What is the chance to get two pairs from a hand of five cards if I have extra card (♠️ king, ♠️ queen and ♦️ queen) added to the standard cards What is the chance to get two pairs if I have extra card (♠️ king, ♠️ queen and ♦️ queen) added to the standard cards, thus I have 55 cards? Can anybody help me !!
Is my solution right?
$$\binom{13}{2}\binom{4}{2}\binom{4}{2}\binom{11}{1}\binom{4}{1} + \binom{13}{2}\binom{5}{2}\binom{5}{2}\binom{11}{1}\binom{5}{1} + \binom{13}{2}\binom{6}{2}\binom{6}{2}\binom{11}{1}\frac{\binom{6}{1}}{\binom{55}{5}}$$
First, I chose $2$ ranks of $13$ and then $2$ suits of $4$ (this is for hearts and clovers). Then I chose $2$ ranks of $13$ and then $2$ suits of $5$ suits (this is for diamonds, but I am not sure). Then I chose $2$ ranks of $13$ and then $2$ suits of $6$ suits (this is for spades, but I am not sure).
Lastly, I divided all on $\binom{55}{5}$
 A: It seems wise to count hands with pairs of kings or queens separately:
Hands with a pair of kings and not a pair of queens:
$$\binom{5}{2}\binom{11}{1}\binom{4}{2}\binom{46}{1}$$
because we have to pick $2$ kings out of the $5$, then one other type of pair not king or queen (so $11$ ranks to choose from), then $2$ cards out of the $4$ available in that rank, and finally, $1$ card out of the remaining $55  - 5- 4 = 46$ cards (subtracting the ranks (kings, and the other one) from the total number of cards we now have). The latter single remaining card I do not see occur in your count at all, which seems fishy to me.
Hands with a pair of queens but not a pair of kings:
$$\binom{6}{2}\binom{11}{2}\binom{4}{2}\binom{45}{1}$$
Which has been simularly computed as the previous type, but now we have $6$ queens to choose $2$ from and now $55-6 - 4 = 45$ final cards to choose from.
Hands with both a pair of queens and a pair of kings:
$$\binom{6}{2}\binom{5}{2}\binom{55-11}{1}$$
Simpler because there are no extra ranks for pairs to choose and more cards left at the end (just kings and queens, $11$ now, are forbidden).
Finally hands with neither pairs of kings nor queens:
$$\binom{11}{2}\binom{4}{2}^2\binom{55- 8}{1}$$
Where we first choose the two ranks to form pairs from and then twice the $2$ pairs themselves, and finally a term for the final card.
Now add these $4$ numbers to get all hands with two pairs and indeed divide by $\binom{55}{5}$ to get the probability (all possible choices of $5$ out of the total expanded deck).
A: Your approach seems fine but there are a few mistakes.
Number of ways to choose two pairs -
i) A pair of queen and a pair of king
$\displaystyle \binom{6}{2}\binom{5}{2}\binom{44}{1}$
ii) A pair of queen and a pair from other $11$ ranks (not king)
$\displaystyle \binom{6}{2} \binom{11}{1} \binom{4}{2} \binom{45}{1}$
iii) A pair of king and a pair from other $11$ ranks (not queen)
$\displaystyle \binom{5}{2} \binom{11}{1} \binom{4}{2} \binom{46}{1}$
iv) Two pairs from other $11$ ranks (other than queen and king)
$\displaystyle \binom{11}{2} \binom{4}{2} \binom{4}{2} \binom{47}{1}$
Now the probability is sum of $(i), (ii), (iii), (iv)$ divided by $\displaystyle {55 \choose 5}$.
