What's the probability of getting a two pair(2 pairs of similar cards) in a 5 hand with the last card being Ace. An example would be 2 6's, 2 3's, and an ace.
I tried,
$$ \frac{13 \cdot \binom{4}{2} \cdot 12 \cdot \binom{4}{2} \cdot 1 \cdot ?? }{\binom{52}{5}} $$
What would be the count of ace there? I thought $\binom{4}{2}$ but we might have already taken ace in while choosing previous cards.
On
I am going to assume that the pairs must be different and neither can include an ace, because otherwise you have 4 of a kind or a full house.
Start by looking at the first 4 cards. To find the number of favorable deals (in order) you can choose any of 48 cards for the 1st, 3 for the 2nd, 44 for the 3rd, and 3 for the 4th. If B is the first card the pairs can be arranged as BBCC, BCBC or BCCB, so multiply by 3. This gives $$48\times3\times44\times3\times3$$ favorable ways of dealing the 4 cards (in order). The total number of ways of dealing the 4 cards in order is $$52 \times51\times50\times49$$ so the probability of getting two non-ace pairs in the first 4 cards is $$\frac{48\times3\times44\times3\times3}{52 \times51\times50\times49}$$ The probability of an ace for the last card then is 4/48, so the answer is $$\frac{48\times3\times44\times3\times3}{52 \times51\times50\times49}\times\frac{4}{48}$$
On
Shorter answer.
There are ${12 \choose 2}$ ways of choosing two ranks that make the two pair (excluding aces). For each of the pairs, there are ${4 \choose 2}$ ways of choosing the suit. There is one way of choosing the rank of an ace out of the remaining 11 rank choices (${11 \choose 1}$ with probability $\frac{1}{11}$), and there are ${4 \choose 1}$ ways of choosing an ace out of the suits.
So the probability is:
$$ \frac{{12 \choose 2}{4 \choose 2}^2 \times 4}{52 \choose 5} $$
On
You want the probability of drawing an ace as the fifth card, times the conditional probability, when given that will happen, that among the drawing of the first four cards gives two non-ace pairs, or a non-ace pair, a non-ace of another rank, and an ace to pair with the fifth ace.$$\tfrac{\tbinom 41}{\tbinom{52}1}\cdot\tfrac{\tbinom{12}{2}\tbinom 42^2+\tbinom{12}{1}\tbinom 42\tbinom{11}1\tbinom41\tbinom 31}{\tbinom{51}{4}}=\dfrac{2376}{270725}$$
PS: This assumes triples do not count as pairs.
There are two possibilities here, depending on whether the hand includes a pair of aces:
Two non-ace pairs among the first four cards drawn, then an ace is drawn with the fifth card: We have to multiply the probability of obtaining two non-ace pairs among the first four cards by the probability of drawing an ace with the fifth card.
There are $\binom{52}{4}$ ways to select four cards from the deck. There are $\binom{12}{2}$ ways of selecting the two non-ace ranks from which the pairs are drawn and $\binom{4}{2}$ ways to draw two of the four cards for each of those ranks. Hence, the probability of drawing two pairs among the first four cards is $$\frac{\dbinom{12}{2}\dbinom{4}{2}^2}{\dbinom{52}{4}}$$ Since all four aces are among the $48$ cards remaining in the deck, the probability of drawing an ace with the fifth card is $$\frac{4}{48} = \frac{1}{12}$$ Hence, the probability that this case occurs is $$\frac{\dbinom{12}{2}\dbinom{4}{2}^2}{\dbinom{52}{4}} \cdot \frac{1}{12}$$
One non-ace pair, one ace, and one card of another rank among the first four cards, then an ace is drawn with the fifth card: We have to multiply the probability of obtaining one non-ace pair, one ace, and one card of another rank among the first four cards by the probability of then obtaining an ace with the fifth card.
There are $12$ ways to pick the rank of the pair, $\binom{4}{2}$ ways to select two cards of that rank, $4$ ways to pick the first ace, $11$ ways to pick the rank of the other card among the first four cards, and four ways to select a card from that rank. Thus, the probability of drawing one non-ace pair, one ace, and one card of another rank among the first four cards is $$\frac{\dbinom{12}{1}\dbinom{4}{2}\dbinom{4}{1}\dbinom{11}{1}\dbinom{4}{1}}{\dbinom{52}{4}}$$ Since exactly one ace is drawn among the first four cards, there are three aces left among the remaining $48$ cards in the deck. Thus, the probability of drawing an ace with the fifth card is $$\frac{3}{48} = \frac{1}{16}$$ Hence, the probability that this case occurs is $$\frac{\dbinom{12}{1}\dbinom{4}{2}\dbinom{4}{1}\dbinom{11}{1}\dbinom{4}{1}}{\dbinom{52}{4}} \cdot \frac{1}{16}$$
Since these two cases are mutually exclusive and exhaustive, the desired probability can be obtained by adding the above probabilities.