I am currently exploring C.S. Peirce's First Principle of cyclical chains, as described by Arthur F. MacTier (2000) in 'Card Concepts' (pp. 70-73). MacTier states that this principle works with any even quantity of cards, and uses this as the basis for his effect 'Fancy a Gamble - I'.

MacTier's effect works fine with 12 cards (the example described in the book). However, it doesn't work with eight cards. I'm struggling to understand why not.

Any thoughts?

Mike

I haven't worked out a general principle for any even amount of cards, but I was able to get Wolfram|Alpha to help see what even numbers will work and which ones won't.

Another way to think of "Fancy A Gamble" is that you're basically performing a reverse Faro, while always keeping the same card on top (I guess you'd call that a reverse out-Faro).

I used the following rather complex looking command:
table[(((y+1)-x)/2) + (((y+1)/2) * (1-(x......, 1, y}]

Just for clarification:
table[] = tell Wolfram|Alpha that the only result you want is a table of numbers (no graphs, or simplified forms, or other extraneous results).

(((y+1)-x)/2) + (((y+1)/2) * (1-(x mod 2))) = the formula for the way the cards are dealt in "Fancy A Gamble I"

{y, 12, 12} = y, in the above formula, is the total number of cards used. Here, the command is basically saying "I want y to range from 12 to 12".

{x, 1, y} = x, in the above formula, represents the position of each card. We want to know what happens to all the positions, so x will range from 1 to the total number of cards (y).

Running the above as is, we get the following result:
{{6, 12, 5, 11, 4, 10, 3, 9, 2, 8, 1, 7}}

What does this mean? It means that whatever card was in position 1 before the deal is now in position 6. Card 2 moved to the 12th position after the deal, card 3 moved to the 5th position, and so on.

The result not only tells us what happens after one deal, but can also be used to determine what happens through multiple deals. How?

We know the 1st card became the 6th. What happens to the 6th card? It becomes the 10th? The 10th obviously becomes the 8th, and so on. So, we can work out that 12 cards gives us the following path:
1, 6, 10, 8, 9, 2, 12, 7, 3, 5, 4, 11 (then back to 1)

With 12 cards, you get a path length of 12, so the card won't appear in the same position until you do 13 deals.

You're having a problem with 8 cards, so let's see what happens to their positions:
{{4, 8, 3, 7, 2, 6, 1, 5}}

Following the paths of each card as we did above, we get 4 different and very short paths:
1, 4, 7 (back to 1)
2, 8, 5 (back to 2)
3 (stays at 3)
6 (stays at 6)

In other words, if the guy chooses 3 or 6, you're going to lose the bet every time. Any other card will give you 2 deals before returning to its original position.

Just for fun, I ran other numbers of even cards, followed their paths, and this is what I found:

2 cards:
{{1, 2}}
1 cards stays in 1st position, 2 card stays in 2nd position
2 cards won't work.

4 cards:
{{2, 4, 1, 3}}
Path = 1, 2, 4, 3 (back to 1)
4 cards will work!

6 cards:
{{3, 6, 2, 5, 1, 4}}
Path = 1, 3, 2, 6, 4, 5 (back to 1)
6 cards will work!

10 cards:
{{5, 10, 4, 9, 3, 8, 2, 7, 1, 6}}
Paths =
1, 5, 3, 4, 9 (back to 1)
2, 10, 6, 8, 7 (back to 2)
10 cards won't work.

14 cards:
{{7, 14, 6, 13, 5, 12, 4, 11, 3, 10, 2, 9, 1, 8}}
Paths =
1, 7, 4, 13 (back to 1)
2, 14, 8, 11 (back to 2)
3, 6, 12, 9 (back to 3)
5 (stays at 5)
10 (stays at 10)
14 cards won't work.

If you change the premise slightly, you might get 10 cards to work. Just tell the spectator that, after they turn up their chosen card on the first deal, you're going to bet on 4 more deals only.

However, if you were really making this bet and doubling each time, the use of 6 or 12 cards would bring in more money.

Wow, thanks Scott for this brilliant analysis. No wonder I was struggling to understand it!

Mike

You're welcome.

Playing around with how I enter the data in Wolfram|Alpha, I found I was able to get presented in a clearer manner.

Here's the new result for 12 cards:
x |
1 |6
2 |12
3 |5
4 |11
5 |4
6 |10
7 |3
8 |9
9 |2
10 |8
11 |1
12 |7

Now the path is easier to follow. Looking at 1, we see 6, so we look at 6 and see 10, and so on.

Seeing how we get the path 1, 6, 10, 8, 9, 2, 12, 7, 3, 5, 4, 11 (and back to 1) is much easier now, isn't it?

If you take the original formula (below), and replace all the 12s with another even number of cards, and then run it through Wolfram|Alpha, you'll get a similarly simple result.
table[(((12+1)-x)/2) + (((12+1)/2) * (1-(x mod 2))), {x, 1, 12}]

For example, here's the table run again with 6 cards instead.