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Re: Entropy: sorted=0 unsorted=237



At 15:48 11 02 2000 , Leigh wrote:
>
>It would be hard to miss the Jokers in this group.

Then at 08:18 PM 2/11/00 -0700, Jim Green wrote:
>
Well, Leigh, for once I was being serious -- except for the 54, which I
assume is a typo.

It wasn't a typo. A standard as-purchased deck as two jokers, which is why
I wrote 54 factorial.

I think that there is a counting problem in the number of states in the
previous post.

Can you be more specific?

Is the number of states really 52!

Yes, if you get rid of us jokers. It's the number of permutations of N
symbols. Do the experiment with N=2 cards, then N=3 cards, etc. until
you're convinced.

and hence S= k lnOmega?

That's right. Or, if you want to measure entropy in units of bits, use
base-2 logarithms and lose the factor of k:
S = log_2(# of microstates)

And in particular 54! = 2.30844E+71
and log_2(54!) = 237.0638
unless my spreadsheet is lying to me.