Episode 128 - Marx's Fave [Discussion]

[marilee]

CHICO MARX'S FAVORITE SCAM! Swindle your friends with this easy rigged contest.

Watch or download this episode now! (http://revision3.com/scamschool/dollarbill)

[greymatters]

If you're reading this thread before watching the current episode, be aware that this week's scam (Chico Marx's favorite) may not seem too challenging if you just watch the video, including the method behind it.

Try a similar challenge online here. (http://mindmagician.org/turnfour.aspx) (Note that there's 9 dots instead of 10 bills, but the principle remains the same.)

Once you've watched the video, that site is also a good place to practice!

Update: I've posted a few other variations, as well as the episode itself, in today's blog post (http://headinside.blogspot.com/2010/08/penny-puzzles-chico-marxs-favorite-scam.html).

[murphy1d]

I'm sure there are relationships between the number of people and the number of bills. In other words, do you always have to have 10 items, or always have 2 items from each person?

I know I could figure this out on my own, but I want to make sure they identify this in the videos. I'd hate to have 4 people and 8 bills and it doesn't work (lol).

[greymatters]

The only relationship that's important in this trick is the relationship of the bills (10, in the video) to the number that's counted each time (4, in the video).

To make the puzzle challenging, you generally want the counted number to be half (or even just under half) than the number of objects used.

The important detail is that the count winds up on 2 different objects if you choose a different direction. If you used 10 objects and counted to 6 each time, you'd wind up on the same card regardless of the direction (try it!), so this would be very easy to solve.

Also, if you counted just 1 each time, it wouldn't take people long to work out how to get 9 bills turned down.

Try this online Java version (http://www.ctkmathgamesforkids.com/Games/SevenCoinsPuzzle.shtml), which lets you change the number of objects and how far you count by clicking on the numbers in the lower-right corner (clicking on the left side of a number decreases it, while clicking towards the right side of a number increases it). The version in the video would be the same as the {10,3} star on this page (This page counts from 0 instead of 1 as in the video, so it's {10,3} instead of {10,4}).

Once you get the hang of using it, that page is a good page to see what combinations of bill amounts and counting numbers will work together. If you set up an arrangement and it doesn't show a pattern, it won't work.

Click on the hint button, and you'll see that all that really happens in this challenge is that you're moving around a circle! However, if you put this in a straight circle, it wouldn't be much of a puzzle, would it?

As with many things Scam School, there's a simple underlying structure, and you're taking advantage of the fact that you're hiding the underlying structure.

If we use the rule of counting from 0 instead of 1 (so, in the video version you'd count from 0 to 3 instead of 1 to 4), then you quickly learn that the best combinations are those where the number of objects and the counting numbers don't have any factors in common. For example: The {10,3} version in the video works well because 10 and 3 don't have any factors except 1 in common (10's factors: 1, 2, 5, and 10. 3's factors: 1, 3). 10 and 4 wouldn't work because 10 and 4 have both 1 and 2 as common factors.

Your 8-bill/counting-to 4 version would work because the {8,3} combination doesn't have any common factors, except 1.

Math lingo: When two numbers have no factors except 1 in common, they are referred to as being mutually prime.

[lokified]

Oh, it's 10 bills! *facepalm* the time I tried it, I had 12 and ruined the effect.