What is the probability of drawing two identical poker hands in a row from 1 52-card deck?

Question

Same ranks, not same suit. Without replacement, 5 card hands, standard 52 card deck (no jokers).

Example:

draw A23QK

draw A23QK (different suits)

Answer 1

$$ \frac{ \binom{13}{5}\!\cdot\!4^5\!\cdot\!3^5+13\!\cdot\!\binom{4}{2}\!\cdot\!\binom{12}{3}\!\cdot\!4^3\!\cdot\!3^3+\binom{13}{2}\!\cdot\!\binom{4}{2}^2\!\cdot\!11\!\cdot\!4\!\cdot\!3 }{ \binom{52}{5}\!\cdot\!\binom{47}{5} } $$ simplified: $$ \frac{486}{5531477}\approx0.008786\% $$ There are three cases:

1. The hand contains no pairs.
2. The hand contains one pair.
3. The hand contains two pair.

For the first case, the probability of drawing a hand of distinct ranks is: $$ \frac{ \binom{13}{5}\!\cdot\!4^5 }{ \binom{52}{5} } $$ From there, the probability of drawing a second matching hand is: $$ \frac{ \binom{13}{5}\!\cdot\!4^5\!\cdot\!3^5 }{ \binom{52}{5}\!\cdot\!\binom{47}{5} } $$ For the second case, the probability of drawing a hand containing one pair is: $$ \frac{ 13\!\cdot\!\binom{4}{2}\!\cdot\!\binom{12}{3}\!\cdot\!4^3 }{ \binom{52}{5} } $$ With the matching hand: $$ \frac{ 13\!\cdot\!\binom{4}{2}\!\cdot\!\binom{12}{3}\!\cdot\!4^3\!\cdot\!3^3 }{ \binom{52}{5}\!\cdot\!\binom{47}{5} } $$ For the third case, the probability of drawing a hand containing two pair is: $$ \frac{ \binom{13}{2}\!\cdot\!\binom{4}{2}^2\!\cdot\!11\!\cdot\!4 }{ \binom{52}{5} } $$ With the matching hand: $$ \frac{ \binom{13}{2}\!\cdot\!\binom{4}{2}^2\!\cdot\!11\!\cdot\!4\!\cdot\!3 }{ \binom{52}{5}\!\cdot\!\binom{47}{5} } $$

Answer 2

I don't know how many jokers are you considering. Let us consider the case with no jokers and then, the case with 2.

1) No jokers:

The poker deck has 52 cards.

In the first hand, you have the next possibilities:

• All cards are different: On the second hand, you have a total of 52-5=47 cards remaining. Then, you have 3 possible options out of 47 for picking a card that is in your hand, on the second, 3 out of 46 of a card in your hand and not repeating and so on. So the probability would be: $$(3/47)*(3/46)*...*(3/43)=3^5/(47!/42!)=9.4295*10^{-8}$$ or $$9.4295*10^{-6} percent$$.

• You have one card repeated: On the second hand, you will need the other two, so you will have to pick the other two cards. The thing here is that it depends on when did you pick those cards because of the first one you pick can be either the first of your hand or the last but one, so the probability is determined by 2/{#cards left} for the first card and 1/{#cards left} for the second (the number of cards left in the second is less than the first). The rest is like explained above.

• You have a set of three cards repeated: You simply cannot draw an equal hand.

• You have two sets of two cards repeated: You have the same problem as with one card repeated, but the procedure is very similar.

2) Two jokers:

The deck has 54 cards.

First-hand options:

• No jokers: Follow the same procedure as above changing 47, 46, 45, ... with 49, 48, 47, ...

• A joker: The probability of picking the other jocker is 1/{#cards left} and the others you need to follow the procedure that I've explained above.

Answer 3

For each of the thirteen ranks, either that rank does not appear, it appears once in both hands, or it appears twice in both hands. There are $4\cdot 3$ ways it can appear once, and $\binom{4}2=6$ ways it can appear twice. Finally, the total number of appearences must be $5$. This translates exactly to finding the coefficient of $x^5$ in the generating function $(1+12x+6x^2)^{13}$. Therefore, the probability is \begin{align} \frac{[x^5](1+12x+6x^2)^{13}}{\binom{52}5\cdot\binom{47}5} &=\frac{[x^5](-5+6(1+x)^2)^{13}}{\binom{52}5\cdot\binom{47}5} \\&=\frac{\sum_k \binom{13}k(-5)^{13-k}\cdot 6^{k}[x^5](1+x)^{2k}}{\binom{52}5\cdot\binom{47}5} \\&=\boxed{\frac{\sum_k \binom{13}k(-5)^{13-k}\cdot 6^{k}\binom{2k}5}{\binom{52}5\cdot\binom{47}5}} \end{align} This agrees with the answer Chameleon gave.