Ch3 Section: Combinations #19

A gin had consists of 10 cards from a deck of 52 cards. Find the probability that a gin hand has

A) all 10 cards of the same suit

  1. Find total number of possible hands of 10 cards from a deck of 52 \[\left (\begin{array}{c} 52 \\ 10 \end{array} \right) = \frac{52!}{10!(52-10)!} \]
total <- factorial(52)/(factorial(10)*factorial(42))
total
## [1] 15820024220
  1. Find the number of possible hands of 10 cards with the same suit. 10 out of 13 cards (times 4 suits) \[\left (\begin{array}{c} 13 \\ 10 \end{array} \right)\]
samesuit <- factorial(13)/(factorial(10)*factorial(3)) * 4
samesuit
## [1] 1144
  1. Probability that a hand has all 10 cards of the same suit = \(\frac{1144}{15820024220}\)
samesuit/total
## [1] 7.231342e-08

B) Exactly 4 cards in one suit and 3 in two other suits

  1. Find the number of possible hands of 4 cards with the same suit. 4 out of 13 cards. (4 possibilities) \[\left (\begin{array}{c} 13 \\ 4 \end{array} \right) \times 4\]

  2. Find the number of possible hands of 3 cards with the same suit. 3 out of 13 cards. (this is done twice). This can only happen twice out of the remaining three suits.

\[\left (\begin{array}{c} 13 \\ 3 \end{array} \right) \times \left (\begin{array}{c} 13 \\ 3 \end{array} \right) \times \left (\begin{array}{c} 3 \\ 2 \end{array} \right)\]

  1. Find the probability \[\frac {\left (\begin{array}{c} 13 \\ 4 \end{array} \right) \times 4 \times \left (\begin{array}{c} 13 \\ 3 \end{array} \right) \times \left (\begin{array}{c} 13 \\ 3 \end{array} \right) \times \left (\begin{array}{c} 3 \\ 2 \end{array} \right)}{\left (\begin{array}{c} 52 \\ 10 \end{array} \right)} \]
n1 <- factorial(13)/(factorial(4)*factorial(9)) * 4
n2 <- factorial(13)/(factorial(3)*factorial(10))
n3 <- factorial(13)/(factorial(3)*factorial(10))
n4 <- factorial(3)/(factorial(2)*factorial(1))
n <- n1*n2*n3*n4

n/total
## [1] 0.04436211

C) A 4, 3, 2, 1, distribution of suits

  1. First suit - 4 out of 13 cards (times 4 suits to choose from)

\[\left (\begin{array}{c} 13 \\ 4 \end{array} \right)\] Second suit - 3 out of 13 cards (times 3 suits to choose from) \[\left (\begin{array}{c} 13 \\ 3 \end{array} \right)\] Third suit - 2 out of 13 cards (times 2 suits to choose from) \[\left (\begin{array}{c} 13 \\ 2 \end{array} \right)\] Fourth suit - 1 out of 13 cards (times 1 suit to choose from) \[\left (\begin{array}{c} 13 \\ 1 \end{array} \right)\]

  1. Find the probability \[\frac {\left (\begin{array}{c} 13 \\ 4 \end{array} \right) \times 4 \times \left (\begin{array}{c} 13 \\ 3 \end{array} \right) \times 3 \times \left (\begin{array}{c} 13 \\ 2 \end{array} \right) \times 2 \times \left (\begin{array}{c} 13 \\ 1 \end{array} \right) \times 1}{\left (\begin{array}{c} 52 \\ 10 \end{array} \right) } \]
n1 <- factorial(13)/(factorial(4)*factorial(9)) * 4
n2 <- factorial(13)/(factorial(3)*factorial(10)) * 3
n3 <- factorial(13)/(factorial(2)*factorial(11)) * 2
n4 <- factorial(13)/(factorial(1)*factorial(12))
n <- n1*n2*n3*n4

n/total
## [1] 0.3145677