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Say I have a subset of the standard scrabble tiles $B$ that contains no blank tiles s.t. $|B|=n$

Also consider I have a target set of letters $L$ s.t. $|L|=k$. These are a set of letters not tiles, so there may be multiple subsets of tiles from $B$ that correspond to $L$.

Say I draw $m$ tiles from $B$ (call it $D\subset B$). What is the probability that $D$ contains tiles corresponding to every letter in $L$ at least once?

Clearly the denominator for this probability will be ${{n}\choose{m}}$, that is the number of possible combinations of tiles $D$ that can be drawn without replacement from $B$.

What I'm missing is the numerator. That is, how many of these possible $D$s contain tiles corresponding to $L$.

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Your model is a bit confusing. If $L \subset B$ and $B$ are tiles, then $L$ must be tiles... –  copper.hat Feb 28 '13 at 17:41
    
Good point, I'll try to clarify. –  CodeFusionMobile Feb 28 '13 at 17:48
    
Since you ask what what the probability $D$ contains tiles corresponding to to every letter in $L$ at least once, I assume $B$ isn't a standard mathematical set, rather its a multiset. So what's the distribution of $B$, or are we supposed to generalize to all possible distributions of $B$? –  DPenner Feb 28 '13 at 18:49
    
I suppose that would be correct. I'm not sure of the exact terminology, but the question is similar to asking the odds of drawing a Spade from a Deck of cards. A deck is a set of 52 unique cards, but multiple cards can have a given property. –  CodeFusionMobile Feb 28 '13 at 19:04
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The probability will depend on how many tiles there are in $B$ with the particular letters you want. If, for each $i=1$, $\dots$, $k$, $B$ contains exactly $v_i$ tiles corresponding to the $i$th letter in $L$, then the number of possible draws which exclude letters #$j_1$, $\dots$, #$j_q$ is $$\binom{n-v_{j_1}-\cdots-v_{j_q}}{m}.$$ So, by the inclusion-exclusion principle, the probability of drawing each of your $k$ selected letters at least once is $$ 1-\binom{n}{m}^{-1} \left(\sum_{1\le q\le k} (-1)^{q-1} \sum_{1\le j_1<\cdots<j_q\le k}\binom{n-v_{j_1}-\cdots-v_{j_q}}{m}\right). $$

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