Some people say that you should always play with 60 cards so that you maximize your chances of drawing a useful card when you need it most. Opponents of this view argue that 60 cards should not be a hard and fast rule. Here are two reasons:

• With more cards, you have more possibilities.
• With more cards, you have more threats.

Since I feel that most people understand the reasoning behind strictly limiting your deck to 60 cards, I won’t further discuss the reasoning here. However, I would like to offer my opinions on using more than 60 cards.

I feel that some of the major goals in designing a deck involve:

1. Tuning it to optimally follow its strategic intent or path(s) to victory.
2. Tuning it to have the optimal mix of answers to the metagame.

In looking at these two deck-designing goals, you may notice that the strategic intent of a deck is affected by the metagame. So for this "debate," let’s just be concerned with the second goal.

Now, I am going to make a suggestion, which many "orthodox" deck-builders may consider to be bunk – an optimally designed deck may have more than 60 cards.

Let’s look at the surplus cards that bring a deck over 60. Firstly, I’ll make an assumption that these surplus cards are the least effective of all of the cards in a deck when considering their role within the metagame. I’ll make another assumption – a deck’s mana-ratios are always optimal.

I’ll be using some algebraic equations to demonstrate the advantage of adding surplus cards. Skip to the bottom for a conclusion if you are not interested in the specifics.

Now for some algebraic representations:

Before I get into the algebra, a brief definition of x_i(M) is warranted. It describes the probability with which a card will be beneficial when top-decked and is affected by f_i(M). For example, Bone Shredder is beneficial against the combined pool of decks with non-black creatures B[å f_i(M)] within the metagame. However, it is virtually a dead draw against the combined pool of black-creature decks and creatureless decks i.e. everything else. Thus the probability modifier of this card takes into account every deck within the metagame, å f_i(M), and is determined by a metagame probability test:

Now I am going to illustrate an under-handed, dirty mathematical trick, which is only meant to be a demonstration to simplify an argument. Let’s compare a 60-card deck to a (60+n) card deck. If we hope to draw a card, which will be deemed beneficial for a given circumstance, let’s look at the algebra.

Here, the argument made by "orthodox" deck-builders for sticking to 60 cards comes into question. The binomial, (b(M)+n), is the debatable term. Specifically, the issue is the quality of the surplus cards, n; normally, a surplus card will not be beneficial in every situation. However, (b(M)+n) assumes that the surplus cards are always beneficial.

To address the contentious binomial (b(M)+n), let’s break down the surplus cards, n, into individual card types, n_i. This will allow the variable of interest, x_i(M), to be incorporated into this melange. Thus, (b(M)+n) can be modified becoming:

[b(M) + å (n_i*x_i(M)]

For example, if the surplus cards are two Diabolic Edict, then n₁*x₁(M) would be equal to 2*x₁(M). Similarly, if the two surplus cards are one Verdant Force and one Overrun, then the Verdant Force is n₁, and the Overrun is n₂, leaving å (n_i*x_i(M) = 1*x₁(M) + 1*x₂(M). Under this modification, let’s now compare the top-decking capabilities of orthodox and unorthodox decks.

To repeat, values for x_i(M) are variable and dependent on the probability that a card is beneficial in relation to the metagame. When orthodox and unorthodox decks are compared in this light for their abilities to top-deck, this equation is then considered.

This equation reveals values for x_i(M) such that an iso-top-deck condition occurs between orthodox and unorthodox decks. In other words, the top-decking potential of orthodox and unorthodox decks is equalized. A rearrangement of this equation reveals that:

While n, b(M) and r are all parameters dependent on the number of cards selected for a deck, the variable of interest, x_i(M), will determine whether an unorthodox deck may improve upon an orthodox deck. What does this equation mean? Let’s look at a situation where:

If the average probability that your surplus cards are beneficial, [å (n_i*x_i(M)) / n], exceeds the probability of top-decking a beneficial card, [b(M) / (60-r)], it would be to your advantage to go with an unorthodox deck. Why? Let’s say that you can determine from the metagame probability test that a specific card will be effective within the metagame. If you add that card to your deck, you increase the probability of drawing a beneficial card because the pool of beneficial cards in your deck will have increased. Remember the under-handed, dirty mathematical trick, (b(M)+n)? Adding a surplus card will improve a library’s top-decking potential when the probability that the surplus card is beneficial exceeds the probability of drawing a beneficial card.

The problem is determining what should be your x_i(M) cards and what modifiers are correct for each card dependent on the foils beating you down inside the metagame. Remember also that you need to be honest when determining those cards in your deck that fall under the definition of surplus cards – they must be a deck’s least effective cards. An application of the metagame probability test on every (non-land) card in your deck could be used to find the cards having the lowest x_i(M).

Robert A.C. Nanka-Bruce