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Comment by randombits0 | Hacker News Reader

What? No, you can store 52! unique states with a pack of playing cards. Just map each state to a data set. /j

`52!` is indeed between 225 and 226 bits of information.

    > 52!
    = 80,658,175,170,943,878,571,660,636,856,403,766,975,289,505,440,883,277,824,000,000,000,000

    > 2^225
    = 53,919,893,334,301,279,589,334,030,174,039,261,347,274,288,845,081,144,962,207,220,498,432

    > 2^226
    = 107,839,786,668,602,559,178,668,060,348,078,522,694,548,577,690,162,289,924,414,440,996,864

I used this fact in an interview ages ago. The interviewer wanted a function, in Java, that shuffled a deck of cards such that every permutation was equally likely. I pointed out this was not possible using the standard library random functions since the seed is a long (akshually... it's 48 bits).

They inexplicably hired my know-it-all ass...

You can show this with a easy one-liner in a lot of languages, e.g. Julia:

  $ julia -E 'log2(factorial(big(52)))'
  225.581003123702761946342444376665612911126819036757601937313805865615023286654

(It's also just math, of course, but it's nice to have a quick way to check these things.)

Since log(xy) = log(x) + log(y), you can simply calculate sum(log2(i) for i in range(1, 53)) :) might be a nicer formulation for when you don’t have arbitrary precision support.

Or you can use the `bc` that's shipped with macOS.

    echo 'l(f(52))/l(2)' | bc -l
    225.58100312370276194868

(`l(x)` is natural log, hence the `/l(2)` to convert to `log2`)

Annoyingly, GNU `bc` doesn't have `f(x)`.

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