This week's Scam School features a classic mathematical trick I haven't thought about for a long time.

I've mentioned “old” tricks before that date as far back as the 1920s or so, but this one can be found in a book that was first published in 1633! That's right, this trick is older than the United States of America.

The best way to start off is to show you the trick. Try and think about how it could be done before you watch the method.



The principle boils down to two simple ideas. If an object is in a corner, it is counted twice, otherwise, it is only counted once. This trick is so simple, in fact, that the presentation becomes all the more important in hiding the method. One of the nicer touches in Brian's approach above is assigning everybody a different side, making it harder to determine that some coins are counted twice.

To really expand the possibilities, there are several variations of the classic version taught in the video. The first variation is to perform the routine by having objects removed instead of adding them. Instead of compensating for an added object by moving objects away from the corners, you compensate for a removed object by adding objects to the corners. Also, instead of starting with a similar number of objects in the corners (such as 4 and 3 as in the above video), you'll generally want to start with a number of objects that's farther apart, such as 4 and 1. This is because the amount of objects in each corner will get closer and closer in number as you proceed.

Another variation is to vary the shape. There's nothing special about the square shape, except for its simplicity. While you could use 5 or 6 sides, larger numbers quickly get challenging to deal with for both the audience and the performer. In Jim Steinmeyer's book Impuzzibilities, he teaches a wonderful version called “Understanding the Bermuda Triangle”, which uses little paper airplanes arranged in a triangle to demonstrate the mysteries of the Bermuda Triangle.

Peter Marucci has a nice version of the Bermuda Triangle concept called Bermuda Runes, which almost makes it seem like the result of some ancient magic ceremony!

It's quite amazing to see the variety of ways performers have adapted this principle for their presentations. Lew Brooks, best known for his Stack Attack DVD, used to have a great version for kids shows involving cookies. He'd be helped by his own kid, or sometimes the birthday kid to whom he'd explain the basics beforehand.

In this version, Lew would arrange plates in a square, and show that 13 cookies were on each side. He would then turn to face the audience, start explaining how to make good cookies, while the kid would sneak up, take a cookie, and rearrange the remaining ones. Lew would turn around and ask the kid if he took any cookies. The kid would shake his head no (since he couldn't speak with a cookie in his mouth), so Lew would count and make sure each side still had 13 cookies, which it did.

As you can imagine, after the 3rd time of a kid taking the cookie, this gets to be really puzzling. Even before the audience realizes how bizarre this is, the tension between the kid trying to hide his cookie stealing and the adult makes it really enjoyable.

Because the principle os so obvious, the more you can keep the focus on the presentation, especially by adding mystery or comedy. What shape would work best? Should you add or remove object? What objects should you use?

In his book Life Savers, Michael Weber has a presentation he calls “To Feed Many”, in which raisins are removed from bowls without seemingly affecting the remaining number, and it's presented as a Buddhist ceremony for preventing hunger. Bill Herz and Paul Harris, in the book Secrets of the Astonishing Executive, on the other hand, use paper clips from the office to demonstrate the simple yet amazing nature of brainstorming at meetings.

Besides just being a fun trick, this routine is an excellent exercise in finding your own way to present and expressing who you are. Play around and leave a comment if you develop any fun and interesting angles!
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Answers to Sunday's puzzles:

1) There are several fallacious arguments in the video. First, getting paid for the day usually means getting paid for the time you worked that day, not for a full 24 hours, so he should still be $365 at this point. The 52 Sundays bit is correct, so we subtract that to get $313. Assuming the agreement is half pay for a half day, we'll subtract $26 for the Saturdays to get $287. Subtracting $14 for a two-week unpaid vacation gives $273. The hour for lunch, assuming it is supposed to be unpaid, is already covered in the time Abbott took for not working 24 hours, so we won't subtract any amount for this. Depending on the holidays mentioned, they may or may not cross over with the Sundays on which Costello didn't work. Unfortunately, this means it is impossible to determine exactly how much Costello should receive without reference to a specific year.

2) Two invalid assumptions are made to make this argument possible. The argument made here assumes that on each living person's family tree, no ancestor appears more than once, and that the same person never appears more than once in any family tree. If a couple has 3 children, then their parents will be parents on 3 different family trees, and the grandparents will appear on 9 different family trees. As stated in the original puzzle, millions of people are counted millions of times, which is where the large amount comes in.

3) The face-down card is the 6 of Hearts. The Ace of Hearts gives the suit, and the other 3 communicate how much greater than the Ace is the face-down card. First, the cards must be considered from low to high, but this gives us 5 as the low, and two 7s which seem to be equal. The suit ranking of CHaSeD must be taken into consideration meaning that the 5 is still low, but the 7 of Clubs is considered medium and the 7 of Diamonds is considered the high. This means the cards are being shown in High/Low/Medium order, which signifies 5. Ace of Hearts plus 5 equals the 6 of Hearts.

4) When calculating the odds of a given 5-card Poker hand, we must take all 5 cards into consideration. With the 4 of a kind, we only took 4 of the 5 into consideration. With any 4 of a kind, the 5th card can be any one of the 48 remaining cards. Multiplying the 13 possible values for the 4 of a kind times the 48 possibilities for a 5th card, we get 624 possible 4 of a kind handss, not 13. The straight flush calculation is correct, so you can see why any one of the 40 straight flushes are harder to get than any one of the 624 4 of a kind hands.