Presh Talwalkar, from the *Mind Your Decisions* blog, recently shared a fun magic trick.

It involves playing cards and dice. Because it's mathematically based, however, you might just fool yourself while performing it. You'll really only fool yourself if you don't analyze the math behind the trick, which is exactly what we're going to do in this post.

First, let's take a look at Presh's video, from this January 2015 post:

There's quite a bit going on here, so let's break this down piece by piece.

**CARDS:** Let's ignore the dice for the time being, and focus only on the playing cards. During the dealing process, cards fall into 1 of 2 categories: Either they're dealt individually, or they're in the remainder that is placed on top of the stack. Let's look at each of these categories separately.

*DEALT CARDS:* We'll start with the first example from the video, where 10 cards were used, 7 of which were dealt. What happens to those 7 cards. The card in position 1 is dealt first, and will obviously become card 10. The card at position 2 will wind up as the 9th card, and so on. Take a look at where these 7 cards end up in the final stack:

```
Starting position Ending position
----------------- ---------------
1 10
2 9
3 8
4 7
5 6
6 5
7 4
```

See the pattern? The starting position plus the ending position always add up to 11, in this example! Why 11? What would happen if we used, say, 18 cards instead? Well, 1 would become 18, 2 would become 17, and so on. In that example, everything totals 19. The resulting total will always be 1 more than the number of cards involved.We can use this to work out a formula for the dealt cards. The total number of cards plus 1, minus the starting position of a dealt card, will give you the ending position. So, in our example with 10 cards, of which 7 are dealt, we can work out the total number of cards (10) plus 1 (equals 11) minus the original position (say, 3, so 11 minus 3 equals 8) gives us the ending location of that card (so, we can easily say that, in this example, the card that started at the 3rd position will wind up at the 8th position).

A simpler way to say this is to use S for starting position, E for ending position, and N for the total number of cards. So, our formula for dealt cards could be written as E = (N + 1) - S

That's OK for the dealt cards, but what about the undealt cards?

*UNDEALT CARDS:*If you have 10 cards with 7 cards dealt, cards 8, 9, and 10 will not be dealt. They are simply placed on top of the dealt cards as a group.

In this case, the original 8th card becomes the first card, the 9th card becomes the 2nd card, and the 10th card becomes the 3rd card. Let's chart these positions and find a pattern:

Starting position Ending position ----------------- --------------- 8 1 9 2 10 3This pattern is even simpler! The starting position, minus the number of dealt cards, gives the new ending position. Using D for the number of dealt cards, and the same variables from the first formula above, we have E = S - D for the undealt cards.

OK, we've got 2 formulas to handle our 2 cases, so let's bring the dice back in.

**DICE:**Since the dice are used to choose random numbers, we'll refer to them with the letter R, and since high, low, and medium numbers are important, we'll use R1 for the dice with the lowest number, R2 for the dice with the middle number, and R3 for the dice with the highest number.

*FOLLOW THE TOP CARD:*As explained, knowing the top card is the key to this trick, so we're only going to follow that particular card.

How many cards are used? In the first performance, the dice rolled with R1=2, R2=3, and R3=5, which means that 10 cards are used. More generally, the dice total determines the number of cards used, so N (total number of cards) = R1 + R2 + R3. In our example, this was 5 + 3 + 2 = 10 cards.

*Step 1:*Note that, when the cards are dealt initially, the top card ALWAYS becomes the bottommost card. So, the starting position for the predicted card is always at the bottom as well. In other words, S (the starting point) = R1 + R2 + R3, as well. So, the predicted card starts at 10 in the first example. In other words, S = 10.

*Step 2:*For the next deal, R2 is removed (3 in the first example), and R1 + R3 dice (D = 7) are dealt. The card as starting position 10 (S=10) is obviously not going to be dealt, so we'll apply the undealt card formula (E = S - D). E = 10 - 7 = 3. So, the predicted card winds up at position 3 in the example.

Let's look at this more generally. The card starts out at position S, which is also R1 + R2 + R3. You're dealing D cards, and D = R1 + R3. So, E = S - D can be re-written as E = R1 + R2 + R3 - (R1 + R3), which simplifies to E = R1 + R2 + R3 - R1 - R3, which further simplifies to E = R2.

In other words, after the first dealing of D cards, the predicted card will wind up at the position denoted by R2, the removed dice! That's interesting and unexpected.

*Step 3:*So, now we want to see what happens to the card at position R2 on the next deal. Because R2 must always be less than the total of R1 + R3 (removing the middle number ensures this), R2 will always be among the cards dealt in this phase. This means we need to follow the dealt card formula from above (E = (N + 1) - S).

In the 10 cards total/7 cards dealt example, we're now tracking the 3rd position, so E = 10 + 1 - 3, which simplifies to E = 8, so our 3rd card winds up in the 8th position.

The current general starting position is, as we already know, is R2. We can turn the formula, then, into E = (N + 1) - R2. Further, since the total number of cards, N, is R1 + R2 + R3, we can change the formula into E = R1 + R2 + R3 + 1 - R2. This simplifies into E = R1 + R3 +1.

Interpreting that general formula, that means the predicted card has now moved to the position denoted by the remaining dice plus 1. Sure enough, in our running 10/7 example, 7 remains, and the predicted card has moved to the 8th (7 + 1) position!

*Step 4:*This should be pretty clear. The card we're following is at position R1 + R3 + 1, and we're going to deal R1 + R3 cards off of it. In our example, the card we're tracking is at position 8, and we're going to deal 7 cards from above it. Either way, the card will be moved to the first position!

**SHORT VERSION:**If you read the above carefully, you can start to see

*WHY*this card trick works. The predicted card starts at the bottom of the pile. Next, it moves to position R2, followed by a move to position R1 + R3 + 1, and finally to position 1.

**CREDITS AND OTHER THOUGHTS:**As noted in Presh's original post, he developed this after reading the Low Down Triple Dealing routine, as found in the book

*Mathematical Card Magic*by Colm Mulcahy.

There's limitless variations to this type of routine, one of which was created by Jim Steinmeyer for use by David Copperfield's

*Fires of Passion*TV special.

If you'd like to play around with this principle without having to constantly deal cards, I've created a formula in Wolfram|Alpha that will effectively deal the cards for you.

Returning one last time to our 10 card pile (N=10) with 7 cards dealt (D=7) example, you simply set N and D, and the calculation will do the rest. The output you get from this run is {8, 9, 10, 7, 6, 5, 4, 3, 2, 1}. The numbers are the starting positions (S), and each number's placement is their ending position (E). 8 being placed first means that it started in the 8th position and moved to the first position. 9 being in the second position means that the cards which started out 9th has been moved to the 2nd position, and so on.

As a matter of fact, since the total cards (N) and the number of dealt cards (D) stay constant in most routines, you can use this one output to track the card through multiple deals. I'll show you what I mean by following through 3 deals of 7 cards from a pile of 10.

What does {8, 9, 10, 7, 6, 5, 4, 3, 2, 1} tell us about the original 10th card? One quick glance tells us it became the 3rd card (because there's a 10 in the 3rd position. Where does the 3rd card go from there? Another quick glance tells us that, since 3 is in the 8th position, the 3rd card must move to the 8th position next. Finally, what happens to the 8th card? It winds up on top, because we can see the 8th card at position 1!

That's how an entire routine with multiple deals can be explored using only one simple mathematical result!

Thanks for creating and sharing this routine, Presh! I'd love to hear about any variations my readers develop in the comments, as well!

## 1 Response to Cards and Dice

NUMBERS 1 - 6-1 does NOT return force card to the top position.

NUMBERS 2 - 6 - 1 does NOT return force card on top

NUMBERS 3 -6-1 does NOT return the force card to the top position.

NUMBERS 4 - 6 - 1 does not return force card on top either.

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