Mana Goldilocks
Bob is building a deck for the trading card game, Arcane: The Congregation™. He's trying to decide how many land cards he should put into his deck. Not enough land cards, and he won't have the mana to cast any of his spell cards. Too many land cards, and his deck will have too few spell cards to be effective.
Bob would like to have 2, 3, or 4 land cards in his opening 7-card hand, which is drawn from his shuffled 40-card deck. Bob builds his deck with a number of land cards so that the chance of this is maximized. To three decimal places, what is the probability that Bob has 2, 3, or 4 land cards in his opening hand?
The answer is 0.795.
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1 solution
Let n be the number of land cards in Bob's deck. Then there are 4 0 − n non-land cards in his deck. The total number of ways to choose a 7-card hand from Bob's deck is ( 7 4 0 ) . The number of ways to choose a hand with exactly k lands is ( k n ) ( 7 − k 4 0 − n ) . Thus, the probability of getting exactly k lands in the opening hand is:
( 7 4 0 ) ( k n ) ( 7 − k 4 0 − n )
So, the probability of getting exactly 2, 3, or 4 lands in Bob's opening hand is:
( 7 4 0 ) ( 2 n ) ( 5 4 0 − n ) + ( 3 n ) ( 4 4 0 − n ) + ( 4 n ) ( 3 4 0 − n )
It's reasonable to guess that the number of lands which mazimizes the chance of 2,3, or 4 lands in the 7-card opening hand is 7 3 × 4 0 ≈ 1 7 . Testing this gives the probability:
( 7 4 0 ) ( 2 1 7 ) ( 5 2 3 ) + ( 3 1 7 ) ( 4 2 3 ) + ( 4 1 7 ) ( 3 2 3 ) ≈ 0 . 7 9 5
Testing other values of n yields lesser probabilities. Thus, Bob puts 17 land cards into his deck, and his probability of drawing 2, 3, or 4 land cards in his opening hand is 0 . 7 9 5 .