Limited Edition - How many C(6, 3) do I need? A limited edition of products came out, with sets of $3$ items pulled randomly from the series of $6$ and sold in $3$-packs. I know that $\dbinom{6}{3} = 20$ different possible outcomes, but I'm having trouble understanding how to find the likelihood of getting all unique items.
Given that the first three will be unique, what is the likelihood that none of the next three will match? For the first item in the second set, I know there's a $50\%$ chance of getting an item from the first set, but the probability changes for the second item in the second set depending on whether the first is a repeat - either $40\%$ or $60\%$.
Is there a simple way to calculate such probability for the whole second $3$-pack? And could such a formula be extended to find out how many $3$-packs are needed for $\gt95\%$ confidence that I'll end up with all $6$ unique items?
 A: Let the 6 products be $A,B,C,D_1,D_2,D_3$, and assume that the first 3-pack chosen is [ABC].
We want to find $P(H_n)$, where $H_n$ is the event that we get all 6 products after n additional selections.
Notice that $P(H_1)=\frac{1}{\binom{6}{3}}=\frac{1}{20}$, since the only way we can have all 6 products after 1 more draw is by choosing $[D_1D_2D_3]$ for the second 3-pack.
Let $T_i$ be the set of selections of n additional 3-packs which do not include $D_i$, for $1\le i\le3$.
Then $\displaystyle P(H_n)=1-P(H_{n}^c)=1-\frac{|T_1\cup T_2\cup T_3|}{\binom{6}{3}^n}=1-\frac{\sum_{i} |T_i|-\sum_{i<j}|T_i\cap T_j|+|T_1\cap T_2\cap T_3|}{20^n}$
$\hspace{.3 in}\displaystyle=1-\frac{\binom{3}{1}\binom{5}{3}^n-\binom{3}{2}\binom{4}{3}^n+1}{20^n}=1-\frac{3\cdot10^n-3\cdot4^n+1}{20^n}=1-\frac{3}{2^n}+\frac{3}{5^n}-\frac{1}{20^n}$.
In particular, $P(H_5)\approx.907$ and $P(H_6)\approx.953$; so 6 additional 3-packs (for a total of seven 3-packs) will give a probability of greater than .95 of obtaining all 6 products.
