I recently collected all 100 of the

series one Animal Crossing Amiibo Cards. Here is a look at my haul (Timmy is probably my favorite)...

Most people use these cards to interact with different video games that the cards are compatible with. Me...I store them in a box. Probably the more important question is why do I collect these cards in the first place? The short answer is that I am a nerd and we can leave it at that. Here is another painful reminder of my nerdiness, as I collect the Amiibo figures as well.

Here is a little background on the cards. The cards come in packs of 6 for *gasp* $5.99. For cards 1 through 15 of the series, you will get 1 card per pack out of the six cards. For cards 16 through 100 of the series, you will get 5 cards per pack out of the 6. Cards 1 through 15 are considered special and are sort of glittery (again...nerdy I know).

The cards are also readily available on eBay in exchange for various amounts of money. Some of the dealers sell

individuals and some make you choose

multiple amounts at a time. Either could be better depending on what you need.

When I started collecting these cards, I did not have a plan. I just started buying packs without thinking. To get all 100 cards, I ended up buying 37 packs (37 * $5.99 *1.07 tax = about $237). Within those 37 packs I didn't even get all of the cards. I ended up with 85 card and 137 doubles (

here are my doubles if anyone is interested...each red box is one card). To get the remaining 15 cards I went on eBay and averaged about $3 per card (cards 1 through 15 are a little pricier) for an eBay total of $45.

My grand total was $282. I did not plan out this experience, but it got me thinking...

**what would be the cheapest way to get all 100 cards?**
$5.99 per pack + tax gets you 1 of cards 1 through 15 and 5 of cards 16 through 100

Most cards 16 through 100 can be had on eBay for $2 per card

Most cards 1 through 15 can be purchased on eBay for $4 per card

Lucky for me, I will get to experience this card buying process again when

series two comes out in the near future. That's right, another series of 100 cards following the same format (it looks like there may be 17 special cards this time instead of 15).

I thought about this problem mathematically from a few perspectives, and I think your students would have a good time with this as well.

If I had purchased the entire first series on eBay...

15 specials * about $4 per card = $60

85 regulars * about $2 per card = $170

Total eBay amount = $230

Clearly I did not get a good deal in spending $282.

Although pretty much impossible, I thought about the fewest number of packs I would need to buy to get all 100:

I would need at least 15 packs to get all the specials and assuming I didn't get any doubles so far, I would be at 15 special cards and 15 * 5 = 75 regular cards for a total of 90 cards. I would then need two more packs to get the remaining 10 regular cards (and I would have 2 special cards as doubles).

17 pack * $5.99 * 1.07 = about $109

I would probably have a better chance of winning the lottery than getting all the cards in 17 packs, so this isn't really feasible, but it does give us a minimum amount of $109 and we already figured out a maximum amount of $230 through eBay.

Now we need to somehow find that sweet spot. When do you stop buying packs and start hunting on eBay? There are a lot of ways you could do this, and it would probably take smarter people than me to figure out an exact answer, but if you were proposing this problem to 6th graders, I would think more about presenting an argument at their level...and depending on the grade level there could possibly be a different argument made using a different math strand.

I am specifically thinking about those always difficult

6th grade statistics and probability standards that focus on data collection and being able to interrupt the data. For the data collection, I think this could be done by the entire class so a lot of data is present.

If you go to

random.org/integers, we can set up a simulation.

For the cards 16-100, it would look like this. The top number would have to be 5 times larger than the bottom number so it sorts it into sets of 5 cards

You will get a list similar to this:

You would have to do the same thing for for cards 1 through 15. The top and the bottom numbers would now need to be the same so you get only one special card per column.

You'll end up with something like this:

Now you can collect your data. How many packs in your situation would it take to get all 100 by just buying packs? If each student did this twice, you would have a lot of data to draw from. From my experience with middle school students, you wouldn't want to have each student do this 100 times, but if you did it twice I feel like they would be excited to see if they could take the fewest packs to get all the cards.

Once you have the data, it would really be up to the student(s) to take what you have most definitely been talking about with all of these data concepts and have them make an argument on what they think would be the best way to get all the cards with spending the least amount of money.

Another factor I will be throwing in myself is selling these cards on eBay to make some of my money back, and this could be a factor you use as well. While doubles are not desirable, it does give you plenty of chances to make some of your money back. Do I try to undersell everyone and chance $1 per card (maybe $2 for special cards)? I won't make all my money back, but they may sell faster. Maybe I will do a normal or higher price and wait out the market? What about shipping? Most of the cards I got were just sent in regular envelopes and placed between two pieces of cardboard making the shipping cost cheaper...but if something happened to the cards during shipping you would get a lot of refund requested and lose inventory. There is also the flip side as well. You could also think about bundling the cards in sets of so many. The possibilities are endless.

However you choose to take this problem (I would have the student discover all of the ideas that I presented before doing anything else), there are many variables involved, which in my opinion makes for a good situation. There doesn't have to be one correct answer, the most important aspect is the level that you can defend your decision.

If nothing else this has caused me to think before I purchase series two. What is that sweetest sweet spot I can find? Maybe your student can help me save some money!

1:1 Mathematical Philosophy

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