## Quiz 1 Solutions

1. In 7 card stud poker, you are dealt 7 cards, and choose the best combination of 5 to make your hand. Give the probability of getting a full house, i.e., you have three cards in one denomination, at least one other denomination has at least two cards, and no denomination has four cards (as then, you could make a better hand than a full house). Hint: Enumerate the possible cases.

Solution: The possible distributions of cards are

• Three cards in each of two denominations and one card in a third denomination. The number of possibilities is
C(13,2)C(4,3)C(4,3)C(11,1)C(4,1) = 54912
Here, we write C(n,k) for the binomial coefficient "n choose k".

• Three cards in one denomination, and two cards in each of two other denominations. The number of possibilities is
C(13,1)C(4,3)C(12,2)C(4,2)C(4,2) = 123552

• Three cards in one denomination, two cards in a second denomination, and one card each from two other denominations:
C(13,1)C(4,3)C(12,1)C(4,2)C(11,2)C(4,1)C(4,1) = 3294720

The probability of a full house is the sum of these three numbers, divided by C(52,7). The probability is approximately .02596

2. How many arrangements of the letters in the phrase
MINNEAPOLIS MINNESOTA

Solution: We first find the number of ways to choose slots for the vowels. There are 9 vowels and 11 consonants. Represent the vowel slots by bars and the consonant slots by stars. Each pair of bars must have at least one star between them. There are 8 gaps between bars, so the number of such arrangements is equal to the number of ways to arrange 11-8=3 stars among the bars with no restriction on where they must go. The number of possible arrangements of 3 stars and 9 bars is C(12,3). Thus, there are C(12,3) ways to choose slots for the vowels.

The vowels consist of 3 I's, 2 E's, 2 A's, and 2 O's, so the number of ways to arrange the vowels in their chosen slots is

9!/(3!2!2!2!)

The consonants consist of 2 M's, 4 N's, 1 P, 1 L, 2 S's, and 1 T, so the number of ways to arrange the consonants in their chosen slots is

11!/(2!4!2!)

Thus, the total number of possible arrangements is

C(12,3)[9!/(3!2!2!2!)][11!/(2!4!2!)] = 691558560000