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Stuart Garside
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Need help and I can't "see" the answer. I have some quest cards which need variables adding to them.

Ember has creatures worth 1 to 13 points, and I want to create at a number of (at least 7) Quest cards which give additional points at the end of the game when you conjure creatures of certain types.

For instance: +1 VP when you conjure: VP 1, VP2, and VP3 creatures.

BUT. How do I make all the cards balanced so they're all absolutely the same values. I'm sure some maths person could do it in a heartbeat.

They all ideally need 2 or 3 creature variables on each card. So card 1 has VPs 1, 4 and 9 (for example) and they all need to be different.

Can anyone help? There must be an easy way to do it, but it'sa case of maximising the numbers (with a cap of about 9 (as opposed to 13)) and having the right number of quest cards...
 
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Jeremy Lennert
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I'm not sure I understand what you're asking.

Do you want a way to create sets of 3 numbers that all add up to the same sum? Like:

3 + 6 + 9 = 18
2 + 5 + 11 = 18
1 + 7 + 10 = 18

If that's what you want, do you just need the 3 numbers on a single card to be different, or are you trying to ensure that the same number is never used on more than one card? (e.g. 3+6+9 and 3+7+8) Do you care what the sum is?

Or are you trying to make it so that the quests are all equally difficult to fulfill? That would require knowing something about the strategy of the game (and "absolutely the same values" is probably not a realistic goal in that case).
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Stuart Garside
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Hi Jeremy

You got it right the first time. I found it quite hard to explain without going into a lot of detail.

You have creatures you can conjure, each with a VP cost from 1 to 13. Each quest should feature a combination of (up to 3) those creature VPs, so every creature they conjure that matches their quest card will give them additional points at the end of the game (think quests in Lords of Waterdeep but instead of quests you've got creatures from VP 1 to 13).

The really high level creatures VP 10+ are probably too hard and rare to make worthwhile, so I was trying to find a way to focus on the lower level creatures.

So your summary of 3+6+9 = 18 was spot on. I'd like each card to feature as many variants of the cards, so no two cards would have the same values

Example:

Quest 1: Gain 1 VP for every 1 and 9 VP creatures you conjure.
Quest 2: Gain 1 VP for every 2 and 8 VP creature you conjure.

I could do it with 2 numbers per quest card, but then it gets weird as there's not a lot of options with just 9 numbers and 7 cards. It would be better with permutations of 3 quests...
 
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It's important to know on how many different cards a specific number can/should appear. For example, if each number should appear at least once, but with no upper limit, your sum could be as low as 16:

1+2+13=16 (this is the only way to use 13)
1+3+12=16 (the only way to use 12)
1+4+11=16
1+5+10=16
1+6+9=16
1+7+8=16
2+3+11=16 (so 11 can be used twice)
2+4+10=16

and so on.

You'll have a huge number of sets to choose from for your quests which have the lowest numbers. So you need to decide whether you want a limit on the number of times a creature appears in your quests. Or a maximum number of quests, so that someone can come up with a fairly even distribution of the creatures.

A minimum on the number of appearances means your sum has to be a bit higher.
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Jeremy Lennert
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To maximize the number of possible combinations, your sum should be close to the average for a random set. Since you're using the numbers 1-9, the average number is 5, so the average of 3 numbers is 15, so you'll get maximum flexibility if you choose a sum of 15.

Also, this may be obvious, but notice that once you've picked the first two numbers of your set, the third number is determined by subtracting them from the target sum. For instance, if you're going for a sum of 13, then once you've picked the numbers 2 and 7, you know the third number has to be 15 - 2 - 7 = 6. Sometimes that formula will give you an invalid result (like 11, -3, or a duplicate of a number you've already used), in which case the first two numbers you picked won't work together, but there will never be a valid third number other than the one given by that formula.

Finally, assuming that order doesn't matter (e.g. 1+6+8 is the same as 8+6+1), then we can always write our sets in ascending order for simplicity.

Armed with that knowledge, you can just kind of start writing sets:

1, 2, 12 (out of range)
1, 3, 11 (out of range)
1, 4, 10 (out of range)
1, 5, 9
1, 6, 8
1, 7, 7 (duplicate number)
2, 3, 10 (out of range)
2, 4, 9
2, 5, 8
2, 6, 7
2, 7, 6 (not in ascending order, which means it's a duplicate set)
3, 4, 8
3, 5, 7
3, 6, 6 (duplicate number)
4, 5, 6

That gives you 8 valid triads for your quest deck.


I also happened to discover that a sum of 13 coincidentally gives you exactly 7 valid triads:
1, 3, 9
1, 4, 8
1, 5, 7
2, 3, 8
2, 4, 7
2, 5, 6
3, 4, 6
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Stuart Garside
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Thank you so much for helping with this and taking the time to post this.

It's just one of those rubbish things I couldn't get my head around.
 
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Stuart Garside
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Antistone wrote:
I also happened to discover that a sum of 13 coincidentally gives you exactly 7 valid triads:
1, 3, 9
1, 4, 8
1, 5, 7
2, 3, 8
2, 4, 7
2, 5, 6
3, 4, 6


THANK YOU! So helpful and spot on useful
 
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