How should we be shuffling these cards?
September 22, 2021 7:39 PM   Subscribe

My mom and I routinely play a version of Canasta that needs six decks of cards. At her house, we shuffle them using a six deck shuffler but at my house, where we play much less often, we don't have a shuffler. How can we best shuffle the cards given that we can't shuffle them all at once?

We find shuffling the decks five times in the shuffler to be a good shuffle. We would like to know how to get the equivalent shuffle when shuffling by hand. Is there math for this?

When we hand shuffle, we can shuffle about one deck worth of cards into another deck woth. So at any given moment, each of us is shuffling two decks and there are two decks on the table. We usually shuffle the same two decks against each other 3-4 times before switching half the cards out for one of the table decks. We continue that until we get really bored of shuffling and then we stop.

How many iterations of shuffle-shuffle-shuffle-switch do we need to do to approximate five full deck shuffles? Or is there a better / more efficient way to shuffle multiple decks?

This isn't exactly an earth shattering question, I realize, but my mom has enough arthritis in her thumbs that if we are wildly over shuffling it would be nice to rein it in a little.
posted by jacquilynne to Sports, Hobbies, & Recreation (14 answers total) 1 user marked this as a favorite
If you gave me that task, I would probably emulate a blackjack dealer. I’m not a pro dealer, but if my memory serves, it looks something like:

Take the six deck shoe, cut it at an arbitrary point, and then put the bottom cards on top.

Take the shoe, and make 6 deck-ish sized piles.

Take pile 1 and 4, and wash, riffle, riffle, riffle, strip, riffle, then use that to start the shoe
Take pile 2 and 5, and wash, riffle, riffle, riffle, strip, riffle, and add to the shoe
Take pile 3 and 6, and wash, riffle, riffle, riffle, strip, riffle, and add to the shoe

Cut the shoe at an arbitrary point, put the bottom on top, and now ready to deal.

If I was trying to accelerate this for home use against family, I might try skipping the wash, since that takes (at least for me) at least as much time as the rest of the shuffle process.

Although having typed that, I see that there are also numerous YouTube videos available for blackjack shuffling that might be easier to understand. (None linked, because I did not watch them, and can’t recommend any specific vids).
posted by whisk(e)y neat at 8:07 PM on September 22, 2021

I'm sympathetic, being an obsessive shuffler myself.

There's a chapter in Shape by Jordan Ellenberg (aka MeFi poster escabeche) that covers the single deck version of this. He says six shuffles on a single deck essentially gets you an infinite 'distance' from a sorted deck: there is no possible order of cards you can't reach in six shuffles. So your instincts are not bad. I'd recommend the book over the paper but the reference is here.

Unfortunately I don't think you're overshuffling in mathematical terms when you do it by hand. You are probably undershuffling if my intuitions about how more cards and the partial deck shuffles affects the math. So there won't be any proof you can do less than you are doing.

But--and this is important--only in the sense that if you know the starting distribution not all ending distributions are equally likely. If you're not carefully counting cards and thinking about distributions carefully, I'd say as long as each section got six shuffles you're in good shape. You won't really notice any patterns. I doubt you'll even be able to draw conclusions like "I just saw three aces of spades in the first 15 cards, that guarantees it will be at least another 130 before I see another one."
posted by mark k at 8:44 PM on September 22, 2021

Response by poster: There's no issue with counting cards and trying to predict where they are. The only card counting we do is when we know X number of a necessary card (2s, 8s, 9s, 10s) have already been played and so there are none left. But that's based on the cards being revealed in the game, not on knowing where they might be in the shuffle.

The real issue is that when you finish a game of canasta, you end up with a lot of those necessary cards split neatly into little piles of 7, while the bad or less good cards, especially 3s but also jokers and 4s and 5s tend to be together in the discard pile. So what we want is a shuffle that integrates these cards back together in a way that's at least somewhat evenly spread apart so we don't end up getting all the threes at roughly the same time, or all the 8s, 9s and 10s early in the game and very few later in the game.
posted by jacquilynne at 5:37 AM on September 23, 2021

If you have a clean flat surface available, “smooshing” for one minute should get you there.
posted by doctord at 6:30 AM on September 23, 2021 [3 favorites]

Best answer: To integrate, you could start with a pile shuffle before adding in the actual randomizing part of process. Something like split one of the groups of 7 into five piles, three w/1 card and two with 2 cards. Then add a layer from the discard pile, then add another from the groups of 7, this time starting with pile three (the first pile short one card). I have no idea of ratio of discard pile to "good" piles so adjust as makes sense. The procedure doesn't have to rigid just equalish. Once all the cards are distributed into the 5 (or whatever) piles, randomly riffle/smoosh them back into larger piles and add some cuts.

This is often used in Magic The Gathering to integrate mana and spells so googling mtg pile shuffle technique might be useful, adding the technique bit because the top hits otherwise will be discussion of the legality of pile shuffling.
posted by Press Butt.on to Check at 6:36 AM on September 23, 2021

A "wash" shuffle or "smooshing" as doctord's link calls it, is easy and effective. Casinos use it. Kids use it. I use it whenever I have a nice table to use. Also, it's arthritis friendly, at least as far as my particular joints are concerned.
posted by mrgoldenbrown at 7:39 AM on September 23, 2021 [1 favorite]

I believe that canasta is similar to hand and foot, which I enjoy playing. We use 5 decks. There's no way I'm shuffling 5 decks. What we do is to each take a deck-sized stack and give it a few shuffles. Then we spread all the cards out on the table. Commence lots of moving of the cards around--shuffling on the table so to say. Then we each pick our 13-card hand and foot. When we pick, we're picking randomly from around the table, so I feel that our hands are truly random. You could pick up the cards and deal if you prefer.

Upon preview, I've learned this is called smooshing.
posted by hydra77 at 9:20 AM on September 23, 2021

Why not just deal each card from a random place in the big deck-o-cards. That way it doesn't matter if the cards-in-order are sufficiently random.
posted by zengargoyle at 4:57 PM on September 23, 2021

Shuffle as easy enough into 6 piles. Take a chunk from the top of each pile and shuffle them to make a 7th pile. Pick the top card of the 7th pile and then 2,3 pick from first pile, 4,5 pick from second pile, etc. Ace is picker's choice of pile.

Shuffle as best you can, pick top card, estimate the number on the card as a 1-13 index into the rest of the cards and pick from around there, slip first card back into random place. Or do colors, or suite, or some combination. Red = top half, Black = bottom half, Spades/Clubs,Hearts/Diamonds = another top/bottom of the current half (a quarter of the deck), number = guess about where to pick the 1-3 or whatever from that quarter of the deck. Put card back in a random place.

You don't need to shuffle and pick from the top down. Create some randomness and use that to pick from the deck. This is probably about the same as smooshing just with less space.
posted by zengargoyle at 5:14 PM on September 23, 2021

Oh, How did the impossible Perfect Bridge Deal happen? - YouTube.

(I'll hush now)
posted by zengargoyle at 5:28 PM on September 23, 2021

Response by poster: With the number of cards we deal with and the space we have available, I think the idea of intentionally creating decks with a mix of good and bad cards in them is probably our most workable solution. Fortunately, the point counting process already creates a large number of piles of 'good' cards with slightly variable numbers of cards in them, so topping them with discards and then assembling them into decks to shuffle should help.

I'm still interested in the math of how to create the equivalent of 5 full shuffles by shuffling sub-decks against each other if anyone has any idea how to figure it out. But it's more of a curiosity interest than a practical need.
posted by jacquilynne at 5:20 AM on September 24, 2021

The math is impossible because RADOM. It's our human failing at perfection that makes shuffling work. It's complicated and even the 'six' shuffles for suitably random is more like a statistics guess. Don't overthink random.

Assumptions made...

You have 6 decks and can shuffle 2 decks at a time. So there are 3 piles of cards.
Both players can shuffle. Each picks a pile and shuffles a bit leaving the 3rd pile on the table.
Each shuffler cuts their pile in half and makes a new deck out of the parts leaving each with one-half of a pile. The old 3rd pile is cut in half and given to each shuffler.

We're back to each shuffler has a pile and one is on the table... Repeat 6 times so that each shuffler sorta gets to fondle each virtual-ish-pile and each gets to pick the top/bottom cut points in the rotation. Then just swap out who gets to stack the ending 3 piles of cards.

This general concept is based on the of folding steel for a a sword or mixing ingredients into dough by folding, extending, and folding, and repeat a few times an even distribution is reached.

Hail Eris!
posted by zengargoyle at 3:10 AM on September 25, 2021

Best answer: So... I sent a Hail Mary email to a mathematician that I've followed their blog for ages and even met them once and got a book signed. They don't remember me (natch). But they did respond with a "not exactly my field but..." sort of answer.


Take the 6 deck stack and split it in half. Take each stack top to bottom and turn into six piles. So on one side you have A (top sixth), B, C, D, E, F (bottom sixth). The other stack is the same G (top), H, I, J, K, L (bottom). Shuffle F and L to make FL, shuffle E and K into EK, ... DJ, CI, BH, AG. Stack them up AG on the top to FL on the bottom. Repeat the cut in half, make sixths, shuffle, stack. That's how an automatic card shuffling machine does it, so you have to do that 5 times if that's what you find satisfactory.

That's a total of 30 single deck shuffles (6x5). If you can do two deck shuffles then split each half into 3 and you end up with 15 double deck shuffles. This is all just mimicking the actions of an automatic card shuffler.
posted by zengargoyle at 3:54 PM on September 25, 2021

Response by poster: That makes sense, zengargoyle, thanks! And 15 shuffles of 2 decks each is actually quite a lot fewer than we normally do, so even if we don't do exactly that method, we could probably dial it back considerably.
posted by jacquilynne at 5:23 AM on September 27, 2021

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