Try all the ways to play - The Tower, The Well, Hot Potato, Poisoned Gift, Triplet
Been playing this with my kids for years. It’s a lot of fun and it’s relatively fair play, too. My 3 y/o keeps up with my 9 y/o and me without us giving her any extra peeks at the cards or other handicaps.
Bonus: the small tin it comes in is free of the thousands of packaging wires and ties that come standard with nearly every other toy.
My family has been enjoying this game for a couple years now. It’s cool. I probably don’t have the skills to fully understand them but I think the maths behind the concept must be fun to work through.
Fantastic for all ages, and in a really small package. Would play it while waiting in line for events at PAX every year. People invariably would see how much fun we were having and hop in.
Just fantastic.
My kids love it, and we have great fun playing together, it’s just a little beyond my 4 year old, but it’s a game were letting him play (his style) works too. Very highly recommended, especially since you only need to match symbols, and understand the rules you are playing with.
It’s a great game to keep on the coffee table for guests to pick up and play on the spur of the moment.
It’s also great for traveling: leave it with your hosts or some other nice person you met along the way. Language differences not a deal-breaker.
We just got this for our kids this past Christmas. Our four-year-old whips our eight-year-old at this.
Come to that, the younger one is playing with her older sister’s snap circuit set and figuring out by trial and error how to get things to work. She just called out to me and beamed about getting the fan and light wired in series. Oh, kid, don’t ever change.
SpotIt Jr. has only 31 cards and can be easily played with younger kids. We got the Alphabet version when working on letters with my kids.
BTW, you may want to correct that “whomever” typo.
I tried to make a version of it on Powerpoint for my English classes, but I couldn’t work out a good way to make sure there was one and only one match between cards. In the end I had to do it by trial and error, but I was annoyed at myself for not being able to work it out properly.
We love this game, and I’ve been trying to work out the math, too. Can anyone show us the algorithm? How do you ensure one (and only one) match in X number of cards with X number of items on each card?
We love SpotIt as well. But probably our most-played game is Eleminis. And you better believe a 4-year-old knows about defensive and offensive (highly offensive, that little scamp!) play. Fast and Fun!
You want to check out combinatorial design theory. Among other things, it tells you if you want a collection of subsets of N elements such that every pair of subsets have exactly one element in common (and every pair of elements is in exactly one set together), then N must be of the form M^2 + M + 1. So it’s not a coincidence that the deck has 31 cards.
There are actually 55 cards and 57 objects altogether, but not all objects have the same number of occurrences. I thought it would have something to do with this, but I didn’t want to review 50+ words each time. My idea was to have a folder full of pictures of vocabulary we had studied in the last week or so, just named “01, 02, 03…” I could then make the game on ppt with the pictures as links rather than importing them, then repeat the game a few weeks later with new vocabulary, without changing the “one and only one match” rule.
This site has a discussion of the maths, and one of the explanations arrives at n^2 - n + 1.
I derived this formula logically but not necessarily mathematically as follows:
I picked a random card and focused on a single image. Assuming eight images per card as are found in this game, this image can only be found 8 times, once on the card you’re holding and 7 more times.
The same holds true for the next image. It can only appear 8 times if it to remain unique - once on the card you are holding and once over each of 7 more cards.
I noticed the trend. Each image appears once on the card you’re holding and requires 7 more cards. So, you need the 1 card you are holding and 7 more per image. Mathematically, I guess that’s: 1card+(7cards×8images). That’s 1+(7×8) or 1+56=57.
Nice idea, but awfully dear!
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