What is Set?

Set is a card game where you try to find "Sets". The more sets you find, the better. First, we need to define a set. A set consists of 3 cards, and must have either every property in common or every property different. So, two cards in a set can't have one property different than the third card in the set. The properties are Color, Shading, Shape, and Number. The Colors can be Purple, Green, and Red. The Shading can be solid, stripes, or blank. The Shape can be an S, a diamond, or an oval. The Number can be 1, 2, or 3. It's a bit easier to explain by using examples.

Shape- All 3 are different
Color- All 3 are different
Shading- All 3 are different
Number- All are the same; they have 2 numbers

Shape- All are the same; oval
Color- All are the same; red
Shading- All are the same; blank
Number- All are different

Shape- All 3 are different
Color- All 3 are different
Shading- All are the same; stripes
Number- All 3 are different


Shape- All 3 are different
Color- All are the same; they are purple
Shading- All 3 are different
Number- All are the same; they have 3 numbers


Shape- All 3 are different
Color- All 3 are different
Shading- All 3 are different
Number- All are the same; they have 3 numbers

So, now that you understand what a set is, we can play the game. Here's an example game below. There are 6 possible sets. The answers are further down.

Here's all of the possible sets:

2-3-8 5-3-1 5-6-8 1-2-9 3-9-6 8-10-11

I want to try and prove what is the largest amount of sets you can get from 12 cards, if all of the cards are different. 

2 of the 4 characteristics can be represented, with the 2 being shading and number. The shape and color stays the same.The first idea that comes to mind is finding the maximum amount with 9 cards; that fits nicely with the 3 different characteristics in each group. I think it would look something like this:









So, if we have these 9 cards, there would be 12 sets. Seems like we will have 12 sets when we have 12 cards, right?
Well, 12 cards will be harder to account for. There aren't 4 characteristics in any of the groups, making it difficult to choose which cards to use for the remaining 3 spots.
And, what's worse, we can't really just add on something like this, because it wouldn't work well with the red.

So, If this were added to the set, you would only be adding 1 set to the group, which is all of the 3 greens together. You can't mix the red and green together, because there will always be 2 red cards and 1 green card, or 1 red card and 2 green cards, which both can't make sets.

I think there is one better option than this;

I believe that this is the best thing you can add on to the 9 cards, it creates 2 extra sets instead of 1. I'll show how the sets are made.

This should be the best 12 card game.


The 2 new sets that would be created are these:

These sets kind of solve the problem of having only red and green cards. We add on a purple card with 3 shapes instead of 2, because if it was purple with 2 shapes, it would only create 1 set.

Now, we can create cards each with 1, 2, and 3 shapes, and different colors.

The reason why we don't have a set with the 1st card being a striped red diamond and the 2nd card being a solid green card is because we don't have enough cards to add on; we only have 3 extra cards to add, even though we really want to add 4. So, if we had a 13 card game, we would be able to get an extra set added on.

So, I believe that a 12 card game can have a maximum of 11 possible sets, if all of the cards are different. 

So, it looks like we're all done!

Well, there are a few more things we could do...

What if a set could have cards be the same as each other? What if every single card was the same??? How many sets could be made?


Here's an example of having every single card be the same:

I know. It looks absolutely crazy hard to understand.
No sarcasm whatsoever at all.





This is actually pretty easy; there are 12 cards to choose from at the beginning, and after selecting your first card, you have 11 cards to choose from, and 10 cards which could represent the last card.

So, really what we're doing is 12*11*10, or just multiplying together all the possibilities of picking the 12, 11, and 10 cards available, resulting with 1320 possible combinations.

There was actually a formula created to do this easier, called the permutation formula. 

This is the formula, where N is the number of cards and R is the number of things you are choosing from.

P(n,r) = n!/(n-r)!

So, if N is 12 and R is 3,
just put it into the formula, and solve.


P(12,3) = (12!)/((12-3)!)

P(12,3) = (12!)/(9!)

P(12,3) = (12*11*10*9!)/(9!)

P(12,3) = 12*11*10

P(12,3) = 1320

Hm... seems oddly familiar, right?

Really what this equation was doing was setting up the 12*11*10.

It uses the "!" factorial to do this, and does it well. 

Basically, it takes the top number 12, and says that it should multiply numbers 12, 11, 10, 9, 8, etc.

So, the 3 tells it to stop at 10, so that it only multiplies 12*11*10, which is what you want.

 Yes we're done.

Becoming 14 (Random Thoughts from a dumb 13 y/o kid)

Well, these are my last few minutes being 13. The start of being a teenager and going through puberty has been... covered with dots. My face looks so disgusting. Wish puberty would be faster, I guess. The last year in the AC was pretty neat. I'll have the luxury of walking around outside and being in high school, but I'll never know what it's like to be an Ackie again. Time goes by, I guess. Life has been rough recently, but I hope it gets better. It had ups and downs, you know? This year has had a lot of firsts, as usual. Especially starting to draw in a Donut style. Seeing my art style change since the beginning of the year surprises me a bit. My new friends have introduced me to a lot of cool new things this year too. I'm too tired to type more. Um, grats on getting older, me in 40 min!! See you soon! Or rather, Be you soon!

Beat Dr. Nim! (Part 1)

YSP is a math summer camp that I go to with some of my friends from school. We learn about lots of different math, more specifically game theory and mathematical investigation around number theory.

Yesterday we looked into a game known as Nim. The setup of the game is putting sixteen objects down, with the last object being different from the others. Some gamblers may want to put money as a bet under the last object.

Here's an example with matchsticks. The top one is the last object.

The game begins with two players, who take turns taking 1-3 normal objects from the pile. The person who picks up the last, special object wins.
The game is seemingly simple and normal, like Tic Tac Toe, but there is a catch: There is a way to win every single game you play.

In fact, Nim has a winning strategy that's so easy to follow, it was even implemented in one of the earliest computer games. It was called "Beat Dr. Nim", and was a simple, plastic board game. It had the same rules as Nim, and let the player go first. But, no matter what you did, "Dr. Nim" would always beat you, as long as you followed the rules (The game did have an option to make Dr. Nim beatable, albeit difficult.)

Here is the back of the box of "Beat Dr. Nim".
It has a picture of the computer game,
 (which you can learn more about -> here <-),
and also a somewhat casually racist
 depiction of "Dr. Nim", although
Nim most likely originated in China
a long time ago.

So, would you like to hear the strategy? It's really simple; As long as you go second, you must take the objects in groups of 4. So, if your opponent starts by taking 3 objects, you take 1, if your opponent takes 2 objects, you take 2, and if your opponent takes 1, you take 3. No matter what your opponent does, you will always be able to group the objects in fours. So, on the last turn, your opponent will have 4 objects left, but will only be able to grab 1-3 of them, guaranteeing you a win. 

Yay you win Nim!!!
Actually, cool fact: NIM upside down looks like WIN.
The makers of "Beat Dr. Nim" noticed this, and you
can see it in the box art for the game a bit easier.

At YSP, they changed the game up a little bit. They called it "Race to 100", in which you had to take turns increasing a number from 0 to 100. You are allowed to add 1-10 to the number, and each player takes turns.
So, if you want to use the same strategy again... if the 1st player can add 1-10, then the number you will always be able to add up to is 11; if they add 1, you can add 10, if they add 2, you can add 11, etc., until you reach if they add 10, you can add 1. So, you can go in steps to reach 100. 

You will end up having the numbers 11, 22, 33, leading up to 99. Wait... if you end with 99, your opponent will just say 1, reach 100, and win the game! And you can't guarantee that you can make a number larger than 99. So, this strategy is flawed. Right...?

(TO BE CONTINUED)

(here's a dramatic sound effect to set the mood)