”

“Predictable. Accelerated collisions cause supercharged pions to decay rapidly into gamma rays, which, according to my theory, cause boson deflection to become unstable. This transforms the higher-massed spade quarks into antimatter, rendering the bulk of those contracts unplayable.”

“You took the words right out of my mouth. Or at least ‘antimatter’ explains Grover pretty well.”

What pair number were Timothy and Grover?

Clarifications: Each spade contract (1 S to 6 S) was doubled but not redoubled. Overtricks may or may not have been made (no restrictions). Only the six scores (not necessarily the contract levels) were in ascending order.

Try it now

Make your best guess and click Verify to find out what, if anything, is wrong. Use the help provided to make corrections and repeat. See how many tries it takes you to solve the puzzle.

Timothy and Grover were pair number
 
Based on the same occurrence and movement, what is the
maximum possible total score any pair could achieve?
 
(Tiebreaker) Your answer        

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Nicholas Greer Wins!

In August 2016 this puzzle was presented as a challenge — with no help provided — inviting anyone who wished to submit a solution. Participation was the third most in this series with 59 persons giving it a try. Only the 20 listed below correctly deduced the pair number of Timothy and Grover.

Trending on the recent Olympics, our gold medal goes to Nicholas Greer, who was the first of 10 to be exact on the tiebreaker. This is Nick’s first win, though it’s hardly a surprise with his high finish almost every month, including a second, two thirds, a fourth, two fifths, two sixths and two sevenths. The silver medal goes to Dan Gheorghiu, our second most prolific winner, and the bronze to our all-time leader, Timothy “sans Grover” Broeken. As they take their positions on the podium, all please rise for “God Save the Queen.”

RankNameLocationPair NumberMax Score
1Nicholas GreerEngland3 NS7410
2Dan GheorghiuBritish Columbia3 NS7410
3Tim BroekenNetherlands3 NS7410
4Sherman YuenSingapore3 NS7410
5Tom SlaterEngland3 NS7410
6Charles BlairIllinois3 NS7410
7Wayne SomervilleNorthern Ireland3 NS7410
8Ivan LoyHong Kong3 NS7410
9Sami MakiFinland3 NS7410
10Jon GreimanIllinois3 NS7410
11Jamie PearsonOntario3 NS7560
12Dennis StuurmanNetherlands3 NS7560
13Hendrik NigulEstonia3 NS6910
14Joseph DiMuroCalifornia3 NS6860
15Dan BakerTexas3 NS6860
16Tina DenleeQuebec3 NS6810
17Richard MorseEngland3 NS8060
18Mike WenbleEngland3 NS6460
19Mike FrentzCalifornia3 NS5860
20Greg GouillartMassachusetts3 NS4960

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Solution

The key to solving this puzzle is to consider the vulnerability sequence faced by each pair in their first six boards. In a Mitchell movement, North-South pairs play board numbers in order, while East-West pairs skip a set each round. The following table shows the first six boards played by each pair and the corresponding vulnerability of their side. (Opposing vulnerability doesn’t matter with all contracts making.)

  Pair  BoardsVul SequencePattern
1 NS1 2 3 4 5 6N V N V V Nunique
2 NS4 5 6 7 8 9V V N V N Nunique
3 NS7 8 9 10 11 12V N N V N Vunique
4 NS10 11 12 13 14 15V N V V N Vunique
5 NS13 14 15 16 17 18V N V N N Vunique
6 NS16 17 18 19 20 21N N V N V Vunique
7 NS19 20 21 22 23 24N V V N V Nunique
8 NS22 23 24 25 26 27N V N N V Nsame 9 NS
9 NS25 26 27 1 2 3N V N N V Nsame 8 NS
1 EW1 2 3 7 8 9N N V V N Vunique
2 EW4 5 6 10 11 12V N V V N Nsame 3 EW
3 EW7 8 9 13 14 15V N V V N Nsame 2 EW
4 EW10 11 12 16 17 18V N N V N Nunique
5 EW13 14 15 19 20 21V N N V V Nsame 6 EW
6 EW16 17 18 22 23 24V N N V V Nsame 5 EW
7 EW19 20 21 25 26 27V V N V V Nunique
8 EW22 23 24 1 2 3V V N N N Vunique
9 EW25 26 27 4 5 6V V N V N Vunique

The six grayed out pairs are unlikely candidates, because their vulnerability sequence is not unique. For example, if Pair 8 NS were able to score 4360, so also could Pair 9 NS; hence the puzzle would have two solutions.

Next organize all the possible results, as in the table below. Each cell shows the two possible scores (nonvul and vul) for that result. Note that 1 S× making one is a special case, yielding the same score vulnerable or not.

ContractMade 1Made 2Made 3Made 4Made 5Made 6Made 7
1 S×160 160260 360360 560460 660560 960660 1160760 1360
2 S×470 670570 870670 1070770 1270870 1470970 1670
3 S×530 730630 930730 1130830 1330930 1530
4 S×590 790690 990790 1190890 1390
5 S×650 850750 1050850 1250
6 S×1210 16601310 1860

One observation is noteworthy: The 10s digit of all scores in each contract from 1 S× to 5 S× is the same, so five of the six scores must end in 60, 70, 30, 90 and 50, which sum to zero (modulo 100). Therefore, to reach a total ending in 60, 6 S× must have been made vulnerable for 1660 or 1860.

Mike Wenble: The aggregate score constrains the 6 S× contract to be vulnerable. Lots of valid combinations sum to 4360, but most require three or four consecutive nonvul hands which is not possible.

Tom Slater: The [60 ending] gives us the vulnerability and position of the small slam, then it’s just a matter of trying some combinations.

Richard Morse: The slam has to be vulnerable to have an answer ending in 60, and the rest is constrained by the rather low aggregate.

Determining all the ways to get 4360 would be tedious by hand, but a breeze by computer, which found 94 different ways. The actual scores, however, don’t matter. All that matters is the vulnerability sequence of the six contracts when the scores are placed in ascending order. In that regard there are just six cases:

CaseWaysVul Sequence  Vuls  
139N N N N N V1
213N N N N V V2
33N N N V N V2
427V N N N N V2
59V N N N V V3
63V N N V N V3

Case 1 can be eliminated since it has only one vulnerable contract, while the minimum faced by any pair is two. All that remains is to compare Cases 2-6 with the vulnerability sequence met by each pair. Lo and behold, there is just one match: Case 6 matches Pair 3 NS. Hello, Timothy and Grover! The three ways they could have scored 4360 are listed below, each having the same first and last result. Vulnerable results are tinted red.

Board 7Board 8Board 9Board 10Board 11Board 12Total
1 S×= 1602 S×= 4704 S×= 5903 S×= 7305 S×+1 7506 S×= 16604360
1 S×= 1603 S×= 5304 S×= 5902 S×= 6705 S×+1 7506 S×= 16604360
1 S×= 1603 S×= 5305 S×= 6502 S×= 6704 S×+1 6906 S×= 16604360

Sherman Yuen: Timothy and Grover were Pair 3 NS. Results were 1 S×= 160, 3 S×= 530, 4 S×= 590, 2 S×= 670, 5 S×+1 750 and 6 S×= 1660.

Nicholas Greer: The double of 5 S on Board 11 may have been a stripe-tailed ape double, but with the theme of this puzzle I’ll assume not, [lest] it have been the oppponents’ best strategy.

Mike Frentz: Last deal must be 6 S×= 1660 (vul)… consistent with seating for NS(3,4,5,6) and EW(1,8,9). Minimum remaining is 2400 (160+470+530+590+650), leaving 300 in vulnerability-overtrick bonuses. Board 1 must be 1 S×, as minimum score for others is 470… At most one of boards 2-5 can be vulnerable, leaving only NS(3,5) and EW(8). Board 1 score must be 160, as 260 makes it impossible for the vulnerable board to come in the correct place. Only solution is NS Pair 3.

Dan Baker: Trick scores, doubled insults, one partscore bonus, five nonvul game bonuses, and nonvul slam bonus add to 3610. Only way to reach a total ending in 60 is to make the slam vulnerable, which gets them to 4060. [Dan similarly deduces Pair 3 NS].

Going for the record

The 4360 scored by Timothy and Grover was a modest total for the occurrence (minimum possible is 3910). For the tiebreaker, the goal was to find the maximum. Choosing the best score for each contract yields a hefty sum of 9060, but it doesn’t wash; all those scores are vulnerable, and no pair is vulnerable on all six boards. For this movement the most vulnerable deals of any pair is four, specifically pairs 4 NS, 7 EW and 9 EW.

Since scores must be ascending, it is better to have vulnerable deals toward the end of the six boards. Pair 4 NS (V N V V N V) seems the best candidate, since pairs 7 and 9 EW each have two vulnerable deals at the start. Working from the high end, a little tinkering shows that 1670 must be the top score so that 1310 (best nonvul by far) can be next, followed by 1250 and 1190 (next highest vuls) then 970 and 960 (best fitting nonvul and vul) for a total of 7210. But can we do better?

Dan Gheorghiu: The Professor’s boson deflection theory was in effect, as I could not find a suitable way to [formulate] the optimal score. Therefore, I turned to intuition and trial and error. Including the highest scores on vulnerable boards and the highest nonvulnerable score (1310) became maximum at boards 16-21.

Yes, even better than being vulnerable four times, is to be vulnerable on the last two deals. The only pair to claim that is 6 NS (N N V N V V) and they alone can achieve the maximum possible aggregate score of 7410. This can be done in three ways as shown below. Note that each result except one (1 S×+5) contains the most possible overtricks (making seven) so it’s hardly realistic, but then, puzzles seldom are.

Board 16Board 17Board 18Board 19Board 20Board 21Total
5 S×+2 8504 S×+3 8901 S×+5 11606 S×+1 13103 S×+4 15302 S×+5 16707410
4 S×+3 8903 S×+4 9305 S×+2 12506 S×+1 13101 S×+6 13602 S×+5 16707410
1 S×+6 7604 S×+3 8905 S×+2 12506 S×+1 13103 S×+4 15302 S×+5 16707410

Tom Slater: Only Pair 6 NS can get the greatest score, being the only pair to finish N-V-V.

Nicholas Greer: The maximum score is achieved by Pair 6 NS making 13 tricks every time in 1, 4, 5, 6, 3 then 2 S×.

Sherman Yuen: Pair 6 NS can score 7410 with results of 5 S×+2 850, 4 S×+3 890, 1 S×+5 1160, 6 S×+1 1310, 3 S×+4 1530 and 2 S×+5 1670.

Double trouble

Ivan Loy: I will never double another spade contract! Why take the risk when 4 S can be cold with zero HCP in both hands?

Wayne Somerville: Everyone at the club seems concerned that I supply the scores for making doubled contracts too fast.

Charles Blair: “If you set every contract you double, …”

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© 2016 Richard Pavlicek