Bunuel wrote:
Playing poker helps to understand the combination and vise-versa.
So, find below problems to master in either of them:
In the poker game Five-Card Draw, each player is dealt a hand consisting of 5 cards from a deck of 52 cards. Each card in the deck has a suit (clubs, hearts, diamonds, or spades) and a value (A, 2,..., 10,J,Q,K).
A. What is the probability that we are dealt a four-of-a-kind?
B. What is the probability that we are dealt full house? (A full house is a hand with both a three-of-a-kind and a two-of-a-kind.)
C. What is the probability that we are dealt two pairs? (Fifth is different)
D. What is the probability that we are dealt three of a kind? (Others are different)
E. What is the probability that we are dealt one pair? (Others are different)
F. What is the probability that we are dealt all five of different ranks?
G. What is the probability that we are dealt hands with every suit?
Bunuel if you could please show this sum by individual probability way it would be very helpful , combinotrics seems a little complex .
I tried the first one :
1. p( four of a kind ): I guess this means exactly four of a kind , i e fifth one cannot be the same kind .
so (13/52)*(12/51)*(11/50)*(10/49) * 4* (39/48)
there are 13 ways to choose the same kind of suit ( spades, hearts , clubs , diamonds) and there are four different kinds .
after we choose the four there are 48 cards left . But the fifth card cannot be of the same suit as the first 4 ( if the first 4 is heart then
the fifth one cannot be a heart) so from 48 we have to remove the remaining hearts i .e 13 - 4 = 9 hence 48 - 9 = 39
so fifth card probability is 39/48 , I would like to stop here and request you proceed .
so do we have to multiply (39/48) with 3 as there are 3 suits remaining :
if anyone could show all these poker problems by individual probability method it would be very helpful . You could be eligible for a lot of Kidos!!