Probability of drawing a pair in a deck of 162 playing cards. I was asked this question by a friend of mine and I found it intriguing: if you are dealt an 8 card hand from a deck of three card decks shuffled together, including jokers, what is the probability of drawing at least a pair?
I thought about it a while and concluded that the best way to do this would likely be finding the probability that you were dealt no pairs and subtract it from 1. But the jokers really make this a bit more of a difficult question. I figured I would let $$c= \text{the probably of drawing an 8 card hand in which there are no pairs and none are jokers}$$ and let $$j = \text{the probability of drawing an 8 card hand in which there are no pairs and one joker}$$ so that my final calculation (I think) would be $$P=1-c-j.$$
Now to calculate $c$ I would have $\binom{13}{8}$ representing choosing the 8 different ranks, $4^8$ choosing the different suit for each card (or would it be $12^8$ as there are $3$ 2 of hearts, $3$ kings of clubs, etc ?), and finally $\binom{162}{8}$ representing the ways to draw a hand of 8 cards from three standard decks shuffled together. Hence $$c= \frac{\binom{13}{8} 4^8}{\binom{162}{8}}.$$ Now calculating $j$ is where I have some questions. How do I express the possibility I am dealt one and only one joker? I was thinking maybe representing each card pull as a possibility on its own. Without loss of generality, suppose my first card pull was the joker. The odds of that happening are $\frac{6}{162}=\frac{1}{27}$ . But then I have to make sure the rest of the card pulls are non- jokers and non-pairs. Here's the best I got it is virtually the same as finding $c$ but with one less card. That is $\binom{13}{7}$ represents the different ways to pull the $7$ different ranks, $4^7$ represents which suit each card is, and $\binom{161}{7}$ are the ways of pulling $7$ cards from the deck (minus one pulled joker). So $$j= \frac{1}{27}\cdot \frac{\binom{13}{7} 4^7}{\binom{161}{7}}.$$
Calculating $j$ is iffy to me. I was thinking alternatively representing a fraction for each card pull. That is $$j = \frac{1}{27}\cdot \frac{156}{161} \cdot \frac{144}{160} \cdot\frac{132}{159} \cdot \frac{120}{158} \cdot \frac{108}{157} \cdot \frac{96}{156} \cdot \frac{84}{155}.$$ Any help or clarification would be much appreciated. My friends are currently in a hot debate about it :)
 A: It should be $1-c-j$, not $(1-c)+(1-j)$.
For $c$ we have $$\frac{\binom{13}{8}12^8}{\binom{162}{8}}$$ and for $j$ we have $$\frac{6\binom{13}{7}12^7}{\binom{162}{8}}$$
A: The probability $c$ you get $8$ distinct $13$ non-joker ranks is
$$\frac{156 \cdot 144 \cdot 132 \cdot 120 \cdot 108 \cdot 96 \cdot 84 \cdot 72 }{162 \cdot161 \cdot160 \cdot159 \cdot158 \cdot157 \cdot156 \cdot155}$$
The probability $j$ you get $7$ distinct $13$ non-joker ranks and one joker (in any of the eight positions) is
$$8 \cdot\frac{6 \cdot 156 \cdot 144 \cdot 132 \cdot 120 \cdot 108 \cdot 96 \cdot 84 }{162 \cdot161 \cdot160 \cdot159 \cdot158 \cdot157 \cdot156 \cdot155}$$
So the probability of drawing a pair or more is $1 - c-j \approx 0.90656$
A: OP hasn't given clarifications whether jokers can be counted as a pair.
Generally, jokers don't count as a  pair, so they could range from $0 \;thru\; 6$ in the hand !
Total number of hands is clear, $\binom{162}8$. From this, we need to subtract hands without pairs to get valid number of hands.
Imagine $13$ blocks of colorful marbles, red, orange, etc, with each block having $12$ marbles of a specific color, plus one block of $6$ black marbles
To eliminate consideration of the black marbles, we shall have to go through case by case. Just as an example, suppose $3$ black are drawn, then only $5$ more need to be drawn from the colorful blocks, and we need to subtract invalid hands accordingly.
The draws with no "regular" pairs to be subtracted from the total ways will be
$$\sum_{k=0}^6\binom6{k}\binom{13}{8-k}\binom{12}1^{8-k} = A,\; say\;$$
And at least one pair = $1- \dfrac{A}{\binom{162}{8}}$

PS:
I understand that when "wild", jokers can also be a pair. Then all that needs to be done to the above formula is to work out for $k = 0 \;to\; 1$
