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I played my first game of Swish this evening and the other player was surprised when I asked if 2 sets of 2 card swishes are a valid 4 card swish. That is - can you pick up 2 pairs of cards that each are 2 card swishes and stack them together into a 4 card swish, so long as every ball swishes into a hoop of the same color and there is no overlapping as you visually see 4 balls in 4 hoops?

The rules don't specifically comment on this, so I am assuming yes. But given the other player's surprise, I wanted to make sure.

Of course, the same question can be extended to 2+3=5, 2+2+2=6, etc. I asked the question with 2+2=4 in order to provide a specific example.

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2 Answers 2

up vote 4 down vote accepted

I decided to email support at Thinkfun.com with:

I played Swish last night for the first time and was unclear about something after both reading the rules multiple times and viewing your video:

The question is whether a pair of 2-card-swishes constitute a valid 4-card-swish, so long as the hoops and balls don't overlap among all 4 cards. The rules don't comment on this situation one way or another so I am assuming the answer is yes (and can be extended equally to 2+3=5, 2+2+2=6, etc., again so long as the hoops and balls don't overlap when you stack the subsets together.

Another way of asking the same question is if you have what seems to be a valid 4 card swish per the rules, is it still valid if the 4 cards can be rearranged into 2 sets of 2 swishes?

I asked this question on Stack Exchange (Board and Card Games) and the person who answered came to the same conclusion I did - that the rules as literally stated indicate this is a valid swish (see link beneath my signature). However, this sits uneasily with me because 2+2=4 swishes are easier to find than 4-swishes that can't be decomposed into 2 sets of 2. Yet the scoring system does not reward finding more difficult swishes.

If you don't know the answer, can you pass this on the game creators to find out their intent?

Here's the answer I received:

Below is the answer to your questions from one of our Production Team Members:

The short answer for his question is no, Two 2-card Swishes do not “count” as a 4-card Swish. But, they do count as a legal move in the game, and the person who finds two 2-card Swishes still wins 4 cards, just as many as a person who makes a 4-card Swish. So, to answer the second point, at first blush this would seem not to make sense, since as Joe points out, the scoring system does not reward the person who found the more complicated Swish. This is actually done deliberately. The reason is that the winner is simply the player who has won the most cards. We tested versions where harder Swishes (higher counts) where worth more, but we decided on this version because it allows multiple players, of multiple skill levels, to play an equally challenging game. As in, when families play, Mom or Dad can be limited to 3 or higher card Swishes, whereas Jr.(s) can go for any number. This way, all players will be challenged, and Jr.(s) have an equal chance to win the game.

We understand that for more experienced gamers, this will seem somewhat “off”. But this was the version that had the greater appeal, and we assume that people that really get into games like this can adjust the rules as they see fit, adding additional points for higher card Swish. I would suggest +1 for a 3-card Swish, +2 for 4-card, and so on. Keep in mind, though, that if you play this way, you have to keep all your Swishes separate as you win them, or else do the math in your head and remember it. This is another minor detail that can add complication, something that we try to avoid.

I interpret the answer as saying it doesn't count as a 4 card swish yet you can collect 2 2-card swishes simultaneously. It's phrased this way so that if an experienced adult is playing a kid using the variant that kids can do 2-card swishes while adults can't, then the adult is not allowed to simultaneously pick up 2 pairs for a 4 card swish, even if there is no ball/hoop overlap.

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Yes, but I agree that the examples aren't completely clear on this matter. It two 2 card Swishes could be overlaid into a 4 card Swish, it doesn't break the rules for "What a Swish is". The examples in the rulebook, and in the video are also no help in this matter, because the rulebook is too brief, and the video only shows examples of 3, 4, and 5 card Swishes being made from cards that need all cards in the set to make the Swish (which wouldn't be the case with a 2+2 Swish = 4 Swish, or 2+3 Swishes = 5 Swish. It is possible this isn't the designers intent, but based upon the rules as written and the fact that an FAQ does not exist that clarifies this question, I believe you should be allowed to claim a 4 card Swish from two independent 2 Card Swishes that can be laid on top of each other.


A Swish is created by layering two or more cards so that every ball swishes into a hoop of the same color. The cards may be rotated and/or flipped but must lay on top of each other in the same orientation, and no hoop or ball can be left unmatched.

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From your reply, I'm not sure you understood my question (maybe I should take pictures of 4 cards to illustrate). Let me state it in reverse. If you have what seems to be a valid 4 card swish per the rules you've quoted, is it still valid if the 4 cards can be rearranged into 2 sets of 2 swishes? –  Joe Golton Jul 12 '12 at 4:37
@JoeGolton, I did slightly misunderstand you. You are asking if 2 sets of two card Swishes that could be scored individually, can be scored all at the same time as a 4 card swish if the hoops and balls don't overlap among all 4 cards. I say the answer is probably yes, but you are correct that the rules don't give an example that makes this perfectly clear. The video doesn't help either youtube.com/… –  user1873 Jul 12 '12 at 5:33
I read over the rules several times and saw the video as well and ended in the same place you did. Probably yes. However, it does seem odd because it is easier to find 2+2=4 swishes than a 4 swish that can not be decomposed into 2 sets of 2. Meanwhile, the scoring treats these two situations identically. –  Joe Golton Jul 12 '12 at 14:19

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