Chris Ferguson card trick at 2004 WSOP

[7n7]

I've been searching the forum for the discussion that took place on Chris Ferguson's card trick at the 2004 WSOP... the one where he knows what card is missing from the deck.

In the thread, there were several examples of how to perform the feat but I can't seem to find it.

Anybody have a link to it or remember what it was titled?

Thanks!

[Hold'me]

What trick are you referring to? Is it the one where he slices fruits & vegetables with his 70+ mph card pitch or where he counts down the cards in a deck and then tells Norman Chad which one is missing? The latter is easier to perform I believe. [img]/images/graemlins/wink.gif[/img]

[7n7]

The card count one. As I said, there was a thread here around the time of the episode that described various ways to do it.

[slickpoppa]

He's Jesus, he can do whatever he wants

[thirddan]

assign each card in the deck a numeric value...
2 [img]/images/graemlins/club.gif[/img] = 1, 3 [img]/images/graemlins/club.gif[/img] = 2...A [img]/images/graemlins/spade.gif[/img] = 52...

add all those values together to get the sum or what the entire deck is...1 + 2 + 3 + .. + 52 = total deck...

as you flip each card over you simply add the numbers together, when you finish the deck subtract your head count from the total count, the answer is the number value of the missing card...hopefully that made sense...

[housenuts]

yup. u need to assign a value to each card.

you can assign values 1-52 or have 2 counts. 1-13 for the ranks and 0-3 for the suits. if you can keep track of 2 counts the numbers stay smaller and the math is easier.

[7n7]

Wasn't there a way to do it by adding or subtracting from 0?

[d10]

Assigning values 1-52 is going to make the trick extremely difficult. You see the Jd come up and first you have to remember that Jd=37, then you have to add 37 to something like 626, then you have to do that flawlessly 51 times in a row. At some point you will add wrong, or you will transpose 2 numbers when you get into the thousands, something will happen and all it takes is one mistake to screw up the whole trick. People usually don't like this trick anyways, as it takes a good few minutes to complete, if you get it wrong at the end you will really piss your audience off.

I don't know if this method has been discussed previously. I thought it up after seeing the trick on the WSOP last year, but I wouldn't be surprised if someone else has also come up with the same idea, or something close to it, since it is almost certainly the easiest way to do the trick. It only involves 1 count which never goes higher than double digits, and you're only adding 1-9 to one of those digits. So if you can add any random number [1-99] with any random number [1-9] then you won't have a problem.

I assign values to the cards as follows:

A=1
2=2
...
T=10
J=20
Q=30
K=40

Assigning large numbers to the face cards works well because face cards are immediately recognizable as special cases, as long as you associate J with 20, Q with 30, and K with 40 you'll never need to think about what value you need to add for the card you're currently looking at, it will be naturally obvious. Also, 20, 30, and 40 is easier to add than 11, 12, 13 I think. If you disagree, feel free to use 11, 12, 13.

That will give you your count. When it rolls past 100, knock that 3rd digit off and keep counting, because it's irrelevant (for example if your count is 95 and you draw an 8, roll the count back to 3). The end count for a 52 card deck is 80 (40+30+...2+1 = 145 x 4 suits = 580, we ignore the 5). So after 51 cards you'll know which value is missing. Try this a few times and ignore the suits until it gets easy. Then you can start to keep track of the suits you see as well.

There are only 4 suits, and 4=2x2. So you can keep track of the suits using anything that you have 2 of, which can be moved into 2 positions. I use my toes, since no one can see what I'm doing if I'm wearing shoes (or even if I'm not, they probably aren't looking at my feet). I can curl my toes up or stretch them out, and I have toes on my left foot and toes on my right. Assign each suit to the following values:

Suit 1: no change 1 - no change 2
Suit 2: no change 1 - change 2
Suit 3: change 1 - no change 2
Suit 4: change 1 - change 2

Use whatever's convenient for you. You probably already have the suits associated in some way in your mind. You can also use the colors to help. For example, I associate the color black as a boring, neutral color. So the actions I associate with black are (change 1 - change 2) and (no change 1 - no change 2). Furthermore I associate spades as having value, while clubs are worthless, probably because the As is a card unlike any other, and what are clubs? So I change on spades, no change on clubs. So when I see a spade, I think "It's black so both my toes will either move or not move, now I see it's a spade so I'm going to move." And then if my toes are flat I curl them up, if they're curled I lay them flat, if one is curled and one is flat then I switch both of them around. For red cards I know I will switch 1 foot but leave the other alone. I switch my left toes on hearts because my heart is on the left side of my body, which just leaves the right side for diamonds. At the end, whatever position my toes are in is what suit the missing card is. For example, at the start my toes are always both straight out. If at the end, they're both curled up, I know the missing card is a spade.

That's my method for keeping track of suits. A nice little base 2 system that I don't have to count. You can make up your own system though based on what you naturally associate the different suits with. You can also use something other than your toes if it makes things easy on you. Shifting toes around can get tiring after 51 shifts, and sometimes I find myself realizing that one of my toes has naturally shifted to an in between position and I have no idea if it's supposed to be flat or curled. You have to keep part of your mind focused on that throughout the trick to avoid those situations, which takes some discipline. Of course if you don't care that it's obvious what you're doing, you could use your hands. Or you might think of something else you can use that I haven't even thought of.

But that, I think, is the easiest way to do this trick. Let both sides of your mind work for you, and keep the numbers manageable. Any idiot can do the trick with this method after a bit of practice.

[sirio11]

You just need to add modulo 13.

For example, 5+9=1 , 9+8=4 and keep adding modulo 13, if the final sum is 6, then you know that the card missing is a 7; or if it's a 2, then the card missing is an 11 (Jack)

Hope this help.

David

[Patrick Duffy]

try writing them all down

[img]/images/graemlins/cool.gif[/img]