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Tako

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From: Memphis, Tennessee, USA

Joined: 2015-08-10

Posts: 6,663

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Algebra problem.

This is the first time in my life I've needed to use some more advanced algebra in everyday life that's not related to school in any way.

Well, it's not exactly algebra, more like advanced problem-solving.

Here's the necessary information you need:

Deck = 30 cards
You have 5 red cards in your deck.
To play a red card, you need to first play 5 light-green cards.
You may have up to 25 light-green cards in your deck.
Your opening hand is 7 cards.

Now here's the problem:

If I need to play a red card on my first turn, what will my deck be composed of? You may use whatever colors of the rainbow you'd like, just as long as there's a strong chance I'll get at least 5 light-green and 1 red on my opening hand.

If you got the answer to that one, you're awesome. Now let me make it more complicated.

Green cards produce one coin per turn. You can have light-green cards, which produce a coin as soon as you play them, or have regular green cards, which produce coins once you're done with your turn. If I wanted to play a red card after two turns, how much green/light green would I need? One red card requires you to pay 5 coins.

If you got the answer to that, you're double the awesome. Now, let me continue the pattern.

If I wanted to play a red card after the third turn, how many green would I need in my deck?

I could probably figure it out, but I can't think thoroughly atm.

Problem: Opening hand = ?1r + ?5lg

7/30 == 1/4.3 (Opening hand / total deck)
30/4.3 == 6.9

6.90 == 1/4 of deck == 1 red + 06 light-green == 07
13.8 == 2/4 of deck == 2 red + 12 light-green == 14
20.7 == 3/4 of deck == 3 red + 18 light-green == 21
27.6 == 4/4 of deck == 4 red + 24 light-green == 28

Scenario deck: 4 red + 24 light-green
Scenario deck test results: 8 wins out of 10 draws

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If anyone wants to take a crack at the solution I'd be grateful.

My ultimate goal is to write an algebraic expression where people fill in variables to get the right deck, eg, "I want to draw 5 green then a card that costs 7 coins before my second turn" and it'll give them the right deck.

Last edited by Tako (Oct 4 2011 3:40:28 pm)

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Yeah, well, you know that's just like, uh, your opinion, man.

Onjit

Member

Joined: 2015-02-15

Posts: 9,710

Website

Re: Algebra problem.

mind = blown.

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:.|:;

noelzkie

Guest

Re: Algebra problem.

Haha. This is too easy.

I dont understand a thing. :'(

Bimps

Member

Joined: 2015-02-08

Posts: 5,067

Re: Algebra problem.

This is the first time in my life I've needed to use some more advanced algebra in everyday life that's not related to school in any way.

Well, it's not exactly algebra, more like advanced problem-solving.

Here's the necessary information you need:

Deck = 30 cards
You have 5 red cards in your deck.
To play a red card, you need to first play 5 light-green cards.
Your opening hand is 7 cards.

If I need to play a red card on my first turn, what will my deck be composed of? You may use whatever colors of the rainbow you'd like, just as long as there's a strong chance I'll get 5 light-green and 1 red on my opening hand.

If you got the answer to that one, you're awesome. Now let me make it more complicated.

Green cards produce one coin per turn. You can have light-green cards, which produce a coin as soon as you play them, or have regular green cards, which produce coins once you're done with your turn.

If I wanted to play a red card after two turns, how much green/light green would I need?

One red card requires you to pay 5 coins.

If you got the answer to that, you're double the awesome. Now, let me continue the pattern.

If I wanted to play a red card after the third turn, how many green would I need in my deck?

Remember that decks have 30 cards -- no more, no less -- and red cards cost 5 coins to be played. Once you play a red card, you lose 5 coins. Only use light-green when you need to

I could probably figure it out, but I can't think thoroughly atm.

Problem: Opening hand = ?1r + ?5l

7/30 ~ 1/5
x1: 1/5 of deck == 1 red + 5 light-green == 6
x2: 2/5 of deck == 2 red + 10 light-green == 12
x3: 3/5 of deck == 3 red + 15 light-green == 18
x4: 4/5 of deck == 4 red + 20 light-green == 24
x5: 5/5 of deck == 5 red + 25 light-green == 30

Scenario deck: 5 red + 25 light-green

7/30 ~

1/6*5/6 = 5/36

Scenario tests results: 8/10 opening hands were a success.

Oh, I know why it wasn't 10/10 -- 7/30 ~ 1/5 was wrong. It's 1/4.

7/30 ~(>) 1/4
30/4 == 7.5
x1: 1/4 of deck == 1 red + 6 light-green == 7.5
x2: 2/4 of deck == 2 red + 13 light-green == 15
x3: 3/4 of deck == 3 red + 19 light-green == 22.5
x4: 4/4 of deck == 4 red + 26 light-green == 30

Scenario deck: 4 red + 26 light-green

7/30 ~

1/7.5 * 6.5/7.5 = 6.5/56.25

2nd scenario test results: 8/10

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Next problem: Opening hand = ?5g THEN ?1r

7/30 ~ 1/5
1/5 of deck == too tired to think.

If anyone wants to take a crack at the solution I'd be grateful.

My ultimate goal is to write an algebraic expression where people fill in variables to get the right deck, eg, "I want to draw 5 green then a card that costs 7 coins before my second turn" and it'll give them the right deck.

Um, 2?

Oturan

Guest

Re: Algebra problem.

Well I might take this on, when I get some time. Because it sounds interesting. Even if it's randomised security calculation, which I hate.

Twipply

Guest

Re: Algebra problem.

Deck = 30 cards
You have 5 red cards in your deck.
To play a red card, you need to first play 5 light-green cards.
Your opening hand is 7 cards.

If I need to play a red card on my first turn, what will my deck be composed of? You may use whatever colors of the rainbow you'd like, just as long as there's a strong chance I'll get 5 light-green and 1 red on my opening hand.

This is far from well defined in my head.   The deck consists of 30 cards, 5 of which are known to be red.   The other 25 are what you want us to decide the colours for?   Can we add more red cards?   Do you want to draw at least 5 light green and 1 red, or exactly that and one other card differently coloured?

Last edited by Twipply (Oct 4 2011 12:02:41 pm)

Cyral

Member

From: United States

Joined: 2015-02-15

Posts: 2,269

Re: Algebra problem.

Hmmm.... Im in algebra, but this confuses me, lol

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Player Since 2011. I used to make bots and stuff.

Tako

Member

From: Memphis, Tennessee, USA

Joined: 2015-08-10

Posts: 6,663

Website

Re: Algebra problem.

This is far from well defined in my head.   The deck consists of 30 cards, 5 of which are known to be red.   The other 25 are what you want us to decide the colours for?

Yes.

Can we add more red cards?

No.

Do you want to draw at least 5 light green and 1 red, or exactly that and one other card differently coloured?

At least 5lg and 1r.

Um, 2?

I'll give you a hint, you need at least 5.

Last edited by Tako (Oct 4 2011 4:16:50 pm)

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Yeah, well, you know that's just like, uh, your opinion, man.

Fuzzyguy132

Guest

Re: Algebra problem.

so, wait, you want the answer to this?

If I wanted to play a red card after the third turn, how many green would I need in my deck?

if so, its 10, cuz you played 5 already in question # 2

Green cards produce one coin per turn. You can have light-green cards, which produce a coin as soon as you play them, or have regular green cards, which produce coins once you're done with your turn. If I wanted to play a red card after two turns, how much green/light green would I need? One red card requires you to pay 5 coins.

and this helps with the answer

To play a red card, you need to first play 5 light-green cards.

and if you played 2 green, you'd need to play 7 cuz

To play a red card, you need to first play 5 light-green cards.

if you want an answer to the bottom, then i have no clue

Tako

Member

From: Memphis, Tennessee, USA

Joined: 2015-08-10

Posts: 6,663

Website

Re: Algebra problem.

Alright, here's another actual problem.

I need to play a card that costs 7 coins on my second turn. How many light-green cards do I need in my deck?
1 red + 04 light-green == 05
2 red + 08 light-green == 10
3 red + 12 light-green == 15
4 red + 16 light-green == 20
5 red + 20 light-green == 25
6 red + 24 light-green == 30

I choose... 3:12, because I need room for other colors.

Test results: 4/10.

Well, let's see. If I have 1:4, then that's like 16% probability of the correct ratio. If I have 6:24, that's 100% probability of the correct ratio. 100% of the correct ratio doesn't mean I'm guaranteed to draw 1 red and 4 light-green, it means I'm at the right probability. Compared to 1:4, which is only 16%, it means I'm not exactly likely to draw it, but it's still the correct ratio, meaning I'm not likely to get 0 red and 1 light-green or something off-beat like that.

So, this means decks have a priority. This red card is quite important, but not the only component, so I'll choose 4:16.

Test results: 4/10.

I should probably be doing more runs than 10, but whatever. Statistically speaking, I'm more probable, even though the data doesn't show it.

Last edited by Tako (Oct 4 2011 8:02:28 pm)

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Yeah, well, you know that's just like, uh, your opinion, man.

Fuzzyguy132

Guest

Re: Algebra problem.

A. 1 red + 04 light-green == 05
B. 2 red + 08 light-green == 10
C. 3 red + 12 light-green == 15
D. 4 red + 16 light-green == 20
E. 5 red + 20 light-green == 25
F. 6 red + 24 light-green == 30

Deck = 30 cards        

NOTE: answer must be under 30

You have 5 red cards in your deck.        

NOTE: red must be under 5.     ELIMINATION: F. is now impossible

To play a red card, you need to first play 5 light-green cards.     

NOTE: green must be over 5     ELIMINATION: A. is now impossible

You may have up to 25 light-green cards in your deck.        

NOTE: green must be under 25

Green cards produce one coin per turn. You need to play a card that costs 7 coins on your second turn.     

NOTE: green must be over 7, for you need to play 7 cards for seven coins.

I need to play a card that costs 7 coins on my second turn.

so, if i understand you correctly, my final answer is B, C, D, and E.

Last edited by Fuzzyguy132 (Oct 4 2011 8:09:00 pm)

Tako

Member

From: Memphis, Tennessee, USA

Joined: 2015-08-10

Posts: 6,663

Website

Re: Algebra problem.

You can have up to 6 red cards.

In that specific problem, I had 5 red cards. In this new problem, I can have as many as I need.

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Yeah, well, you know that's just like, uh, your opinion, man.

EENinja

Guest

Re: Algebra problem.

This is a good one

I had a brief read over of it. It's probably to advanced for me though...

I've just started year 11 math...

I can ask my math teacher if you want

Tachyonic

Guest

Re: Algebra problem.

Ask Singing Banana, he's a great mathematician!

Tako

Member

From: Memphis, Tennessee, USA

Joined: 2015-08-10

Posts: 6,663

Website

Re: Algebra problem.

Ask Singing Banana, he's a great mathematician!

I'll send him a message

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Yeah, well, you know that's just like, uh, your opinion, man.