Faro coincidence

Boy the world has changed! When I was young if you wanted to learn magic you had to hang out at a magic store. A real brick and mortar building. To get to the real good stuff you had to make friends with the magician behind the counter and prove your worth. If you were lucky they might show you something more than “Scotch and Soda”. They might recommend a really good hardback book on magic, but those books weren’t cheap. You had to pay your dues, so to speak.

Today you have the Internet. The brick and mortar stores are almost completely gone. I miss them.

Anyway, while I was surfing the web I found this self-working coincidence trick that looked interesting. Now like a lot of these tricks on the Internet, this one had several “kickers” to the coincidence effect. Go ahead and take a moment to look at it. It is at https://youtu.be/9KhQrR5uqN8?si=2rnxeBWteVBaXC7m .

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Another Lie Detector

This is another lie detector plot. I have another that I also perform which you can read about at https://robertjwallace.com/lie-detector/. This one is easier.

Start by having a card selected and controlled to the bottom of the deck. The convincing control is a good method which you can learn here: https://www.youtube.com/watch?v=78aZh6eUneI&pp=ygUSY29udmluY2luZyBjb250cm9s

False shuffle and/or cut keeping the card on the bottom.

Set the deck down and ask the spectator to cut around two thirds of the cards to the side, to eliminate them. This cut is key. The bottom part of the deck must contain at least 15 cards, but not more than 23.

Pick up the remaining cards and explain that you are going to use the deck to determine the card by asking the spectator to name their card, but allowing the spectator to either lie and name a different card, or to tell the truth and name their card.

Demonstrate by dealing out the cards, spelling the phrase “your card is the”, dealing one card for each letter face down in a pile. Drop the remaining cards on top of those and ask the spectator what their card was. They can lie or tell the truth. Pick up the pile of cards and spell the name of the card they named, one card per letter.

For example if they say the “eight of clubs” you would deal out “e”, “i”…”s” with one card per letter down to a pile, then drop the remaining cards on top.

Now pick up the pile and deal the cards in the same manner, spelling out the phrase “did you lie or not” (14 letters). Turn over the last card dealt (the “t”), and it will be their card, showing whether or not they lied.

So why does the pile need to be between 15 and n cards? Well this is because of the number of letters in the names of the cards. Cards can be spelled out with from 10 to 15 letters. So to spell out the longest card name without running out of cards you need at least 15 cards. Now after the first deal of “your card is the”, the position of the chosen card is equal to the number of cards in the packet minus the number of letters in “your card is the”. The number of letters is 13. That means the chosen card will be at the 2nd from the top if the packet has 15 cards, for example. For the trick to work, that position must be within the number of letters in the name of the card that the spectator names. This is so that the chosen card is dealt by on the second deal. Since the spectator’s smallest name contains only 10 letters, the chosen card must be within the top ten cards after the first deal. The largest number of cards for that to happen is 23 (23 – 13 = 10).

To visualize this algebraically:

n = number of cards in the packet

c = the location of the chosen card

At the start with c at the bottom of the packet, c = n.

After the first deal of “your card is the”, c = n – 13.

After the second deal, c = n – c + 1. Since, from the last step, c = n – 13, this is the same as c = n – ( n – 13) + 1. This will always equal 14 for n = 15 to 23.

Then after the third deal of “did you lie or not”, c with be the last card dealt.

A slightly modified Erdnase Stack

The one aspect of the Erdnase Stack that I have never liked is the undercut and throw during the second part of the stack. Below is the description of the four card stack for a five handed game with that step in bold:

FOR any game in which cards are dealt singly. Three of the desired cards are placed on top, one on bottom. Under-cut about one-third deck, injog top card, run two less than twice number players, out-jog and shuffle off to last card, so that it will be left on top. Under-cut to out-jog, forming break at in-jog, run one less than number players, throw to break, run one, in-jog running one less than twice number players, out-jog and shuffle off. Under-cut to in-jog and throw on top. Under-cut to out-jog, run one less than number players and throw balance on top. This gives the four desired cards to the dealer in four rounds.

Erdnase, Expert at the card table

Normally during an overhand shuffle you don’t simply cut a block off and throw it to the top. Instead, what I do is what was done in the first halve of the stack, namely, undercut to the out-jog and form a break at the in-jog. Then as I start the second part of the shuffle, I let the cards below the in-jog drop onto the top of the deck as I run one less than the number of players. This effectively hides the throw and reduces the right hands shuffling motion.

I think it makes the shuffle more deceptive.

Faro stack for 3, 4, 5, 6, 7, or 8 hands, four aces

This was a “problem” I gave myself. Obviously you can stack four aces for a four handed game with two faro shuffles, but what if you want to deal a five handed game. Using my faro shuffle simulator I found a pattern that once the aces are in the correct locations a single faro out shuffle will stack them. (I used my faro shuffle simulator to work out the pattern)

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