Friday, October 31, 2014

Certain Glory Marks Martin Gardner's Centennial

Card Colm started in October 2004, the first one ("Low Down Triple Dealing") marking the 90th birthday of Martin Gardner (1914–2010). With this one, we therefore complete a decade's worth of explorations of the kind of fun to be had when mathematical principles and a deck of cards get together.

We also salute an extremely prolific writer and top notch communicator of mathematics, with something brand new—courtesy of an old friend of his—that we are confident Martin would have found quite delightful.

What follows also makes a great Halloween Trick and Treat.

In the past month or so, fans new and old of Martin's written legacy have been attending and hosting Celebration of Mind events, all over the world, to mark the 21 October 2014 centennial of his birth. More events will take place from now till the end of the year.

Why not grab a deck of cards and a few receptive people, and try out out the first effect below, or one of the dozens from earlier? (The period 2004–2011 is documented here.) Please register here so we know where and when Martin is being honored.

Donkey Wontons

First, pull out any Ace (value 1), 2, 3, 4, 5 from a deck of cards, then turn away, and ask a random spectator to select one of those cards and show it to the rest of the audience, before replacing it. Turn back. For the sake of argument, assume these are the cards used. The rest of the deck is set aside. The suits play no role in what follows.

Now, place some of those five cards face down on the table, make a true statement about each one, and ask the spectator if the card picked is one of those.

Next, gather up those cards, recombine them with the others, and place a second set of some of the five face down on the table. Again, make a true statement about each of them, asking if the card picked is one of those now on the table.

Should it be possible to tell which card was picked, if both questions were answered truthfully? First instincts suggest that two Yes/No answers can distinguish any one card from four, but perhaps not from five. However, one needs to consider the kinds of statements made about the cards placed on the table, as well as how many cards are involved each time.

For instance, imagine that two cards are placed on the table the first time, and it's correctly claimed, "The first is even-valued, and the second is 4 or higher." When the spectator is asked if the selected card is present, the answer is an unambiguous "No" provided that the selected card was the Ace or 3. However, if that card was the 2, 4 or 5, the spectator doesn't know for sure, and the only honest answer is, "I Don't Know."

It seems that we need to broaden our horizons and accept three types of answer: Yes, No, and I Don't Know, the last being equivalent to Maybe. This means that, after asking the usual question about two displays of face-down subsets of the original five cards, about which certain statements were made, there are nine possible combinations of answers that could be truthfully given. Presumably this is sufficient to narrow which one of the five cards was the one selected, if appropriate cards were displayed and appropriate statements made.

At first glance, it's tempting to think that the "I Don't Know" or "Maybe" option yields no information, but that's not the case at all. In the example above, for instance, such an answer tells us that the selected card was definitely the 2, 3 or 5. From there, one more clever "display and declare" decision, which we henceforth refer to as a query, should narrow down the selection to a single card. Such adaptive querying, where the second one depends on the answer received the first time around, is indeed possible every time.

Let's be more ambitious, and seek an approach that uses two pre-determined queries that will be used in all cases.

In addition, let's relax the assumption that the spectator is guaranteed to be a truth teller. Perhaps the spectator is a compulsive liar instead. We have strayed into the realm of ternary logic, where we agree that the negation of Yes, No, and Maybe, are, respectively, No, Yes and again Maybe. (It's one of these latter answers that a liar would unhesitatingly give.)

This seems a tall order. We seek two queries which will be asked of a spectator, who could be either a consistent truth teller or a consistent liar, such that the answers received will permit the certain identification of the selected card.
This whole conundrum, and the surprisingly simple solution presented below, comes to us (and now to you) courtesy of mathematician and puzzle master Jerry Farrell, who was kind enough to share it as part of the Martin Gardner CentennialCelebrations.

Here are two queries that work in all cases. Remember all cards are displayed face down.
Query 1. The 5 is placed on the table, and beside it either the Ace or 3. The statement made reflects this perfectly "The first card is the 5 and the last one is also odd-valued."

Query 2. The 2 and 3 are placed on the table, followed by either the Ace or 5. The statement again reflects this reality: The first card is the 2, the next card is the 3, and the last one is odd-valued."
The ternary tree below tracks the possible answers to the two queries; note that the options are always taken in the order Yes, Maybe, No, from left to right. The best part about this scheme is that it works regardless of whether the spectator is a truth teller or liar.

(tree courtesy of John Miller of Geom-e-Tree)

For instance, either two Yes or two No answers lead to the same conclusion: the selected card was the 4. Similarly, either a Yes or a No the first time followed by a "Maybe" the second time means that the card was the 5. But there's more, as the "+" or "-" signs attached to those values reveal: in most cases we can also say whether the spectator was lying or not. Note that "+" denotes truth telling and "-" indicates lying. The bottom line is, well, pleasantly symmetric. Also, this tree is easily memorized, avoiding the need for a crib sheet in performance.

Only in the case where the selected card was the Ace—corresponding to getting the answer "Maybe" twice—are we unsure whether we've been dealing with a truth teller or a liar all along. In all other cases, that information is revealed automatically, and it's prudent to announce the card picked while slipping in a casual comment such as, "I'm glad you told the truth" or "Despite your lies, I figured out your card."

If the spectator is lying, that person will probably then reveal that by denying that the selected card has been identified. Since that card was shown around earlier, you have nothing to worry about, the audience will be on your side. Furthermore, when the card was the Ace, a simple announcement of that in the form of the question, "Your card was the Ace, right?" will force both truth tellers and liars alike to self-identify.

In performance, when a participating spectator is decided on at the outset, stress that he or she must either lie all the way, or be totally truthful. Also, carefully explain the three options for each question asked, as well as the convention that the negation of Maybe is still Maybe, whether people agree with that logic or not!

Exercise 1: rework the above example with more interesting cards, such as four of a kind and a random fifth card, or a high straight, or even a killer straight flush.

Lying Reactor

Now imagine that the spectator picks a card from a much larger set, such as the fourteen displayed below. This time, suits play a major role.

These can be stacked at the top of the deck, in any order, and kept there throughout some sloppy false shuffling, before being dealt face down to the table and really mixed up. Finally, one is peeked at (and shown around). This gives the illusion that the cards used are totally random, which makes the outcome below all the more miraculous.

It turns out that with three carefully planned queries, posed to a truth teller or liar under the same conditions before, enough information will be gained to determine the selected card.

Here are three queries that work in all cases, as suggested by Jerry Farrell.
Query 1. The 2D, 4H, 6H, 8D, and X are placed on the table, where X is any card from the highest straight. The statement made reflects this, naming the first four cards and saying that the last one has value 10 or greater (Ace is high here).

Query 2. The 2D, 7S, 9S, 10C, and QS are placed on the table, followed by any Heart not used already. The statement made names the first five cards, and indicates that the last one is a Heart.

Query 3. The 3C, 4H, 9S, and 10C are placed on the table, followed by any Red card not used already. The statement made names the first four cards, and says that the last one is Red.
The resulting three-level tree, corresponding to the two-level one shown earlier, accounts for all 27 possible combinations of answers YYY, YYM, YYN, YMY, ..., NNN. The bottom line there is also elegantly symmetric, as can be checked. Even more delightfully, it points to these cards, in the order listed:
Joker, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K, Ace, K, Q, J, 10, 9, 8, 7, 6, 5, 4, 3, 2, Joker
That is to say, three Yes or three No answers guarantee the Joker, whereas two Yes answers followed by one Maybe points to the 2, as do two No answers followed by one Maybe, and so on.

Exercise 2: Sketch the three-level tree that ties all of this together, and practice decoding the various possibilities for the spectator's answers.

Exercise 3: Investigate the distribution of "+" and "-" in the extended bottom line above, does it work out as neat and tidy as it did in the earlier two-query case?

Exercise 4: Can the above be modified to switch the roles of the Ace and Joker? Would this make the Joker selection the only case where the truthfulness of the spectator would have to be teased out as suggested earlier, rather than being revealed automatically by the three answers received?

Exercise 5: Take it to the next level: if four queries are allowed, then from how many cards could a spectator be invited to choose one? Is it possible to devise queries leading to another symmetric bottom line in the resulting level-four tree?

These wonderful creations of Jerry Farrell's tie together several of Martin's favorite themes in mathematics and magic: logic, the interplay between numbers and language, the use of ternary methods, the convivial/contrary (or knight/knave) bystander motif, and the freedom to choice between lying or telling the truth in interactions with the performer. We are very grateful to Jerry for graciously allowing us to share these with everyone, via this column. As usual, we couldn't resist throwing in some anagrams, in the section headers.

Martin was a lifelong fan of wordplay, and contributed frequently to Word Ways: The Journal of Recreational Linguistics, a quarterly magazine now edited by Jerry Farrell, which was "established in 1968 at the behest of Martin Gardner." The Journalof Recreational Mathematics was also started, coincidentally in 1968, as a result of a suggestion of Martin's.

Don't forget the wise words of Persi Diaconis, "Pick up anything Martin Gardner wrote. You'll smile and learn something."

Thanks for Dana Richards for essential clarifications, and Vicki Powers for additional insight.

And finally, thank you for reading some or all of our ruminations here over the past decade.

"Certain Glory" and "Lying Reactor" (like "Lying Creator") are anagrams of "Ternary Logic." "Donkey Wontons" is an anagram of "Yes, No, Don't Know." Finally, "True Thrill Alert" is an anagram of "Truth Teller, Liar" (as is "Ultra Three Trill").

Friday, August 29, 2014

Trim Hex Anagrams

As Martin Gardner fans new and old gear up to attend or host centennial Celebration of Mind events on or around 21 October 2014, including a special screening of long-lost video footage of the legendary mathematics popularizer—at the MAA Carriage House, Washington, DC—we add to the celebrations with another round of card considerations spawned by his always entertaining writing.

Between 1977 and 1986, Martin penned 111 "puzzle tales" for his friend Isaac Asimov's Science Fiction Magazine. They all resurfaced in three collections, the first and last of which are much more well known than the middle one. His words in the foreword to one of these books apply equally to all of them:
Each chapter poses a problem answered in the First Answers section. The solution in turn raises another problem that is solved in the Second Answers section. This may suggest a third question, and in several instances there is a fourth.... Many of the problems lead into nontrivial regions of mathematics and science.
As a result, a quick glance through any of these volumes soon suggests interesting things to explore with a deck of cards.

The first puzzle tale collection was Science Fiction Puzzle Tales (Potter, 1981), more familiar today in its MAA incarnation Mathematical Puzzle Tales from 2000. Just six months ago, "Postage Stamp Issue" (February 2014's Card Colm) drew inspiration from something found there.

Back in June 2005, "A Little ErdÃ¶s/Szekeres Magic" took advantage of an observation from the final collection Riddles of the Sphinx and Other Mathematical Puzzle Tales (MAA, 1987). That two-person card effect actually first appeared in "(Smells like)Team spirit" from Mathematical Card Tricks (AMS Online, October 2000).

This leaves the comparitively forgotten middle collection Puzzles From Other Worlds: Fantastical Brainteasers from Isaac Asimov's Science Fiction Magazine (Vintage, 1984), issued with two different covers, which includes several topics perfectly suited to card implementations. (What a wonderful MAA reissue this would make!) Below, we merely skim the surface of possibilities suggested by one of its puzzles, in a later iteration in the book.

Considered Yet Given

Consider the magic hexagram below:

(The image here is borrowed from Harvey Heinz's Magic 9x5 Hexagrams.) It appears on page 75 of Puzzles From Other Worlds, and unlike magic squares, is unfamiliar to most audiences. This pattern is doubly magic: each five-number horizontal, vertical, or diagonal row has a sum of 46, whereas each pair of opposite vertices sums to 7.

On page 165, Martin reveals that he was the first to find it, in response to a challenge issued by Harold Reiter in Mathematics Teacher in March 1983. He also found its "complement in 20"—obtained by subtracting each entry from 20—which has magic sum 54. The Heinz link above reveals that more have been discovered since, although far fewer than 57 varieties.

Note that 46 + 7 = 53, which is one more than the number of cards in a full deck. By taking advantage of an idea used in "A Magic Timepiece Influenced By Martin Gardner" (October 2007's Card Colm), we can put all of this to work.

Fan the faces of a jumbled-looking stack of 19 small index cards, on which the numbers 1 to 19 have been written, and deal them out face down on a large table into the above pattern. Turn away and ask for any five-card row or diagonal to be selected, and left in place. Have the remaining cards gathered up and set aside (the technical term for this is "destroying evidence").

Turn back. Quietly note to yourself if the selected row or diagonal includes the central 13 or not. If it doesn't, simply ask that the five index cards be turned over and their numbers added; the total will be 46. Indicate a deck of playing cards sitting on top of a folded page on the corner of the table, and have the person who selected the index cards count down to the 46th playing card. You wrote its value and suit on the page earlier before you folded it, so when the 46th card is arrived at, you can have the paper opened to confirm that it was totally predictable.

If the selected index cards include the central one, you can have even more fun. Pick them up, mix them a little, and hand them out. Ask for the total or the total of the two smallest numbers. (The mixing is to hide the fact that the two smallest numbers are always at each end of the row.) Either 46 or 7 will be reported. If 46, proceed as before, but if 7, hand over the 52-card deck face up and request that the 7th card from the deck face be located, as you mix in the other three index cards in with the rest. The card arrived at is the same as the 46th one from the other end, and hence will match the written prediction.

Here's a less cluttered magic hexagram, which is a type of magic star; it has magic constant 26 for sums of four values in a row.

(The image is from from Wikipedia.) This too can be put to magic use, in a predictable way, one might say. This time, the intial display could consist of any Ace to Queen from a deck, with the forced or predictable card being 26 card from the top of the rest of the deck.

Electric Solvents

The hexagrams above had their summands at vertices. Another type has it summands in its cells. Consider the (unique) magic hexagrams below:

(from Weisstein, Eric W. "Magic Hexagram," MathWorld–A Wolfram Web Resource http://mathworld.wolfram.com/MagicHexagram.html)

Each (multicell) row or diagonal in the hexagram on the left sums to 33, and each corresponding row or diagonal in the hexagram on the right sums to 32. Note that these hexagrams are complementary: given one, the other may be obtained by subtracting its entries from 13. Using an idea implemented in "Sixy Alpha Omegas" (August 2007's Card Colm), we can fudge the 32/33 coincidence to magical effect as follows.

Draw two large blank hexagrams of the type just displayed, in which each cell is large enough to house a playing card. Take out a black-backed deck and show the faces of the cards to demonstrate how "random" they are. Deal out face-down cards from left to right into the rows of one of the hexagram, working from top to bottom, to match the left hexagram depicted above. The deck has of course been set up, only the values (Jack = 11, and Queen = 12) matter, so use a random mix of suits. Next, deal the next dozen cards into the other blank hexagon to match a rotated version of the right hexagram depicted above. Explain that in a moment you will turn away while an audience member picks a row or diagonal from either hexagram of cards. Have a few cards turned over briefly in both hexagrams to "verify" that they are different and unrelated.

Turn away and ask for any five-card row or diagonal to be selected, in either hexagram. Once again, have the rest of the cards gathered up and set aside. Now, have the values of the five chosen cards summed. From a red-backed deck that's been sitting in plain view all along have that number of cards dealt to the table. If 32 cards have been dealt, say, "The next card is yours," whereas if 33 cards have been dealt, say, "The last card is yours." Ask what it is, and have a prediction envelope—also on display from the outset—opened to confirm that you foresaw the outcome, value and suit. All that needs to be done to get this right is to start by jotting down the name of the card originally 33rd from the top.

Celebrations of Mind

John Peterson has written that "when Isaac Asimov was asked whom he thought was the most intelligent person on earth, he answered without hesitation, 'Martin Gardner.'" Centennial celebrations this autumn for this prolific and far-ranging author include numerous special articles and tributes. There's a wealth of Mathematics, Magic, & Mystery to explore at the associated Mathematics Awareness Month site too.

"Trim Hex Anagrams" is an anagram of "Martin's Hexagram." So is "Arrange Maths Mix". (There's another mirthful one we can't print here.) "Considered Yet Given" is one of a huge number of anagrams of "Distroying Evidence," and "Electric Solvents" is an anagram of "Cells Not Vertices."

Monday, June 30, 2014

Dicey Cards

Throughout his long and very productive life, mathematics popularizer Martin Gardner was intrigued by mathematical magic. He wrote extensively about it, and one of his 100+ books served as partial inspiration for the theme of 2014's Mathematics Awareness Month. To mark Martin's centennial year, recent Card Colms (Foregone Outset and Postage Stamp Issue) have drawn on his writings to explore new card diversions, and this month's romp continues that trend.

Entrant Vision

Consider a three-card game where a face-up Ace, 2, and 3 are on offer, and you invite a friend to select any card. If she picks the Ace, you pick the 2 and say, "2 is bigger than 1, so I win." If she picks the 2, you pick the 3 and say, "3 is bigger than 2, so I win," and of course, if she picks the 3, you pick the Ace and say, "Ace beats a 3, so I win." Your friend wouldn't be too impressed, even if you only played one round of this game. If multiple rounds were played, she'd probably complain that you changed the rules as you went along. "The Ace can't be both high and low," she might well cry, "You can't have it both ways." And yet we do have it both ways when playing poker: both Ace, 2, 3, 4, 5, and 10, Jack, Queen, King, Ace are considered to be straights, examples of rare five-card hands.

Wrap-arounds like that of course lead to cyclic arguments; the one above can be summarized as 1 < 2 < 3 < 1 < ... ad infinitum. Cycles are hardly a new concept in mathematics, though there is shock value in seeing them arise in the context of strict inequalities, as we are much more familiar with transitive relations than non-transitive or intransitive ones. Some well-known games exhibit cyclic logic: for instance, rock-paper-scissors. Surprisingly, the three-card swindle above is essentially what's at the heart of much more subtle paradoxes involving dice. We survey some of these before switching our attention to playing cards.

Non-transitive phenomena of this stype first came to the public's attention via Martin Gardner's "Mathematical Games" column in Scientific American, in December 1970. There, he discussed a set of four dice A, B,C, D, discovered by statistician Bradley Efron, for which which A > B > C > D > A ..., in the sense that each die beats the next one listed with probability 2/3. See chapter 22 of Martin's The Colossal Book of Mathematics (Norton, 2001), or Ivars Peterson's "Tricky Dice Revisited."  As Gardner notes there, Karl Fulves published applications of the Efron dice to card effects as early as 1971. Gardner provides several other card incarnations.

Twisted Mortice

We're going to focus on sets of just three dice, for which the margin of victory is generally smaller. There are sets of three non-transitive dice close to ordinary dice for which the margin of victory is very small indeed, but we prefer to focus on those associated with English toy collector Tim Rowett of Grand Illusions. He suggests colored dice on which are the following numbers:

Red = {1, 4, 4, 4, 4, 4},
Green = {2, 2, 2, 5, 5, 5},
Blue = {3, 3, 3, 3, 3, 6}.

Assume the dice are fair, meaning that each of the six sides comes up with the same probability, and consider the game of rolling any two of the dice together, over and over. There are the 6 x 6 = 36 equally likely outcomes. In the case of Red and Green dice, the number on the Red one is less than the number on the Green one 6 + 3 + 3 + 3 + 3 + 3 = 21 times. So 21/36 (or about 58%) of the time, on average, Red loses to Green. Similarly, the Green die loses to the Blue one 6 + 6 + 6 + 3 = 21 out of 36 times, so again about 58% of the time, on average. In conclusion, Green beats Red and Blue beats Green, on average.

The big surprise is that not only does Red beat Blue, on average, violating one's deeply ingrained expectations of transitivity, it does so by an even larger margin. Red in fact beats Blue 5 + 5 + 5 + 5 + 5 = 25 times out out 36, or about or about 69% of the time, on average. Hence we arrive at the circular conclusion:

Red < Green < Blue < Red < ...

Note the similarity to the 1 < 2 < 3 < 1 < ... seen earlier, also bearing in mind the lowest values on each of the three colored dice. Unlike in that case, which required the Ace to be considered low in one context, and high in another, the pairwise comparisons here seem quite legitimate. For the record, all three dice have a mean of 21/6.

The standard way to take advantage these dice is a game where you invite a friend to select any one of the dice, following which you pick another. Decide on a fixed number of throws, such as a dozen, and roll the two selected dice that number of times. If you've picked your die wisely, you should win more often than your friend. Of course, if she picks the Red die, you pick the Green, if she picks the Green you pick the Blue, whereas if she picks the Blue, you pick the Red.

Amusing Strength

For a terrific kicker, play this a few times over, finally revealing your secret technique, then invite your friend to try to beat you. This time, you offer to select your die first, then have her pick one to beat yours. Once it's clear that she has mastered the game, produce a second set of such dice, which she can inspect to verify is identical to the first set. Announce that you'll continue to go first, only this time each of you selects two dice of the same color. The pairs are rolled, over and over, and the totals of the numbers obtained by each of you is used to decide on the winner. The strategy she has just learned will backfire badly on her: If you start by selecting the two Green dice, she will confidently select the two Blue ones, only to find that on average she will lose. Astonishingly, the new cycle of victory reverses the former one:

Red < Green < Blue < Red < ...

but

Red + Red > Green + Green > Blue + Blue > Red + Red > ...

Secondary Dice (Crayons Decide)

How might all of this work with cards, bearing in mind that a deck only has four cards of each value? The basic idea is to replace each die with a packet of six cards, from which one is randomly selected (with replacement) in between repeated shuffles.

One possibility is to first double each of the values used for the dice, yielding even numbers from 2 to 12 inclusive, then bump a few of them up by 1 to cut down on excessive repetitions. This also neatly sidesteps the issue of whether Aces are low or high.

Red = {2, 8, 8, 8, 9, 9},
Green = {4, 4, 5, 10, 10, Jack},
Blue = {6, 6, 7, 7, 7, Queen}.

Here the colors refer to the card backs for three decks. Red and Blue are standard, find a deck with a different color to represent Green. Suits are irrelevant. (Alternatively, you may opt to do all of this using cards from a single deck, as long as you don't get confused as to which packet is which; perhaps separate the packets on the table with large gaps, and use colored markers or crayons as guides.)

Start with three such face-down packets of six cards, and ask a new friend to pick one of the packets. You pick another one, remembering that Red < Green < Blue < Red < ... as before. The cards in each selected packet are thoroughly mixed and the rolling of dice is replaced by the random selection of one card for each of you from each packet. Record whose card has the highest value, replace the cards in their respective packets and mix them again, and continue. With ten or twelve rounds, you should come out ahead on average, as in the dice case.

Tawdry Codices

Here's another card incarnation with an additional element of randomization, based on some well-known "magic square" dice. Consider the columns of the 3x3 magic square as shown:

2 7 6
9 5 1
4 3 8

Imagine a corresponding packet of face-down cards in mixed suits, running 2, 7, 6, 9, 5, Ace, 4, 3, 8, from the top down. Dealt from left to right, face up, results in this display.

In practice, all dealing is done face down, and the columns overlap a little vertically facilitating order-preserving pickups. Now gather the three columns in any order–maintaining the card order within each column, and deal out again into three piles, from left to right. Repeat over and over, until the cards seem well mixed. Have a friend pick one column (pile) for himself, with or without first looking at the card faces. Turn the other two piles over so they are face up, and casually pick one of them for yourself. If you choose wisely, you will on average win in a game of "best of a dozen" as before, where this time the card packets have size three, simulating 3-sided dice. Once more, your guiding light is 1 < 2 < 3 < 1 < ...: Simply scan the six visible card faces for the Ace, 2, and 3. You will always see exactly two of those. The missing one is in your friend's pile, so pick for yourself the pile that's "bigger and better"—remembering that Aces are both low and high!

Why does this work? Certainly it's easy to check in the case when the three piles are {2, 9, 4}, {7, 5, 3}, {6, 1, 8}, as they are at the outset: then "Pile 2" (the one with the 2) beats "Pile 1" (the once with the Ace) about 56% (=5/9) of the time on average, and "Pile 3" beats "Pile 2" and "Pile 1" beats "Pile 3" by the same margin. This property of the columns of the standard 3x3 magic square (as displayed above) is well known, and is the basis for a set of corresponding non-transitive 6-sided dice where each of the three key values is used twice.

But what about the collection of the three columns in random order, in between repeated dealings from left to right? Since there are six (=3!) ways in which to gather the cards each time, one might expect as many possibilities for the resulting three piles. It turns out that there are in essence only two things that can happen. If the number of rounds of dealing and collecting is even, one ends up with something equivalent to the initial display, from the perspective of competing piles: the order in which the piles occur is irrelevant, as is the internal card order within each pile. For instance, after four or six rounds of dealing and collecting, one could end up with:

While not a conventional magic square, it retains enough magical charm for our purposes: It respects the partitioning of 1–9 into {1, 6, 8}, {2, 4, 9}, {3, 5, 7}, as does the result after any even number of rounds of dealing and collecting.

On the other hand, after any odd number of rounds of dealing and collecting, we get something like:

This is a representative of a class of equivalent arrays—the partitioning of 1–9 into {1, 4, 9}, {2, 6, 7}, {3, 4, 8}—which are in turn transposes of the kind we saw above after even numbers of rounds of dealing and collecting.

Even better, the 1 < 2 < 3 < 1 < ... mantra again holds here for the new "Pile 1," "Pile 2," and "Pile 3"! That's why the stunt suggested works no matter how many times the piles are dealt out and collected.

Indeed, the two partitions toggled between above are two of five existing for 1–9 that give rise to non-transitive dice, as documented in The On-Line Encyclopedia of Integer Sequences entry Sequence A121228.

There are many more variations on all of the above worth digging out. For instance, M. Oskar van Deventer came up with a set of seven dice such that for any two chosen dice there is a third one that beats them both.

"Recent Divinations" and "Card Intentions Vie" are two of many amusing anagrams of "Non-transitive Dice," and "Entrant Vision" (like "Star Invention") is an anagram of "Non-transitive." "Twisted Mortice" is an anagram of "Tim Rowett's Dice," and "Amusing Strength" is an anagram of "Strange Sum Thing." "Secondary Dice" and "Crayons Decide" are anagrams of "Dicey Cards One," and "Tawdry Codices" is an anagram of "Dicey Cards Two."