I'm playing board games and I'm trying to calculate the risk of a certain strategy. To do so I need this probability.
There are 9 different cards and 3 copies of each card for a total of 27 cards. What is the probability that when drawing 9 cards from 27 we get all 3 copies of at least 1 of the 9 unique cards?
Here's my current attempt at solving the problem but I think something is wrong:
n= number of identical cards being drawn = 3 k = total draws = 9 m = copies of each card = 3 a = number of unique cards being considered for n = 1 c = highest number of triples you can draw= 3 c*m = total cards = 27
f(a) = (m choose n)^a ((m(c-a)) choose (k-an)) = ((m(c-a)) choose (k-an)) g(a) = Sum(a=1 to 3)(-1)^(a+1)(9 choose a)f(a) (-1)^(1+1)(9 choose 1)(3(9-1) choose 9-1(3))+(-1)^(2+1)(9 choose 2)(3(9-2) choose 9-2(3))+(-1)^(3+1)(9choose3)(3(9-3)choose9-3(3)) =1211364-47880+84
1163568/(27 choose 9)~=.25
The probability that all $3$ copies of $k$ particular cards are drawn is
$$ \frac{\binom{27-3k}{9-3k}}{\binom{27}9}\;, $$
so by inclusion–exclusion the probability that all $3$ copies of at least one card are drawn is
$$ \sum_{k=1}^3\binom9k(-1)^{k+1}\frac{\binom{27-3k}{9-3k}}{\binom{27}9}=\frac{9\binom{24}6-36\binom{21}3+84\binom{18}0}{\binom{27}9}=\frac{387856}{1562275}\approx0.25\;. $$
Now that I’ve done the calculation, I can discern that this is exactly what you calculated. Next time, please use MathJax (see the link in the comments) to make your post legible.