poker probability question- HELP PLEASE :)

[rachey21]

In a version of poker, a player is dealt 5 cards. They are allowed to look at two cars while the other three remain hidden. Given that the player can see exactly one ace, what is the probability that she has 4 of a kind?

= 624 <- equation without seeing the two cards

 

[soroban]

Hello, rachey21!

In a version of poker, a player is dealt 5 cards.
The player is allowed to look at two cards while the other three remain hidden.
Given that the player can see exactly one Ace,
what is the probability that she has four-of-a-kind?

The player sees an Ace and an Other.
There are 50 cards remaining in the deck.

What is the probability that the other three cards are
. . the 3 remaining Aces or the 3 remaining Others?

There is 1 way to get the three remaining Aces
. . . and 1 way to get the three remaining Others.
Hence, there are 2 ways to get four-of-a-kind.

There are: .\(\displaystyle \displaystyle{50\choose3} \,=\,19,\!600\) possible outcome.

Therefore: .\(\displaystyle P(\text{4-of-a-kind}) \:=\:\dfrac{2}{19.600} \:=\:\dfrac{1}{9.800}\)