Probability of seeing each face except one in $n$ throws of a die with 4 faces

Question

Imagine that we throw a die with 4 faces $n$ times. The faces have probabilities of $(0.1, 0.4, 0.2, 0.3)$ of showing up.

What is the probability that we see all faces except a specific, pre-determined face $i$ (say $i = 1$) in $n$ throws?

PS: This isn't homework but a puzzle that came across that I couldn't solve analytically. Thanks.

Answer 1

Any formula for this is going to be messy, but I can explain to you how to do the problem. I'm assuming that the forbidden face is specified in advance. If you mean to say all but one (unspecified) face, do the problem for each specific face and add up the answers.

The probability that face $i$ doesn't show up on a single toss is $(1-p_i)$ and the probability that it doesn't show up in $n$ tosses is $(1-p_i)^n.$ However, this includes cases where some other face doesn't show up either, so we must subtract $$\sum_{j\neq i}(1-p_i-p_j)^n$$ where it is understood that $1\leq j\leq k$. (Sorry, I don't know how to get two lines in the subscript in MathJax.) This isn't the end of the story. Suppose in some sequence of tosses that doesn't include $i$ neither $j$ nor $m$ has shown up. We've subtracted the probability of this sequence twice, and we only want to subtract once so we need to add it back. That is, we must add $$\sum(1-p_i-p_j-p_m)^n$$ where the sum is over all $2$-element sets $\{j,m\}$, with $1\leq j,m\leq k$ with $i\neq j,\ i\neq m$.

Now of course, we have to consider sequences where $3$ faces other than $i$ don't show up. We continue as above, alternately, adding and subtracting, by applying the principle of inclusion and exclusion.

Answer 2

It wouldn't win any beauty contests, but you could imagine a transition matrix between the sixteen possible states of combinations of faces that would come up. For instance, if you imagined the problem with flipping a coin with a $\frac13$ probability of heads and a $\frac23$ probability of tails, your matrix would look like this:

$$ M=\begin{pmatrix} 0 & 0 & 0 & 0 \\ \frac13 & \frac13 & 0& 0 \\ \frac23 & 0 & \frac23 & 0 \\ 0 & \frac23 & \frac13 & 1 \\ \end{pmatrix} $$

where the first column represents not seeing any results, the second representing heads only, the third tails only, and the fourth both heads and tails. Then you could just calculate $$M^n\begin{pmatrix}1\\0\\0\\0\end{pmatrix}$$ for any $n$ that interestsed you and read off the probability for the individual final state that interested you. Your problem would invove a $16\times16$ matrix, but it wouldn't be hard to create it and load it into Mathematica or wherever to churn out the probability you wanted.