Playing cards probability

if being dealt only 3 cards out of a complete deck of cards (52 cards), what is the probability that you will be dealt one 3, one 5, and one 7 in any particular order? This is not for homework, but rather a card game that some friends of mine play. I would also love to know how to solve the problem so as to better understand how the answer was derived.


Well-Known Member
I will give you an easier example only as a tool on how to calculate your question.
If you have 4 cards a,b,c,d and you take out 2 cards and you want it to be a,b
n=4 , k=2

Combinatipons of all the cards:
You write all the combinations 4! = 24. [a,b,c,d] , [a,b,d,c] , [a,c,b,d],[a,c,d,b], [a,d,b,c],[a,d,c,b] ...[d,a,b,c].....[d,c,b,a]
now you need to count how many combinations contains a,b in the first 2 places. [a,b,c,d], [a,b,d,c] [b,a,c,d], [b,a,d,c]
combinations of a,b 2! [a,b] and [b,a]
combinations of c,d (4-2)! [c,d] and [d,c]
total combination of a,b in the first 2 placedes: 2!*2!=4

the probability of getting the first 2 cards as [a,b] or [b,a] (no mean for the order):
(number of combinations that meet your criteria) divided by (number of combinations)
4/24 = 1/6

Another option, using all the combinations to take two cards with no meaning to the order:
number of combinations: 6 [a,b] , [a,c], [a,d], [b,c], [b,d], [c,d] n!/(k!*(n-k)!=4!/(2!*(4-2)!)=6 {n=4, k=2}
combination that meet your criteria: 1 [a,b]
the probability: 1/6
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