Mathemagic: what happens at the intersection between magic and Math? Talk by: Kush Fanikiso Email: [email protected] Advised by: Professor Steven J. Miller.

Mathemagic: what happens at the intersection between magic and Math?

Talk by: Kush FanikisoEmail: [email protected] by: Professor Steven J. Miller

Key ideas

The mathematical basis of the trick is that for most orderings of the deck most secret numbers produce the same final key.

The truth is that I am not trying to read you, I am playing the game myself and my last card is probably your last card as well.

The full technical analysis of this trick uses Geometric distributions and Markov chains to calculate probabilities.

Main idea behind the full proof

Lagarias, Rains and Vanderbei use coupling methods for Markov chains to obtain upper and lower bounds on the coupling probability, which represents the failure probability of the Kruskal Count trick.

They imagine that they have two independent Markov chains on a finite state space and then find reasonable bounds on the likelihood of those two chains intersecting.

The last thing they do is they run 106 simulations to see how reasonable their calculations are.

What is the probability that I succeed?

Lets make the assumption that the two chains are equally likely to intersect at any of the 52 cards with a constant probability p2.

Let q = (1 – p2) be the probability that the two chains don’t intersect at any of the 52 cards.

Let I ∈ {1, 2, 3, … , 52} be the random variable which is the card position at which they intersect.

This gives us the standard geometric probability distribution:

P(I = r) = qr-1p2 for r = 1, 2, 3, …

Useful result to keep in mind

€

ark=0

∞

∑ k= ark=1

∞

∑ k−1=a

1− r

We have that:

€

P(I > r) = (1−p2)x=r+1

∞

∑ x−1p =(1−p2)rp

1− (1−p2)= (1−p2)r = qr

The probability that I succeed is thus

P = 1 – P(I > 52) = 1 – q52

= 1 – (1 – p2)52

Now lets find a readsonable estimate for p

For each of the two chains the average step size is:

This naturally generates a probability of 13/70 and we call p this probability.

€

1+2 + 3+⋅⋅⋅+5 +5 +513

=70

13

So we now have

P(succeed) = 1 – P(I > 52) = 1 – [1 – (13/70)2]52

= 1 – (4731/4900)52

= 0.8388…

Table of Theoretical probabilitiesPicture cards

count asValue of p Magician

starts with any of the first 10 cards

Magician starts with first card

11, 12, 13 1/7 65.77 70.05

10, 10, 10 13/85 70.80 74.67

5, 5, 5 13/70 83.88 86.41

1, 1, 1 13/58 93.15 94.40

Letter count 1/4 96.51 97.21