The next few days

[soullimbo]

Hey guys...I think I've found a way. We each buy one of the extra packs. We can then exchange cards between ourselves.

In theory, once the extra decks have been purchased, we will have this between us:

3 basic decks
1x blue
1x red
1x green

That's more than enough to make a decent "deck of 20" each which includes only "multiplication" symbols of one colour, and only numbers from the other colours.

Let me know what you think, not long to decide...

[switza]

I'm not seeing anything about color?

I was going to buy a 2 * and see what that gives.

[soullimbo]

The extra decks that you can buy with the garnets are all different colours.

The pack that costs 2 garnets is BLUE
The pack that costs 4 garnets is RED
The pack that costs 6 garnets is GREEN

In other words, if we buy the 3 decks between us, we can arrange the cards ourselves from the 6 packs (our original 3 grey ones and the 3 different colours)

So when we submit our 20 cards and he shuffles them, we won't see which cards they are, but we will wee which colour.

For example:
I take the 4 multiplication signs from our combined grey decks, and take only numbers from the other 3 colours
Tomcat takes the 4 multiplication signs from the green deck, and numbers from the remaining decks
Adam takes 4 multiplication from the Blue/Red decks , and numbers from the grey/green decks

If we divide the numbers fairly and evenly, we can all come out of this ok. We just need to agree for all 3 of us to be online at a particular time to sort it all out before the challenge starts

[switza]

My new deck is:

3, 3, 4, 4, 5, 5, 6, 6, 8, 9, -, -, -, ÷, ÷, +, +, +, x, x

[soullimbo]

Did you buy the pack for 2 garnets?

[switza]

The middle one- 4 garnets

[soullimbo]

I'll buy a pack for 2 garnets then ok

[Tomcat Murr]

The challenge is making she we all get something we at least won't lose, though.

3 basic decks = 6 multiplication symbols
blue = 2 multiplication symbols
red = 2 multiplication symbols
green = 4 multiplication symbols (?)

14 multiplication symbols.

So if I have cards from green, then symbols from blue and red.
Pete has cards from blue, and symbols from basic and green.
Adam has cards from red, and symbols from basic and blue.

In that example, Pete would have the lowest value cards overall, so it would make sense to have more symbols. Unless we give the most symbols to whoever has green to assure them the win, then the others to at least assure they don't lose - to make sure Nyx doesn't have the same idea (she will). Which is part of the catch here. I can buy the 6 garnet pack and see what happens. But the important thing is making sure we have enough multiplication signs to get everyone enough.  The problem is that no matter how you cut that, we'll each have number cards from the same sets as the multiplication cards, too - or from a different set, but requiring lower numbers (which is fine - still enough to not lose, if we can get it to alternate each open/pass).

Also, Adam - does your bought deck show as a different color after buying them? (or is that just in the description?)

[switza]

Mas can confirm but I'm thinking they're red. Hopefully my colorblindness doesn't adversely affect this.

[soullimbo]

ok, I bought a deck for 2 garnets, they are blue. So here is what we have so far :



0, 1, 2, 3, 4, 5, 6, 7, 8, 9, -, -, -, ÷, ÷, ÷, +, +, x, x

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, -, -, -, ÷, ÷, ÷, +, +, x, x

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, -, -, -, ÷, ÷, ÷, +, +, x, x



0, 0, 1, 1, 2, 2, 3, 4, 7, 8, -, -, ÷, ÷, ÷,+, +, +, x, x


3, 3, 4, 4, 5, 5, 6, 6, 8, 9, -, -, -, ÷, ÷, +, +, +, x, x

[Tomcat Murr]

This is my concern... Either you have ten x, or your have to use symbols and numbers of the same color - no matter how you cut it, that's unavoidable. Which means the risk of getting a hand where, say, only 1 or 2 symbols are used is a real possibility. And if your hand starts with an x, with 0 before it, you're screwed. I guess using + becomes more necessary. Also, 0 cards are dangerous, because if you get it in your hand multiplied by anything, you're suddenly at 0.

I asked Mas if we could flip the cards, like Jinho did, and Mas said no. Which is unfortunate, because that would be the most sure way to win here.

So, I guess (I'm lazy right now), it would have to be a mix of x and +, and just hope for the best from there. But ideally, to keep from losing *hard*, numbers and symbols should be separate colors (albeit I forgot about Adam's colorblindness - but we can easily look over his shuffled hand before he starts opening/passing and tell him the colors). Then there's the matter of making sure none of us have a 0 in our number hand. If anyone's not feeling lazy and wants to do all the math (how many + and x signs, and how to split them to not repeat that color with numbers in each person's hand)/what decks are needed and how to split them, by all means. It's not a sure way to win (maybe not even a likely way to win), but it is a *reasonably* sure way to not lose. I could buy a green deck, but I could also buy a red and a blue deck for the same price (or even 3 blue decks), if the goal is to to keep enough number/symbols available without overlapping the color for each.

There is a 99% sure way to win, but I think it would cost 14 garnets for just one person's hand, not taking into account each others' hands. There's also the option of trading with Misty and Rachael (I'm sure Nyx wouldn't fall for it) - for example, offering them a 0 and a 9 in exchange for 2 3s (ie: getting rid of the 0) - but that's still down to what cards someone opens.

[soullimbo]

No , you only need 4 "x" in your hand. You get 3 chances to have them shuffled. Which means that at least 1 will be ok i.e where one of the x isn't the first card. Here's the example I propose:

These are the cards I would submit
GREY : x x x x
BLUE : 8 7
RED : 9 8 6 6 5 4 4
GREEN : (not sure what's in that deck, but it doesn't matter for this exercise) - - ÷ ÷ + +

As you can see, it's the grey/blue/red cards that are useful. The green ones are there as "throwaways"

You (tomcat) could go with this:
Green : xx
Red : xx
Grey : 9 9 9 8 8 8 7 7 7
blue : - - - ÷ ÷ + +

Adam could go with:
Blue : x x
Grey : x x
Red : - - - ÷ ÷ + + 3
Green : 8 numbers

Of course, this is assuming there are 2 "x" in the green deck. If I remember the show, the most exepnsive deck also included "10"s . As you can see, each one of us has "throwaway" cards, cards that we know to avoid at all cost. The only problem is if 2 "x" are next to each other after being shuffled, but like I said , you have 3 chances of re-shufling, so you would be very unlucky to have them next to each other 3 times in a row!

[soullimbo]

So here's what I think it might look like in my case, after the cards are shuffled:
B=x
B= numbers
B= numbers
B= Throwaways

B B B B B B B B B B B  B B B B B B B B B

These are the required conditions for my scenario to work:
The first card can't be "x", and there has to be at least one B/B before the first x
Must be at least one blue/red card between "x"s
"x" can't be next to each other


So in this scenario, I have to open the first card, then pass the next 2 cards, then open the first x. After that, I need to make sure that I open 2 numbers next to each other in the middle, because at the end, I will have to open the last 2 cards x/B


I've underlined which cards I would open


So My 10 chosen cards would be
B x B  x B x B B x B = 

[Tomcat Murr]

Ahhhh... throwaways didn't occur to me as a way to fill out the deck. Good thinking.

The shuffling is an issue. And you have to make sure the shuffle doesn't require you to pass 10 too early, being stuck with the remaining order/cards.

[Tomcat Murr]

3 star deck – 7, 7, 7, 8, 8, 8, 9, 9, 10, 10, -, -, ÷, ÷, +, +, +, x, x, x

And yes, Mas posted it in green.