Probability of 4 of a kind in 7 cards, if first dealt 2 different cards? Question: In Texas Hold 'Em, you choose your hand from among seven cards: two that are dealt to you, plus five more that are available to all players.

Suppose the two cards dealt to you are of different denominations. What is the probability that you'll be able to make four of a kind out of all seven cards available to you? 



I'm trying to use Bayes conditional to solve this:

Prob (4 of a kind in 7 cards after first being dealt 2 different denominations ) = ( Prob(Being dealt 2 different denominations first (in 2 cards), given 4 of a kind in 7 cards) * Prob(4 of a kind in 7 cards) ) / Prob (Being dealt 2 different denominations first (in 2 cards))

I'm looking for clarity on how to think about this problem (and, I believe, how I should avoid overthinking it as I've spent too much time pondering "what ifs"). 

I can't solve "Prob (Being dealt 2 different denominations first (in 2 cards), given 4 of a kind in 7 cards)", so I'm not sure if I'm stating the possibilities properly. What if I'm dealt Queen, King, and 3 more Kings in the 5 cards. Or what if there are 4 Jacks in the 5 cards and the 4-of-a-kind is made completely in the 5? The location of the 4 of a kind (if linked with one of the initial 2 cards or not), along with how to separate the probabilities of the initial 2 cards from the 5 is confusing me.
I would recommend doing this directly rather than using $Pr(A\mid B)=\frac{Pr(A\cap B)}{Pr(B)}$

You are dealt two cards of different ranks.  Without loss of generality, suppose that the first two cards dealt to you are the Ace of spades and the King of spades.

Now... in order to have a four of a kind, among the next five cards one of the following three situations must occur:


Among the next five cards, the remaining three aces appear
Among the next five cards, the remaining three kings appear
Among the next five cards, four of the cards are all of the same rank (not an ace and not a king)


Note that these cases do not in fact overlap.

There are $\binom{50}{5}$ possible ways to deal the next five cards (where order doesn't matter).  $\binom{47}{2}$ of those ways have all three remaining aces.  Similarly $\binom{47}{2}$ have all three remaining kings.

As for counting the final case, first pick which rank it is that appears four times.  There are $11$ options.  Then, pick which the final remaining card is.  There are $46$ options here.  This gives $11\cdot 46$ such choices.

All in all, this gives the probability as:

$$\dfrac{\binom{47}{2}+\binom{47}{2}+11\cdot 46}{\binom{50}{5}}$$



If you insist on using $Pr(A\mid B)=\frac{Pr(A\cap B)}{Pr(B)}$ where $A$ represents getting a four of a kind and $B$ represents the first two cards are of different ranks, I recommend using the sample space whose elements are of the form $(\text{first card},\text{second card},\text{set of remaining cards})$, which we recognize being equiprobable so we can use counting techniques.

You will get $Pr(B)=\dfrac{52\cdot 48\cdot \binom{50}{5}}{52\cdot 51\cdot \binom{50}{5}}=\dfrac{48}{51}$

Then, in calculating $Pr(A\cap B)$, we do so similarly to the first method, except having to deal with the extra step of deciding the specific cards used in the first two slots:

$Pr(A\cap B)=\dfrac{52\cdot 48\cdot \binom{47}{2}+52\cdot 48\cdot \binom{47}{2}+52\cdot 48\cdot 11\cdot 46}{52\cdot 51\cdot \binom{50}{5}}$

Completing the arithmetic arrives at the same answer as before.
Let's say the first two cards are "A" and "B".  To get "four of a kind", you the next three cards must be of type "A" or of type "B".  There are 50 cards left to draw from, 3 of type "A" and 3 of type "B".  The probability the third card matches either "A" or "B" is 6/50= 3/25.  After that we must match the third card drawn.  On the fourth draw there are 49 cards left, 2 of the correct type.  The probability the fourth card matches the third card is 2/49.  There are then 48 cards left, just 1 of the correct type.  The probability the fifth card matches the third and fourth cards is 1/48.

The probability of four of a kind given that the first two cards are different is (3/25)(2/49)(1/48).