The above is one of my favorite card tricks as it is not based on sleight of hand, but rather, mathematics. It was created by Martin Kruskal.

The above is one of my favorite card tricks as it is not based on sleight of hand, but rather, mathematics. It was created by Martin Kruskal.

# Google+ Page // Facebook Page // Twitter Page

## 19 Comments

## Leave a comment

Comments temporarily disabled.

# Google+ Page // Facebook Page // Twitter Page

New to Spiked Math?

View the top comics.

View the top comics.

**New Feature:**Browse the archives in quick view! Choose from a black, white or grey background.**Top Math Comics**

(Ranked by SM-CRA)

Other Sites:

haha. now they're right saying 'you just got lucky'

I came across this 'trick' a number of years ago now (albeit with face cards = 10). Something I was never able to do was calculate the convergence rate. Of course, I could tell a computer to run a million trials and spout out some statistics, but an analytic calculation always seemed to be too hard for me ;-)

One approach using Markov chains.

I like to consider myself a magician (I'm not one, I just like to think I am) and I must say that trick is ingenious!

Why is the trick named after Kruskal? I only know that there's a statistical test named after this person...

an unrelated idea to the comic but it would be cool if the comics had tags so we could find specific ones easier

Sure thing. Let me see if I can set it up.

We (@Universität Stuttgart) use a variation of this as a game to show visitors (particularly potential new students) a bit of spectacular math. We give them 50 dices, let them be rolled and arranged in a line. Then the visitor has to start at one end hopping like described above (e.g. first number is a 3 -> count three steps forward etc.). Leftover dices are cut off. Next step: re-rolling the first dice. You will end up at the last dice again (with a prob ~ 95%). It's fun to see the astonished faces (and easy to explain, just mark all dices hit and go for another 'run').

We estimate the probability as follows: Each dice gets you ~3.5 steps forward -> 50/3.5 ~ 14 steps. Prob to not hit a dice from the old track: <5/6. Not hitting 14 times: (5/6)^14 ~ 0.08 -> 92% as lower bound to hit the old track and therefore end on the last dice.

damn, html screwed last paragraph...

Prob to not hit a dice from the old track: lower than 5/6.

Not hitting 14 times: (5/6)^14 ~ 0.08

92% as lower bound to hit the old track and therefore end on the last dice.

Should be fixed now, I replaced the < with <

Hahahaha (Or disappointed that you are wrong) XD

I like how the volunteer is referred to as the victim.

I am a bit gutted that you told the world this, I enjoy using this trick on unsuspecting people now I will have to find another trick that uses logic not trickery.

I only told the math geeks who visit this site ;-)

There is another very nice trick which doesn't require sleight of hand and appears to be super amazing! Basically you deal x piles of x cards each, and have your victims (up to x victims allowed) each pick a distinct pile and to memorize a single card in their chosen pile (you should have x>=6 to make it more interesting). Then have the victims put their piles in front of them and you then gather up the piles in a random order (have the victims tell you the order to gather them in order to make it even more amazing - if you are comfortable with false shuffles/cuts then throw one of those in, but this isn't required).

Now redeal the deck into x piles of x cards each. Show each pile one by one asking the question "do you see your card". The victims will say yes/no depending on whether they see their card or not. You only have to show/ask for x-1 piles (again you can show in any order). Then declare "I know all of your cards!" and proceed to pull out the victims cards at lightning speed from each pile. Then to amaze them even more, give each victim their chosen card and they will be surprised that not only did you pull out all the chosen cards but you were able to declare which one belonged to which victim... TADA!!

If you can figure out the trick then bonus points to you :P Otherwise, let me know if you need the secret. Hint: Consider x=2, then x=3.

Who says magicians are not good mathematicians! :P

I'd love to claim the extra credit by revealing your method, but I've known a variation of that trick so long I don't recall if I figured it out, had it explained to me, or had the mechanism spelled out with the effect.

I

ad libmy own favored "magic" display from the fact that the ratio of terms in any Fibonacci style sequence quickly converges to .5*(sqrt(5)-1).That's easy! You first deal cards in a row, the deal then going by the columns. You can think of this as "transposing" a matrix of cards, if you will...then, as you know the column of the normal version and transposed version of the matrix of cards, you can point the exact card!

Credit points for me! YEAAAAAAAAAAAAAAAH!

I think I know it, but I guess you have to re-deal them in an special order and it calls for some memorizing(person x has his cards on deck y before getting all the piles togheter), right :o?

Actually, according to this article, the best strategy involves always choosing the first card, i.e. "guessing one," not because the victim will likely choose 1, but because this gives the highest probability of hitting the victim's sequence later on.