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I am playing a card game called Magic The Gathering.

In my opening hand, I draw seven cards from a 100 card deck. Half of the cards in the deck are called land cards. The other 50 cards in the deck are creature cards.

What is the probability of drawing the following card types in my opening hand:

0 land cards, 7 creature cards
1 land card, 6 creature cards
2 lands cards, 5 creatures cards
...
7 lands cards, 0 creature cards

I was trying to conceptualize this problem with marbles, but can't figure it out.

I would like to generalize the solution to card decks that are not evenly split between lands and creatures.

the_prole
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2 Answers2

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With a 100 card deck, 50 lands, and 50 creatures, drawing a seven card hand without replacement, letting $X$ represent the random variable counting the number of lands in the opening hand the probability of getting exactly $k$ land cards (and $7-k$ creatures) will be

$$Pr(X=k)=\frac{\binom{50}{k}\binom{50}{7-k}}{\binom{100}{7}}$$

This generalizes of course:

Given an $N$ card deck with $l$ land cards and $c$ creature cards (where $l+c=N$), when drawing $d$ cards (without replacement) the probability of drawing $k$ land cards and $d-k$ creature cards will be:

$$Pr(X=k)=\frac{\binom{l}{k}\binom{c}{d-k}}{\binom{N}{d}}$$

These are calculated by direct counting methods. We may assume that each card has a unique identification in order to assist in constructing an equiprobable sample space. As such, each of the $\binom{N}{d}$ $d$-card hands are equally likely to occur. To count the number of hands with exactly $k$ lands, pick which specific $k$ lands were selected from the $l$ available and then pick which $d-k$ of the creatures were selected from the $c$ available.

Here, $\binom{n}{r}$ denotes the binomial coefficient, counting the number of ways to select a subset of $r$ things from a set of $n$ distinct objects. One has $\binom{n}{r}=\frac{n!}{r!(n-r)!}=\frac{1\cdot 2\cdot 3\cdots n}{(1\cdot 2\cdots r)(1\cdot 2\cdots (n-r))}$

JMoravitz
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What you're looking here is a random variable $X$ that follows the hypergeometric distribution that defines your probability to open up with $k$ lands. In this case, $$P(X=k) = \frac{{{50}\choose{k}}{{50}\choose{7-k}}}{{100}\choose{7}}$$

The probability is characterized intuitively by thinking the amount of possible combinations of successes $k$ and failures $7-k$ and dividing them with the amount of possible ways to pick $7$ cards from $100$.

Forgive my abuse of notation; here are the values to your question: $P(X=k) = [0.006, 0.04, 0.16, 0.28, 0.28, 0.16, 0.04, 0.006]$

So with half and half you're most likely to get $3$ or $4$ lands, and the distribution is nicely symmetric.

You can simulate the more "orthodox" $40\%$ land composition with the general hypergeometric distribution, and you can also calculate the probabilities for mulligan(s). For the "orthodox" $40\%$ composition you will see that the probability of getting $3$ lands is $0.30$ and the probability of getting $4$ lands is $0.195$, so a starting hand with $3$ or $4$ lands will occur once in two games.

Eemil Wallin
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