What's the probability for a 14-card hand to contain at least one full-house? (This question is related to my previous question: When drawing 14 cards from a set of 52 cards, is it more likely to have a full house or two consecutive pairs?).
Problem: What is the probability for a 14-card hand drawn from a 52-card deck to contain at least one full house?
Based on the very nice and verbose answer I received for the two consecutive pairs I approached the problem in a similar way:
Let $a_k$ denote the number of ways to draw at least 3 cards of one particular rank and at least 2 cards of $k-1$ particular other ranks, then
$$a_k=\sum_{i_1=3}^4\sum_{i_2=2}^4\cdots\sum_{i_k=2}^4\binom4{i_1}\cdots\binom4{i_k}\binom{52-4k}{14-i_1-\cdots-i_k} $$
Computing the values with Sage gives the following results:
\begin{array}{r|r} k&a_k\\\hline 2&21227073582\\ 3&3670707300\\ 4&455417394\\ 5&33893088\\ 6&982368 \end{array}
For every particular $k$ ranks that I choose there are $13$ ways of selecting the rank with at least 3 cards drawn from it and $12\choose k-1$ ways of selecting the $k-1$ remaining ranks. Using the inclusion-exclusion principle I came up with this expression:
$$ \sum_{k=2}^6(-1)^{k}13\binom{12}{k-1}a_{k} = 1246348340952 $$
The probability for having a full-house amongst 14 randomly drawn cards from a 52-card deck, would therefore be:
$$ \frac{1246348340952}{\binom{52}{14}}\approx70.456306\% $$
So, now my question is, whether I applied the inclusion-exclusion principle correctly. I somehow have the feeling that I am still counting certain events twice, by distinguishing between the rank of which at least 3 cards have been drawn and the ranks from which at least two cards have been drawn, since e.g. selecting one particular rank $A$ from which at least 3 cards have been drawn and another rank $B$ from which at least 2 cards have been drawn overlap in the cases, where 3 or more cards have been drawn from rank $B$ as well. I just find it very hard to picture, how all these events overlap, and to verify my answer.
Can anybody give my some advice on how to check the correct use of the inclusion-exclusion principle for this and maybe comparable problems I might try to solve in the future?
 A: I think this calculation is perhaps best carried out as follows:
First find the number $a_k$ of hands that have at least $3$ cards of each of $k$ particular ranks:
$$
a_k=\sum_{i_1=3}^4\cdots\sum_{i_k=3}^4\binom4{i_1}\cdots\binom4{i_k}\binom{52-4k}{14-i_1-\cdots-i_k}\;.
$$
The results are
\begin{array}{r|r}
k&a_k\\\hline
0&1768966344600\\
1&96921517368\\
2&3149345628\\
3&46618572\\
4&170592\\
\end{array}
Then by inclusion–exclusion the number of hands that don’t have at least $3$ cards in any rank is
$$
\sum_{k=0}^4(-1)^k\binom{13}ka_k=741424639488\;.
$$
Of the remaining hands that have at least $3$ cards in at least $1$ rank, the only ones that don’t contain a full house are the ones with $3$ or $4$ cards in exactly $1$ rank and the remaining $11$ or $10$ cards spread out over the remaining $12$ ranks; and those are easily counted. So the number of hands with a full house is
$$
\binom{52}{14}-741424639488-\binom{13}1\binom43\binom{12}{11}\binom41^{11}-\binom{13}1\binom44\binom{12}{10}\binom41^{10}=1024024781208
$$
and the probability to draw a full house is
$$
\frac{1024024781208}{\binom{52}{14}}=\frac{3282130709}{5669763925}\approx57.888\%\;,
$$
in agreement with the computer results.
A: For this, we have two cases. In the first case we have at least three cards in each of at least two ranks. This is similar to what you have already done, so I'll leave out the details. The result should be $219349055640$.
In the second case we have at least three cards in exactly one rank and exactly two cards in at least one other rank. 
$$13\binom{4}{4}\sum_{k=1}^5\left(\binom{12}{k}\binom{4}{2}^k\binom{12-k}{10-2k}\binom{4}{1}^{10-2k}\right)\\+13\binom{4}{3}\sum_{k=1}^5\left(\binom{12}{k}\binom{4}{2}^k\binom{12-k}{11-2k}\binom{4}{1}^{11-2k}\right)\\=804675725568$$ 
For the total of 
$$\begin{align}219349055640&\\804675725568&\\=1024024781208&\end{align}$$
