In the card game Magic: the Gathering, spells are cast using mana produced by "land" cards. Suppose your deck has 60 cards, 24 of which are lands.

Question

In the card game Magic: the Gathering, spells are cast using mana produced by "land" cards. Suppose your deck has 60 cards, 24 of which are lands.

(a) What is the probability that your opening hand of 7 cards has no lands in it?

(b) What is the probability that your opening hand of 7 cards has exactly 3 lands?

(c) What is the probability that your opening hand of 7 cards has 5 or more lands?

Answer 1

Since a hand of cards is selection WITHOUT replacement (the same card CANNOT be chosen more than once), it would be easiest to do this using a Hypergeometric Distribution. We have a population of size 60 (the total deck size) with 24 "successes" (land cards), and are choosing a sample of size 7. The probability of exactly k "success" in the sample would be given by the formula:

1. Prob(k) = C(24, k) * C(36, 7-k) / C(60, 7)

where C(a,b) = a!/(b! * (a-b)!) is the number of Combinations of a things taken b at a time (often written in formulas as a Binomial Coefficient.

I'm using technology (Geogebra, for the record) to compute the results below and rounding to four decimal places.

a) No lands means zero "successes", P(0) = .0216

b) Exactly three lands is P(3) = .3087

c) Five or more lands would mean either 5, 6, or 7 lands (since the hand size if 7), one would need to compute P(5) + P(6) + P(7) = .0828 using the above formula.