We are playing a version of Canasta.

Tonight we had 6 players and 4 decks. Everyone playing individually. The goal is to get the most points which is most easily achieved by getting a “canasta” - i.e. 7 of the same number including wild cards (2s and jokers). As you play, you lay down your sets (3 or more cards). So you typically know which card number someone is targeting for his/her canasta.

The question is whether adding a 5th deck speeds up the likelihood that someone will get a canasta faster.

Two competing hypotheses (which may both be wrong):

  1. it does not change anything because the odds for which cards you pull should not change when you add another deck that has the same odds. It will be more likely someone gets a canasta with more decks but it won’t change how fast it happens.
  2. it does change because there are more of each card (going from 4 decks to 5 decks means going from 16 to 20 of each number). Having more of each number means that more than one person can successfully get a canasta for the same number.

Does anyone have a model for how to think about this?

  • A couple of things to remember is that a lot will depend on the shuffle and deal and more importantly having more cards to legally play means that more people can attempt to complete a canasta on the same value card which can have an impact. For example with your current setup 4 players can get a canasta in a single value (4 natural + 3 wild) but with each deck added another player is able to complete a canasta in that value. – Joe W Feb 10 '19 at 21:00

Adding an extra deck definitely increases the chance that someone will get a Canasta because cards are drawn without replacement.

The point is easiest to see via a drastic oversimplification. Take a number of decks, take out six cards that are the same from those decks, shuffle the remaining cards together, and then try to draw a seventh card to match the six. Each deck you add contributes 54 cards in general (because of Jokers), 10 of which match (because 2s and Jokers are wild), but you have to subtract 6 from both the card total and number of matching cards for the six cards you have taken out to start. The probability that the next card you draw matches the six you have of that with various deck sizes is as follows:

1 Deck: (10 - 6) / (54 - 6) = 8%

2 Decks: (10*2 - 6) / (54*2 - 6) = 14%

3 Decks: (10*3 - 6) / (54*3 - 6) = 15%

4 Decks: (10*4 - 6) / (54*4 - 6) = 16%

5 Decks: (10*5 - 6) / (54*5 - 6) = 17%

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  • Of course, there will never be a case of only one deck! – user45266 Feb 12 '19 at 6:56
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    @user45266 That was just an extreme case to highlight how choosing without replacement works probability-wise – Zags Feb 12 '19 at 11:53

Increasing the number of decks should create a Canasta sooner, but I can't give you precise numbers.

When you add more decks, there will be more of each card – and since the decks are randomized, there will be at least one rank that ends up concentrated in the top half of the deck. So while all of the aces might be at the bottom, and all of the fours evenly distributed, all of the tens might be at the top – and if not the tens, then perhaps the kings. Something is likely to end up concentrated near the top of the deck, which will make it more likely that someone plays a Canasta.

I wouldn't expect this to make a big change in game length, but it should shorten the average game a little.

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