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    • @Factotum I was thinking about trying to have a small card set example. Suppose there is no synergy and you have 6 cards with their card value equal to their number, 1, 2, ... 6. You could create a deck with 5 cards D1, and you could create a deck with 6 cards D2. So the logical way to build D1 is to put cards 2 through 6 in the deck, and leave out card 1 which has the lowest value.

      In this hypothetical situation we start with 0 cards in our hand. Then the draw potential for each deck would be:

      D1 draw potential = (2 + 3 + 4 + 5 + 6) / 5 = 4
      D2 draw potential = (1 + 2 + 3 + 4 + 5 + 6) / 6 = 3.5

      If you could generalize this to where you are comparing a deck of size n vs n+1, where the deck of n+1 has all of the exact same cards, plus one more which is guaranteed to be smaller, we want to prove something like:

      sum(n): the sum of value in deck 1 (avg value * n)
      c: an n + 1 card that is lower value than the lowest value in the deck
      of n cards.
      prove: draw potential for the larger deck is worse than the smaller deck.

      1: sum(n) / n > (sum(n) + c) / (n + 1)
      2: (n + 1) * sum(n) > (sum(n) + c) * n
      3: n * sum (n) + sum(n) > n * sum(n) + c * n
      4: sum(n) > c * n
      5: sum(n) / n > c

      Seen in this form, we can see that as long as the card c has a value less than the average value of the smaller deck, then it reduces the average value of the larger deck. Meaning lower draw potential.

      I think this can be generalized to include synergy as well, it just makes things more complicated.

      Edit: Math

    • @Eric Thanks. I think this is similar to the original reasoning of just cutting the "worst" card from your oversized deck. This is definitely correct for card games without synergy.

      In the meantime, I've thought about your last post and my response to it. One thing led to another, and now I have this crazy formula for the "Average Deck Value"... I think I've got some explaining to do! :D