My buddy and I were discussing different ways of deciding how much mana to include when constructing a deck. Without thinking much about it I used a method of finding the average converted mana cost for a deck, and he has taken issue with it. My intuition is that in general this is a quick way to get an idea rather than adding every card and dividing by the total. My method is to take one of every named card with a mana cost and average them, rather than averaging every single card with a mana cost. Since you can't have more than 4 of card (lands aside), it would look something like this:

  • 4 x Card A with cost 1
  • 3 x Card B with cost 1
  • 4 x Card C with cost 2
  • 1 x Card D with cost 2
  • 3 x Card E with cost 3

and so on. My way is then to take one of each card A through E and average just those single examples. So $(1+1+2+2+3)/5 = 1.8 $instead of $((4\times 1)+(3\times 1)+(4\times 2)+(1\times 2)+(3\times 3))/15=1.733$. My gut says that in most cases this works well enough, especially as more cards are included as they would be in an actual deck and examples we made to test it seem to back it up. I'm not asking if this is necessarily a good way to do it, but rather is there a term for what I'm actually calculating here?

  • 3
    $\begingroup$ An 'approximation'. $\endgroup$
    – Ian Miller
    Mar 27 '16 at 16:13
  • $\begingroup$ So you are throwing out the "weights" associated to the fact that you have different numbers of copies of things in your deck? That seems a bit weird, what if your deck is a swarm of tiny creatures, 4 of each kind, intended to keep you alive until you have drawn a single copy of a M mana cost card that instantly wins you the game (where M is huge, say 10+)? (Not saying such a deck can actually be built in real MTG, but you can imagine it.) $\endgroup$
    – Ian
    Mar 27 '16 at 16:16
  • $\begingroup$ Note the particular discrepancy: if you have 23 lands, 36 1 cost cards from 9 types, and 1 card of cost M, then your scheme gives $\frac{M+9}{10}$ while the actual average is $\frac{M+36}{37}$. These will differ by quite a bit if $M$ is big enough. $\endgroup$
    – Ian
    Mar 27 '16 at 16:18
  • $\begingroup$ It might be easier to just count the number of cards with $1$ cost (regardless of type) times the number of those cards; the number with $2$ cost times the number of those cards; and so forth, and at the end you have just a few numbers to add up, giving the total cost of the deck, then divide by the number of cards. $\endgroup$
    – David K
    Mar 27 '16 at 16:24

What your friend is doing, and what the right thing to do here, is called a weighted average. Every type of card has a cost, and that card occurs a fixed number of times: the second approach takes both of these into account. What you are doing is an approximation that works well enough, as long as the number if cards if each type are almost equal in number. If there were 100 cards of type E and only 5 cards each of the other types, your method would be very inaccurate. You can try this.

  • $\begingroup$ A premise of the question is that there can be only four cards (at most) of type A. But one of Ian's comments on the question points out how you can still get a very inaccurate result. $\endgroup$
    – David K
    Mar 27 '16 at 16:29
  • $\begingroup$ @DavidK edited. It was just illustrative, anyway. $\endgroup$ Mar 27 '16 at 16:31
  • $\begingroup$ It wasn't just type A that was limited; I just pointed that out as an example. You can only have four of any given type. But even limited to four of each kind, you can still get a quite noticeable inaccuracy. $\endgroup$
    – David K
    Mar 27 '16 at 16:35
  • $\begingroup$ I recognize it's not extremely accurate, it's just easier to do in my head while I'm building it and in most cases we came up with for a realistic deck the difference was in the decimal, but since you can't add fractional mana to the deck it seemed close enough. I had never really formalized what I was doing, like I said I was just kind of following intuition. But when he argued it couldn't work, and I showed him how close it tended to be outside of extreme cases I really thought it through and realized I wasn't sure what exactly I was calculating, if there was a term for it, or what I'd lose. $\endgroup$ Mar 27 '16 at 16:50

You might say you are finding the cost "averaged over the types of cards in the deck."

I think most people, not just your buddy, would consider this an incorrect way of averaging. It's fine if you have the same number of every kind of card, but gets worse as the relative frequency of high-value cards differs from the relative frequency of low-value cards. Your method biases the average toward the types with the fewest number of cards in the deck.

Biased might in fact be the best word to describe your method.

Consider your example if you have not yet put the cards of type C in your deck and have only put in one copy of type E. You then have:

  • $4$ x Card A with cost $1$
  • $3$ x Card B with cost $1$
  • $1$ x Card D with cost $2$
  • $1$ x Card E with cost $3$

The accurate average is $$ ((7 \times 1) + 2 + 3)/9 = 1.333\ldots .$$

Your method gives $$ (1 + 1 + 2 + 3)/4 = 1.75.$$

Considering that just by taking the range of values you know the average has to be between $1$ and $3$, your method seems rather inaccurate in this case; it's only a little better than just taking the midpoint of the range, $(1+3)/2 = 2$.

You might be better off just lightly shuffling the deck, drawing a few cards out at random, and averaging those. That would get rid of the bias, at least in a statistical sense.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.