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We are playing a game with $n\ge 3$ i.i.d.r.v's $X_1,\dots X_n $ where $X_i \in \{A,B,C,F\}$

Let $P(X_i=A)=\alpha,P(X_i=B)=\beta, P(X_i=C)=\gamma, P(X_i=F)=\eta$

so $\alpha,\beta,\gamma,\eta>0$ with $\alpha+\beta+\gamma+\eta=1$

We define a "win" to be when all three of $A,B,C$ appear in at least one of our draws.

So if we define $S:=\{ A,B,C\}$, what is $P(S\subseteq \{X_1,\dots,X_n\})$ ?

[The actual situation I want uses n=5 and $\alpha \ne \beta \ne \gamma $ ]

My attempt: If $n=3$, we have $P(S\subseteq \{X_1, X_2,X_3\}) $

$= P(X_1=A)P(X_2=B)P(X_3=C) \cdot $ [number of words you can make with $ABC$]

$=6\alpha \beta \gamma $

If $n=4$, we have four possibilities $\{A,B,C,A\}, \{A,B,C,B\}, \{A,B,C,C\}, \{A,B,C,F\}, $

$P_A:=P(\{X_1,X_2,X_3,X_4\} = \{A,B,C,A\})= \alpha^2 \beta \gamma \cdot \frac{4!}{2!1!1!}=12\alpha^2 \beta \gamma$

Similarly, $P_B=12\alpha \beta^2 \gamma, P_C=12\alpha \beta\gamma^2 $

$P_F:=P(\{X_1,X_2,X_3,X_4\} = \{A,B,C,F\}) = \alpha \beta \gamma \eta \cdot \frac{4!}{1!1!1!1!} = 24\alpha\beta\gamma\eta$

So $P(S\subseteq \{X_1,X_2,X_3,X_4\} ) = 12\alpha^2 \beta \gamma +12\alpha \beta^2 \gamma +12\alpha \beta \gamma^2 + 24 \alpha \beta \gamma \eta $

$=12\alpha \beta \gamma ( \alpha + \beta + \gamma + 2\eta) = 12\alpha \beta \gamma (1+\eta)$

Now for the bad stuff, if $n=5$ we have 16 possibilities:

$P_{AA} = P (\{X_1,X_2,X_3,X_4,X_5\} = \{A,B,C,A,A\}) = \alpha^3 \beta \gamma \cdot \frac{5!}{3!1!1!} = 20\alpha^3\beta\gamma$

and similarly $P_{BB} = 20\alpha \beta^3 \gamma$ and $P_{CC}=20\alpha \beta \gamma^3$

$P_{AB} = P (\{X_1,X_2,X_3,X_4,X_5\} = \{A,B,C,A,B\}) = \alpha^2 \beta^2 \gamma \cdot \frac{5!}{2!2!1!} = 30\alpha^2\beta^2\gamma$

and similarly $P_{AC}=30\alpha^2\beta \gamma^2 $ and $P_{BC}=30\alpha \beta^2\gamma^2$

$P_{AF} = P (\{X_1,X_2,X_3,X_4,X_5\} = \{A,B,C,A,F\}) = \alpha^2 \beta \gamma \eta \cdot \frac{5!}{2!1!1!1!} = 60\alpha^2\beta\gamma\eta$

and similarly $P_{BF}= 60\alpha \beta^2 \gamma \eta$ and $ P_{CF}=60\alpha \beta \gamma^2 \eta$

$P_{FF}= P (\{X_1,X_2,X_3,X_4,X_5\} = \{A,B,C,F,F\}) = \alpha \beta \gamma \eta^2 \cdot \frac{5!}{2!1!1!1!} = 60\alpha \beta \gamma \eta^2$

$P_{AA}+P_{BB}+P_{CC}+P_{AC}+P_{AB}+P_{BC}+P_{AF}+P_{BF}+P_{CF}+P_{FF}$

$=20\alpha \beta \gamma (\alpha^2+\beta^2+\gamma^2)+30\alpha \beta \gamma (\alpha\beta+\alpha\gamma +\beta \gamma)+60\alpha \beta \gamma \eta (\alpha+\beta+\gamma+\eta)$

$=10\alpha \beta \gamma (2\alpha^2+2\beta^2+2\gamma^2+3\alpha\beta+3\alpha\gamma +3\beta \gamma+6\eta )$

$=10\alpha \beta \gamma [ 2(\eta^2+\eta+1)-\alpha\beta-\alpha\gamma - \beta \gamma ]$

While the question asked for the specific case of $n=5$, i do not see how i would realistically do this for larger $n$. Is there a more elegant solution, or maybe I made an error? I would appreciate some help.

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It's as easy as PIE.

Use the Principle of Inclusion and Exclusion.

Let $I_A, I_B, I_C$ be the events indicating that $A, B, C$ respectively have occured at least once among the $n$ samples.

You seek $$\def\P{\operatorname{\mathsf P}}\begin{align} &\qquad{\P(I_A\cap I_B\cap I_C)} \\[1ex] &=~ {1-\P(I_A^\complement\cup I_B^\complement\cup I_C^\complement)} \\[1ex] &=~ {1-{\P(I_A^\complement)}-{\P(I_B^\complement)}-{\P(I_C^\complement)}+{\P(I_A^\complement\cap I_B^\complement)}+{\P(I_A^\complement\cap I_C^\complement)}+{\P(I_B^\complement\cap I_C^\complement)}-{\P(I_A^\complement\cap I_B^\complement\cap I_C^\complement)}} \\[1ex] &= {1-{(1-\alpha)^n}-{(1-\beta)^n}-{(1-\gamma)^n}+{(1-\alpha-\beta)^n}+{(1-\alpha-\gamma)^n}+{(1-\beta-\gamma)^n}-{(1-\alpha-\beta-\gamma)^n}}\end{align}$$

Where $\P(I_A^\complement\cap I_B^\complement)$ is the probability that neither $A$ nor $B$ occured among the $n$ samples, that being, $(1-\alpha-\beta)^n$, and so forth.

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