# 100 Years of Martin Gardner!

October 23, 2014 8:19 AM Subscribe

In Honor of the Centennial of Martin Gardner's birth (October 21, 1914),

*we've lined up Thirty-One Tricks and Treats for you: Magazine articles, new and classic puzzles, unique video interviews, and lots more.*✤ The Nature of Things / Martin Gardner [46min video] ✤ The College Mathematics Journal, January 2012 dedicated to Gardner with all articles readable online.His

posted by Rock Steady at 9:00 AM on October 23, 2014 [6 favorites]

*Annotated Alice*is a real treat.posted by Rock Steady at 9:00 AM on October 23, 2014 [6 favorites]

This year I finally got around to implementing Matchbox Educable Noughts And Crosses Engine (MENACE), a primitive machine that learns how to play tic-tac-toe. I read about it in Gardner's Mathematical Games column. It seems to date from the March 1962 issue, so I must have been working my way through back issues as I would have read it sometime in the 70s.

When I first contemplated doing it I couldn't figure out how to fit all the data into the 16K of RAM available to me; this year's version lets me tweak the program and run 20,000 rounds in less than a minute. It was a great idea and fun to think about. I'm so glad I had Gardner to introduce me to it.

I don't know of anyone who has replaced him.

posted by benito.strauss at 9:04 AM on October 23, 2014 [1 favorite]

When I first contemplated doing it I couldn't figure out how to fit all the data into the 16K of RAM available to me; this year's version lets me tweak the program and run 20,000 rounds in less than a minute. It was a great idea and fun to think about. I'm so glad I had Gardner to introduce me to it.

I don't know of anyone who has replaced him.

posted by benito.strauss at 9:04 AM on October 23, 2014 [1 favorite]

When I was a kid I constantly confused Martin Gardner with Gardner Dozois. That was a pretty cool imaginary person.

posted by bq at 9:24 AM on October 23, 2014 [3 favorites]

posted by bq at 9:24 AM on October 23, 2014 [3 favorites]

IO9 started posting weekly puzzles a bit back (example). Every one I've seen this far have made me go "oh, that's easy, that's the one from that book by Martin Gardner I had as a kid".

posted by effbot at 9:31 AM on October 23, 2014

posted by effbot at 9:31 AM on October 23, 2014

*his year I finally got around to implementing Matchbox Educable Noughts And Crosses Engine (MENACE)*

Personally, I'm willing and happy to devote a crab nebula-sized space to playing a perfect game of Go... I'm not sure why they're describing that as infeasible.

posted by Wolfdog at 10:44 AM on October 23, 2014 [1 favorite]

*This year I finally got around to implementing Matchbox Educable Noughts And Crosses Engine (MENACE)..*

You should read one of my favorite SF novels, The Adolescence of P-1. Basic plot: a hacker implements MENACE on a IBM/360 and it becomes sentient.

I grew up reading Martin Gardner, I think someone first gave me a stack of old SciAms when I was about 7 years old and I was hooked. All I can recall from that early age is my astonishment that anyone could apply such intense, persistent mental effort to inscrutable puzzles with no apparent useful purpose. It was probably unintentional that Mathematical Games column was paired with the eminently practical Amateur Scientist column. Together, they established an approach to science that encompassed the extremes, from abstract to concrete.

posted by charlie don't surf at 10:47 AM on October 23, 2014 [1 favorite]

I'm not sure if I should be happy about this post. A lot of these "31" links are to large collections themselves--exponential growth of Gardner related stuff and I have work to do...

Speaking of scale, the calculation which gives us

seems to be a very rough approximation, found by assuming that the same reducing factor of 1.5 (which allows tic-tac-toe positions to be considered only up to symmetries) also applies to the 19x19 Go board. Any idea how accurate that is?

posted by TreeRooster at 11:25 AM on October 23, 2014 [1 favorite]

Speaking of scale, the calculation which gives us

*..crab nebula-sized..*seems to be a very rough approximation, found by assuming that the same reducing factor of 1.5 (which allows tic-tac-toe positions to be considered only up to symmetries) also applies to the 19x19 Go board. Any idea how accurate that is?

posted by TreeRooster at 11:25 AM on October 23, 2014 [1 favorite]

*I'm not sure if I should be happy about this post. A lot of these "31" links are to large collections themselves--exponential growth of Gardner related stuff and I have work to do...*

Well my work day is absolutely shot. I started going through a list of MG columns, I came to a dead stop when I remembered reading this when I was about 8 years old:

*1966 Oct Can the shuffling of cards (and other apparently random events) be reversed*

It is online but paywalled. It is cheaper to go down to the library, find the originals and make copies with my iPhone camera, especially since I will need copies of a whole bunch of different columns.

But as I read through the topics he covered, I came to a realization. Recently I have been working as a mathematician (albeit at a very low level) and I work with younger mathematicians with degrees in math from BA through PhD. Since I do not have a degree in math, I am presumably less educated in those topics, certainly my knowledge is less rigorous. But occasionally I see the "real" mathematicians doing things that astonish me. Hey what are you doing, don't you recognize Goldbach's Conjecture? Why are you trying to prove that? Nobody has succeeded so far, but don't let me stop you from trying. What do you mean, you never heard of Flatland? Hey don't you recognize this problem? It's a Bin Packing problem, just find a solution that's 75% efficient and you're doing better than average. Now seriously, you have never heard of the Sieve of Eratosthenes? Or the Towers of Hanoi?

I looked through the list of Gardner's columns, and now I realize these are all concepts I learned from him. Then I realized, all my mathematician coworkers were educated well after Gardner's retirement. They do not know what they missed.

posted by charlie don't surf at 11:55 AM on October 23, 2014 [3 favorites]

I've already decided my next tattoo will be an illustration from

posted by mykescipark at 12:07 PM on October 23, 2014

*Aha! Gotcha*, one of my favorite books growing up. (Replacing "1982" with my birth year, of course.)posted by mykescipark at 12:07 PM on October 23, 2014

I had never heard of him, but I think my kid would be crazy about his work. Is there an anthology of some sort?

posted by latkes at 12:15 PM on October 23, 2014

posted by latkes at 12:15 PM on October 23, 2014

latkes, The Colossal Book of Mathematics is the most complete collection of Gardner stuff I could find. But it's very big – 700+ pages – which could either delight or intimidate a kid, depending on personality. If you want to start with smaller chunks, the books in The New Martin Gardner Mathematical Library look good. They're reasonable in paperback on Amazon: here, here, and here.

I don't know how old your sprout is, but Gardner does require some maturity of mind. I probably wouldn't give him to a kid under 12 years old.

posted by benito.strauss at 12:35 PM on October 23, 2014 [1 favorite]

I don't know how old your sprout is, but Gardner does require some maturity of mind. I probably wouldn't give him to a kid under 12 years old.

posted by benito.strauss at 12:35 PM on October 23, 2014 [1 favorite]

I haven't seen this CD-ROM complilation of Mathematical Games books mentioned yet here (but in all likelihood I learned of it elsewhere on MeFi previously).

posted by AbnerRavenwood at 1:17 PM on October 23, 2014 [2 favorites]

posted by AbnerRavenwood at 1:17 PM on October 23, 2014 [2 favorites]

Searchable? Ooooh! A hearty welcome to MetaFilter, Abner.

And latkes, if your kid is more in the 7-11 year range, I have fond memories of "536 Puzzles and Curious Problems", by H. E. Dudeney, as one of my earlier word/logic/geometry puzzle books. In fact, the linked Amazon page is the very edition that I had, and it's only my resolution to stop stockpiling books I don't really need that stops me from nostalgia buying it.

posted by benito.strauss at 1:25 PM on October 23, 2014 [2 favorites]

And latkes, if your kid is more in the 7-11 year range, I have fond memories of "536 Puzzles and Curious Problems", by H. E. Dudeney, as one of my earlier word/logic/geometry puzzle books. In fact, the linked Amazon page is the very edition that I had, and it's only my resolution to stop stockpiling books I don't really need that stops me from nostalgia buying it.

posted by benito.strauss at 1:25 PM on October 23, 2014 [2 favorites]

The CD collection is the 15 anthologies (all published by the MAA, I believe) which include nearly all of the SciAm

posted by Wolfdog at 1:31 PM on October 23, 2014 [1 favorite]

*Mathematical Games*columns as well as addenda to them, with each book as a separate PDF file. It's quite possibly the best use a CD has ever been put to.posted by Wolfdog at 1:31 PM on October 23, 2014 [1 favorite]

benito.strauss:

Too late, too late...

Also, if you haven't read the Annotated Alice, you're missing a right treat.

posted by RedOrGreen at 2:21 PM on October 23, 2014 [2 favorites]

*It's only my resolution to stop stockpiling books I don't really need that stops me from nostalgia buying it.*Too late, too late...

Also, if you haven't read the Annotated Alice, you're missing a right treat.

posted by RedOrGreen at 2:21 PM on October 23, 2014 [2 favorites]

I still remember the moment I read about Martin Gardner's death. It was on the blue and I burst into tears. Thanks for this post. Any day I think about Martin Gardner is a good day.

posted by town of cats at 2:45 PM on October 23, 2014 [1 favorite]

posted by town of cats at 2:45 PM on October 23, 2014 [1 favorite]

I might have managed to forget enough from the Annotated Alice that it could be worth re-reading. Gardner could also be snarky, too.

posted by benito.strauss at 6:35 PM on October 23, 2014

posted by benito.strauss at 6:35 PM on October 23, 2014

OK, Will Shortz's challenge:

I've come up with 4 class sizes that work and expect there are more (perhaps an infinite number) - am I missing something? (Only one size is about a typical class size - maybe that's the "right" answer?).

posted by Bort at 12:21 PM on October 24, 2014

*The following challenge is based on a puzzle from a Martin Gardner book, that may not be well-known. Out of a regular grade school classroom, two students are chosen at random. Both happen to have blue eyes. If the odds are exactly 50-50 that two randomly chosen students in the class will have blue eyes: How many students are in the class?*I've come up with 4 class sizes that work and expect there are more (perhaps an infinite number) - am I missing something? (Only one size is about a typical class size - maybe that's the "right" answer?).

posted by Bort at 12:21 PM on October 24, 2014

I'm not too good at this...but I get that 4*(number of pairs of blue-eyed kids) = n(n-1), where n is the class size. We know there's at least 1 pair of blues, but we need n(n-1) to be divisible by 4 and of course n>(number of blues).

posted by TreeRooster at 5:06 PM on October 24, 2014

posted by TreeRooster at 5:06 PM on October 24, 2014

*** SPOILERS ***

Well, I figure if B == Number of Blue Eyed Kids and T == Total Number of Kids, then picking 2 at random will produce BlueEyed/BlueEyed 50% of the time when B/T * ((B-1)/(T-1)) = 0.5 - Or when T(T-1) = 2B(B-1). Using a google spreadsheet, I found the following 4 values:

B T

-- --

3 4

15 21

85 120

493 697

For T up to about 1000ish

posted by Bort at 8:41 PM on October 24, 2014

Well, I figure if B == Number of Blue Eyed Kids and T == Total Number of Kids, then picking 2 at random will produce BlueEyed/BlueEyed 50% of the time when B/T * ((B-1)/(T-1)) = 0.5 - Or when T(T-1) = 2B(B-1). Using a google spreadsheet, I found the following 4 values:

B T

-- --

3 4

15 21

85 120

493 697

For T up to about 1000ish

posted by Bort at 8:41 PM on October 24, 2014

This is a Martin Gardner puzzle, it's a trick. There is always a short cut.

Answer: There are three students, two with blue eyes.

Here are the combinations. B for Blue and X for nonblue.

(B, B) Match

(B, X) Nonmatch

That's all you get, two pairs of colors, with equal odds of being picked at random.

It doesn't matter that there's a third pair, (X, B) since that is the same as (B, X), a nonmatch.

Proof of this answer using the Binomial Theorem is left as an exercise for the reader.

posted by charlie don't surf at 11:57 PM on October 24, 2014

Answer: There are three students, two with blue eyes.

Here are the combinations. B for Blue and X for nonblue.

(B, B) Match

(B, X) Nonmatch

That's all you get, two pairs of colors, with equal odds of being picked at random.

It doesn't matter that there's a third pair, (X, B) since that is the same as (B, X), a nonmatch.

Proof of this answer using the Binomial Theorem is left as an exercise for the reader.

posted by charlie don't surf at 11:57 PM on October 24, 2014

*It doesn't matter that there's a third pair, (X, B) since that is the same as (B, X), a nonmatch.*

Channeling Walter Wagner there, I see.

As for Gardner tricks, the "regular grade school classroom" bit strongly indicates that the expected answer isn't three or four students :)

posted by effbot at 5:56 AM on October 25, 2014

I think you need to revisit probability class, charlie. :)

For 3 kids, where 2 have blue eyes, the probability is 1/3 of getting BB:

First B = 2/3 chance. 2 B, 3 Kids.

Second B = 1/2 chance. 1 B, 2 Kids.

Total Prob = 2/3 * 1/2 = 1/3.

posted by Bort at 6:34 AM on October 25, 2014

For 3 kids, where 2 have blue eyes, the probability is 1/3 of getting BB:

First B = 2/3 chance. 2 B, 3 Kids.

Second B = 1/2 chance. 1 B, 2 Kids.

Total Prob = 2/3 * 1/2 = 1/3.

posted by Bort at 6:34 AM on October 25, 2014

*As for Gardner tricks, the "regular grade school classroom" bit strongly indicates that the expected answer isn't three or four students :)*

That is classic Gardner, it is literally a magic trick, it's a misdirection. The problem asks for a solution, this strongly indicates there is one solution.

*For 3 kids, where 2 have blue eyes, the probability is 1/3 of getting BB.. = 1/3.*

You calculated

_{3}C

_{2}= 3!/((3-2)! * 2!) = 3. There are three combinations, the odds of each combination is 1:3. But two of the combinations are the same, B,X and X,B, a multiplicity of 2, so there are only two

*distinct*sets of two students that can be picked: one with matching eye colors and one that does not match. So the odds of picking a pair that does not have matching eye colors is 1:2. And the odds of picking a pair that matches is also 1:2.

cf. Gardner's The Two Children Problem

posted by charlie don't surf at 8:45 AM on October 25, 2014

That's incorrect. You can confirm 1/3 by doing the following experiment:

Take 3 quarters.

Mark 2 of them in some way (to indicate blue eyes).

Put them in a bag, or cup, or giggle in your hand for a bit.

Pick/drop 2.

Count how many times you get the 2 marked.

I got 9 for 30.

Did it again - 9 for 30 again.

Look at it this way, say there are 100 kids and 2 have blue eyes. Would you still argue there are 2 distinct sets so the odds are still 1:2?

What odds do you calculate for 3 blue out of 4 kids? How about 15 out of 21? Those are both 1:2 -

3/4 * 2/3 = 0.5

and

15/21 * 14/20 = 0.5

posted by Bort at 9:23 AM on October 25, 2014 [1 favorite]

Take 3 quarters.

Mark 2 of them in some way (to indicate blue eyes).

Put them in a bag, or cup, or giggle in your hand for a bit.

Pick/drop 2.

Count how many times you get the 2 marked.

I got 9 for 30.

Did it again - 9 for 30 again.

Look at it this way, say there are 100 kids and 2 have blue eyes. Would you still argue there are 2 distinct sets so the odds are still 1:2?

What odds do you calculate for 3 blue out of 4 kids? How about 15 out of 21? Those are both 1:2 -

3/4 * 2/3 = 0.5

and

15/21 * 14/20 = 0.5

posted by Bort at 9:23 AM on October 25, 2014 [1 favorite]

Another way to see it is to just enumerate the options.

Abe, Bill, and Chris are in the class. Abe and Bill have blue eyes. Chris has brown.

Options:

(A,B) - 2 blue

(A,C)

(B,A) - 2 blue

(B,C)

(C,A)

(C,B)

You get 2 blue 1/3 of the time.

Eliminate position:

(A,B) - 2 blue

(A,C)

(B,C)

Again, blue 1/3 of time.

posted by Bort at 9:53 AM on October 25, 2014

Abe, Bill, and Chris are in the class. Abe and Bill have blue eyes. Chris has brown.

Options:

(A,B) - 2 blue

(A,C)

(B,A) - 2 blue

(B,C)

(C,A)

(C,B)

You get 2 blue 1/3 of the time.

Eliminate position:

(A,B) - 2 blue

(A,C)

(B,C)

Again, blue 1/3 of time.

posted by Bort at 9:53 AM on October 25, 2014

*...there are only two distinct sets of two students that can be picked: one with matching eye colors and one that does not match. So the odds of picking a pair that does not have matching eye colors is 1:2. And the odds of picking a pair that matches is also 1:2.*

Ok, you're reading "two students are chosen at random" as "chose a student at random, write down his eye color on a piece of paper, ask him to return to the class, repeat the procedure once, and compare what you wrote" rather than "chose two students at random and look at their eyes"? I don't think the next sentence, "Both happen to have blue eyes," supports that interpretation.

(and it's a pretty absurd interpretation even without that sentence; I mean, if you have a box of chocolates and your kid asks "can I have two pieces?", and you give them one, take it back, and give it to him again, and say "that's two!" I'd strongly suspect that way more than one in two kids would just go "oh, not again" and point you to the dad joke thread. )

(and no, I don't agree that this is nearly as ambiguous as the puzzle in boy/girl paradox.)

posted by effbot at 9:53 AM on October 25, 2014

To illustrate the difference, here's a short snippet of Python that prints the two sets:

posted by effbot at 9:57 AM on October 25, 2014

>>> list(itertools.combinations(("blue1","blue2","other"), 2)) [('blue1', 'blue2'), ('blue1', 'other'), ('blue2', 'other')] >>> list(itertools.combinations_with_replacement(("blue1","blue2","other"), 2)) [('blue1', 'blue1'), ('blue1', 'blue2'), ('blue1', 'other'), ('blue2', 'blue2'), ('blue2', 'other'), ('other', 'other')]The first one has one matching pair out of three combinations, the second where you can pair kids with themselves has three matching pairs out of six.

posted by effbot at 9:57 AM on October 25, 2014

I don't think you guys are giving enough credit to Mr. Gardner. If this problem could be calculated in an obvious way, it wouldn't be a puzzle. I am certain we will

posted by charlie don't surf at 7:44 PM on October 25, 2014

*all*be proven wrong, the problem will depend upon some paradoxical factor that will be blindingly obvious once it is revealed, and will probably depend upon a careful interpretation of the exact wording of the problem. And there will be at least one misdirection, some item in the problem that appears to be crucial but is irrelevant. But perhaps I am overestimating the problem, just due to the source being Gardner. We will find out soon enough.posted by charlie don't surf at 7:44 PM on October 25, 2014

Point to Bort, but I am disappointed. Where is the Gardnerian twist?

posted by charlie don't surf at 6:47 PM on October 27, 2014

posted by charlie don't surf at 6:47 PM on October 27, 2014

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