Question

Assume that the 3 players are dealt (one after another) 7 cards each from the same well-shuffled deck of 52 cards. What is the probability that at least one of the players will get four spades? (any order) What is the probability that none of them gets four spades?

Answer 1

n = total number of ways to choose 7 cards from a pack of 52 playing cards.

= ⁵²C₇

A : The player A will get 4 spades.

n(A) = ¹³C₄ * ³⁹C₃

P(A) = ¹³C4 * ³⁹C3 / ⁵²C7 =0.0488

B : The player B will get 4 spades.

n(B) = ¹³C₄ * ³⁹C₃

P(B) = ¹³C₄ * ³⁹C₃ / ⁵²C₇ =0.0488

C : The player B will get 4 spades.

n(C) = ¹³C₄ * ³⁹C₃

P(C) = ¹³C₄ * ³⁹C₃ / ⁵²C₇ =0.0488

A ,B and C are independant event

P (A B ) =P(A) *P(B) = 0.0488 *0.0488 = 0.002381

P (A C ) =P(A) *P(C) = 0.0488 *0.0488 = 0.002381

P (B C) =P(B) *P(C) = 0.0488 *0.0488 = 0.002381

P(A B C ) =P(A) *P(B) *P(C) = 0.0488*0.0488*0.0488 = 0.000116

a) P ( At least one of the player will get four spades ) = P( A B C)

P( A B C) = P(A) + P(B) +P(C) - (P (A B ) +P (A C) +P (B C) ) + P(A B C )

= 0.0488+0.0488+0.0488 - ( 0.002381 +0.002381 +0.002381 ) + 0.000116

= 0.1464 - .007143 +0.000116

= 0.1393

P( At least one player will get four spades) = 0.1393

b) P ( none of three players will get four spades ) = P( A' B' C' ) =  P( A B C)' = 1 -  P( A B C)

= 1 - 0.1393

= 0.8607

P( None of three players will get four spades) = 0.8607