this post was submitted on 01 Jan 2025
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[–] [email protected] 21 points 1 month 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 month ago* (last edited 1 month 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 month ago

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

[–] [email protected] 10 points 1 month 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 month ago

No. Surprisingly much of the order is preserved.

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

Only if you pick them up with your eyes closed

[–] [email protected] 9 points 1 month ago (1 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] 13 points 1 month ago (1 children)

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

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

I got 226 bits? log2(52!) right?

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

Yeah looks like I did something wrong. Must have just been looking at the numbers raised by powers of 2. Good find! Thank you

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

No problem! It only stood out to me because it'd be an amazing coincidence if 52! fit snuggly into 2^8. Being a nerd, I have many powers of 2 memorized!

With a little playing around, it seems like 57! is the last whole number that fits (254.5 bits needed) so you could add a couple of jokers in or even use a tarot deck's minor arcana (56 cards) and have room with a 256-bit wide BigInt, with a few bits to spare!

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

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

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

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

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

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

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

This is oddly satisfying to know. Thank you.

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

I've always found this factoid pretty dubious. First off it's statistical so there are no absolute truths or facts, only tendencies.

But do this experiment for yourself, I just did a few times: take all the kings and aces out. Order them (I used Spade, Club, Heart, Diamond, Aces first then kings) and put them on top of the deck. Now do 7 riffles and look through the deck.

What I found is that the 8 chosen cards are still weighted towards the top of the deck (a little less than half moved to the bottom half, but none to the bottom 1/3rd). The suits got shuffled a little but all the spades and most clubs were still in the top 1/3rd of the deck. Most of the aces and kings stayed together in close pairs or triplets, within 5 cards of each other.

So no, 7 riffles isn't realistically good enough for fair play at actual tables. You need to mix in some cuts and pool shuffles to break up the structure of the deck and properly distribute the cards.

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

I do think cuts are necessary, and I use them each time. If each riffle includes a cut I feel this factoid is undoubtedly true, period.

Therefore, I think the definition of riffle must include a cut.