No two card shuffles will EVER be the same.

HungLikeJesus · 02-05-2008, 10:30 PM

I'd better present my logic to see if anyone disagrees.

(n*(n-1))/(2*52!) = P; solve for P, where P = probability of 2 or more matching shuffled decks in n shuffles.

(n-1)^2 < n*(n-1) < n^2, so

(n-1)^2/(2*52!) < n*(n-1)/(2*52!) < n^2/(2*52!),
so the first equation can be simplified (for large values of n) as

n^2 ~= P*2*52!

n = square_root(P*2*52!)

If P = 0.5 (50%), n = 8.98e33
If P = 0.1 (10%), n = 4.0e33
If P = 0.01 (1%), n = 1.27e33

I just noticed a flaw in this logic. Can you see it?

02-05-2008, 10:30 PM	#11
HungLikeJesus Only looks like a disaster tourist Join Date: Feb 2007 Location: above 7,000 feet Posts: 7,208	I'd better present my logic to see if anyone disagrees. (n(n-1))/(252!) = P; solve for P, where P = probability of 2 or more matching shuffled decks in n shuffles. (n-1)^2 < n(n-1) < n^2, so (n-1)^2/(252!) < n(n-1)/(252!) < n^2/(252!), so the first equation can be simplified (for large values of n) as n^2 ~= P252! n = square_root(P2*52!) If P = 0.5 (50%), n = 8.98e33 If P = 0.1 (10%), n = 4.0e33 If P = 0.01 (1%), n = 1.27e33 I just noticed a flaw in this logic. Can you see it? __________________ Keep Your Bodies Off My Lawn SteveDallas's Random Thread Picker.

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