Combinatorics Gurus helpppp (1 Viewer)

[HeroWise]

For a poker hand (5 cards) what is the probability of getting a single pair

I got: (13C1)(4C2)(48C3)/52C5


But the solution in cambridge says other wise.

(13C1)(4C2)(12C3)4^3/52C5

I can understand where they are cming from But when i do 48C3 how am I over counting? This question is confusing me atm

 

[fan96]

does not exclude the possibility of having another pair.

i.e. the question (probably) asks for the probability of exactly one pair, but you calculated the probability of having at least one pair.

Doing (and then choosing suits afterwards) guarantees that all of the cards will have a different value.

 

[HeroWise]

What how does that include more pairs? Lets say i picked 2 queens Thats set. I cant have another 1 or 2 queens cause it will give more than a pair. Thats why i have 48 which is 52-4

Does that mean I cant have 3 Aces included with the original 2 queens

 

[fan96]

What how does that include more pairs? Lets say i picked 2 queens Thats set. I cant have another 1 or 2 queens cause it will give more than a pair. Thats why i have 48 which is 52-4

Does that mean I cant have 3 Aces included with the original 2 queens

If your first pair were queens then your could choose 2 kings, 2 aces, 2 jacks etc. which would be more than 1 pair.

 

[HeroWise]

makes sense ty ty