this post was submitted on 01 Jan 2025
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Autism

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[–] CountVon 20 points 1 week ago (2 children)

After thoroughly shuffling, the exact order of the deck is one of 52! (52 factorial, or 52 * 51 * 50 * ... * 2 * 1) possible combinations. That is such a large number that it's possible, even likely, that the exact ordering of your deck has never existed before and will never exist again.

[–] [email protected] 8 points 1 week ago* (last edited 1 week ago) (1 children)

True, but due to the Birthday Paradox, the chance of any two people shuffling a deck the same way at some point is a lot higher than you might think.

[–] [email protected] 4 points 1 week ago

In the future you can denote your factorial with 51! through single inline backticks for clairity!

[–] [email protected] 10 points 1 week ago (2 children)

It’s also not sufficient to randomize a deck of cards using a 32-bit seed as was once common in software.

Indeed, even with a 64 bit seed, it is not sufficient.

[–] [email protected] 12 points 1 week ago

Some quick math tells me you need 256 bits. Big numbers are wild

[–] mindbleach 1 points 1 week ago

x = new Array(52).fill(0).map( (v,i,a) => i+1 )

x.reduce( (a,b) => a + Math.log(b) ) / Math.log(2)

227.02369816459176

So it'd take 228 bits.

[–] [email protected] 10 points 1 week ago (2 children)

What if I just tossed the cards into the air and then picked them up? Would one toss be sufficient? 🤔

[–] [email protected] 8 points 1 week ago

No. Surprisingly much of the order is preserved.

[–] [email protected] 5 points 1 week ago

Only if you pick them up with your eyes closed

[–] [email protected] 7 points 1 week ago (2 children)

With or without cutting the deck in-between each shuffle?

[–] Chef 6 points 1 week ago (1 children)

That’s called a “box.” I believe the box/riffle action counts as one shuffle.

[–] [email protected] 1 points 1 week ago

And how many boxes make a riffle? I always do three or four between each riffle, and wonder how much it changes.

[–] [email protected] 6 points 1 week ago (2 children)

I always thought riffle shuffles were super ineffective. Most of the cards remain in each others vicinity. What is a better way?

[–] [email protected] 24 points 1 week ago (2 children)

You're essentially splitting the deck and recombining the two halves imperfectly multiple times in a row. Like if a riffle was perfect, you would get the cards from both halves equally distributed, but nobody can do it perfectly, so they actually end up properly randomized. After 7 imperfect riffles, the entire deck is unpredictable.

After 4 perfect ABAB riffle shuffles, you would end up with the same order as you started with. If your shuffles are imperfect, your deck becomes more random every time.

[–] [email protected] 11 points 1 week ago

I read ABAB as “Assigned Bitch at Birth” and immediately was like “hey that’s me”

[–] [email protected] 2 points 1 week ago* (last edited 1 week ago)

I don't know much about card tricks, just that many appear to use non-random cuts and those ABAB shuffles to get cards where they need to go. This one 'The Hotel' might even be easy enough for me to learn:

https://youtu.be/P-6gCH1hRGs

[–] [email protected] 2 points 1 week ago

It's the best way, period. Even as inefficient as it may seem, inexpertly done, seven shuffles and you're good.

[–] [email protected] 5 points 1 week ago

Numberphile has a bunch of videos on it, and yes 7 the accepted number because more shuffling doesn't increase the randomness in an effective manner.

https://www.youtube.com/watch?v=AxJubaijQbI

[–] [email protected] 3 points 1 week ago

I never learned how to properly shuffle cards that way. My hands just fail at the basic mechanics. Perhaps coincidentally, I would be mortified if something like that were done to the vast majority of the games in my collection. That ain't no $2 Bicycle deck, mi amigo.

[–] [email protected] 3 points 1 week ago

This is oddly satisfying to know. Thank you.