Cake
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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