6/17/11

MGWCC #159 -- Friday, June 17th, 2011 -- "Going Shopping"

Good afternoon, crossword fans -- welcome to Week 159 of my contest. If you're new to the contest and would like to enter, please see the site FAQ on the left sidebar for instructions.

LAST WEEK'S RESULTS:

Toughish and somewhat inelegant Week 2 meta: 198 solvers found the contest answer, which is on the low end of normal for a second week, but about a quarter of those guessed well or didn't fully grok what was going on.

The five theme entries consisted of four playing card spoonerisms, plus one nudge towards the meta. The spoonerisms were:

SPACE OF AIDES A♠
HEN OF TARTS 10♥
DEVON OF SIMONS 7♦
CLICKS OF SUBS 6♣

The first meta nudge came at 44-a: {Magician's secret, or what you need to find in the grid} yielded HIDDEN CARD.

Two more nudges in the clues told solvers that they didn't need the hidden card's SUIT (55-d) and that an ACE should count as 13 for the meta (59-d). Solvers were then supposed to take the four card values used in the theme answers (6,7,10 and 13 for the ace, as instructed), then find those four numbered boxes on the top row -- and lo, they indeed spell out the contest answer, JACK. And, it should be said, the large majority of solvers who submitted JACK did just this.

But here's where the meta inelegance enters the picture: why, many solvers wondered, was an ace to be counted as 13? In games where the ace has a numeric value it's usually 1 or an 11, as in blackjack, or 15 (as in gin rummy). If you count from 10, then a jack would be 11, a queen 12, a king 13, and an ace 14. There's no well-known game (that I'm aware of) where an ace is 13. This oddity threw some solvers off, like Joel Alderson:

Why not just count the ace as 11, like blackjack?


This is an especially easy solution since there already happens to be a K at the 11 square (the first K in HICKOK).

So how did that happen? Truth is, I just miscounted. I play a lot of cards but while making this puzzle the number 13 was in my mind for an ace and I never challenged it. A solver pointed out that I must have been thinking that there are 13 cards in a suit and the ace is the highest, therefore A=13.

So I could have fixed the entire meta by simply having the clue at 59-d reference 11 instead of 13, but Jay Giess found something even better -- an extremely clever idea I wish I had thought of:

Matt, Did you consider KITSCH for 1-down, FRAPPE for 2-down, and CPAS for 23-across? Granted it would mean CSO (Chicago Symphony Orchestra ?) for 21-down, but then the ace could be 1,11, or 13!

Beautiful! Kicking myself for not noticing this possibility, especially since the grid fix would have been so simple.

The one saving grace of this inelegance is that it didn't really affect the play of the meta, just left those solvers who noticed it scratching their heads a bit and unsure that JACK was correct. But considering both the ace=13 problem and missing Jay's lovely meta improvement, I'm glad this puzzle escaped with a 4-star rating at Crossword Fiend!

Joel Horn noted a coincidence:

Cute to follow the B-52s puzzle with a cards puzzle. What's next--a puzzle about weeks in a year?

This week's winner, whose name was chosen at random from the 198 correct entries received, is Larry Spencer of Ocala, Fla. Larry has selected as his prize an autographed copy of Sip & Solve Hard Crosswords.


TIP JAR HAUL:


The MGWCC tip jar overflowed last week as 184 solvers chipped in $5,386.84. That's almost double last year's total ($2,875), a trend I can certainly live with. Big thanks to all who contributed -- you paid for Portugal with quite a few euros left over!

THIS WEEK'S INSTRUCTIONS:


This week's contest answer is the opposite of the five-letter word referenced by the theme entries. E-mail it to me at crosswordcontest@gmail.com by Tuesday at noon ET. Please put the contest answer word in the subject line of your e-mail.

To print the puzzle out, click on the image below and hit "print" on your browser. To solve using Across Lite either solve on the applet below or download the free software here, then join the Google Group (1,523 members now!) here. To solve with friends at Team Crossword, click here.





Solve well, and be not led astray by words intended to deceive.

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