Magic the Gathering is a popular card game. Cards can be land cards, or other cards. We consider a game with two players. Each player has a deck of 40 cards. Each player shuffles t... Magic the Gathering is a popular card game. Cards can be land cards, or other cards. We consider a game with two players. Each player has a deck of 40 cards. Each player shuffles their deck, then deals seven cards, called their hand. (a) Assume that player one has 10 land cards in their deck and player two has 20. With what probability will each player have four lands in their hand? (b) Assume that player one has 10 land cards in their deck and player two has 20. With what probability will player one have two lands and player two have three lands in hand? (c) Assume that player one has 10 land cards in their deck and player two has 20. With what probability will player two have more lands in hand than player one? Read more

Steps to Solve

1. Calculate Player A's probability of drawing exactly 3 lands

Player A has a 60-card deck with 24 lands. We want to find the probability of drawing exactly 3 lands in a 7-card hand. This can be calculated using combinations:

$$P(A) = \frac{\binom{24}{3} \binom{36}{4}}{\binom{60}{7}}$$

where $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ represents the number of ways to choose $k$ items from a set of $n$ items.

$\binom{24}{3}$ is the number of ways to choose 3 lands from 24 lands. $\binom{36}{4}$ is the number of ways to choose 4 non-land cards from the remaining 36 cards. $\binom{60}{7}$ is the total number of possible 7-card hands from the 60-card deck

2. Calculate Player B's probability of drawing exactly 3 lands

Player B has a 40-card deck with 16 lands. We want to find the probability of drawing exactly 3 lands in a 7-card hand.

$$P(B) = \frac{\binom{16}{3} \binom{24}{4}}{\binom{40}{7}}$$

where $\binom{16}{3}$ is the number of ways to choose 3 lands from 16 lands, $\binom{24}{4}$ is the number of ways to choose 4 non-land cards from the remaining 24 cards, and $\binom{40}{7}$ is the total number of possible 7-card hands from the 40-card deck.

3. Compute the combinations and probabilities

Calculate the values for each combination and then compute the probabilities for Player A and Player B.

For Player A:

$\binom{24}{3} = \frac{24!}{3!21!} = \frac{24 \times 23 \times 22}{3 \times 2 \times 1} = 2024$

$\binom{36}{4} = \frac{36!}{4!32!} = \frac{36 \times 35 \times 34 \times 33}{4 \times 3 \times 2 \times 1} = 58905$

$\binom{60}{7} = \frac{60!}{7!53!} = \frac{60 \times 59 \times 58 \times 57 \times 56 \times 55 \times 54}{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1} = 38608020$

$P(A) = \frac{2024 \times 58905}{38608020} = \frac{119226720}{38608020} \approx 0.3088$

For Player B:

$\binom{16}{3} = \frac{16!}{3!13!} = \frac{16 \times 15 \times 14}{3 \times 2 \times 1} = 560$

$\binom{24}{4} = \frac{24!}{4!20!} = \frac{24 \times 23 \times 22 \times 21}{4 \times 3 \times 2 \times 1} = 10626$

$\binom{40}{7} = \frac{40!}{7!33!} = \frac{40 \times 39 \times 38 \times 37 \times 36 \times 35 \times 34}{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1} = 18643560$

$P(B) = \frac{560 \times 10626}{18643560} = \frac{5950560}{18643560} \approx 0.3192$

4. Compare the probabilities

Compare $P(A)$ and $P(B)$ to determine which player is more likely to draw exactly 3 lands.

Since $P(B) \approx 0.3192$ and $P(A) \approx 0.3088$, Player B is more likely to draw exactly 3 lands in their opening hand.