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

Mathemagic : what happens at the intersection between magic and Math?. Talk by: Kush Fanikiso Email: kf2@williams.edu Advised 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.

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

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  1. Mathemagic: what happens at the intersection between magic and Math? Talk by: Kush Fanikiso Email: kf2@williams.edu Advised by: Professor Steven J. Miller

  2. 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.

  3. 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.

  4. What is the probability that I succeed?

  5. 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.

  6. 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, …

  7. Useful result to keep in mind

  8. We have that:

  9. The probability that I succeed is thus • P = 1 – P(I > 52) = 1 – q52 • = 1 – (1 – p2)52

  10. 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.

  11. So we now have • P(succeed) = 1 – P(I > 52) = 1 – [1 – (13/70)2]52 • = 1 – (4731/4900)52 • = 0.8388…

  12. Table of Theoretical probabilities

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