Probability of some dice rolls

Question

Question

A fair six-sided die is rolled $12$ times. Find the probability that each of the six possible outcomes ($1$, $2$, $3$, $4$, $5$ and $6$) come up at least once.

My working

Let $X_n$ be the random variable denoting the number of times $n$ appears, out of $12$ rolls. For example, $X_1$ will be the random variable denoting the number of times $1$ appears, out of $12$ rolls.

Then, $$X_n \sim B(12, \frac 1 6)$$ and the required probability is

\begin{align} P(X_1 \geq 1)P(X_2 \geq 1)P(X_3 \geq 1)P(X_4 \geq 1)P(X_5 \geq 1)P(X_6 \geq 1) & = [P(X \geq 1)]^6 \\[5 mm] & = [1 - P(X = 0)]^6 \\[5 mm] & = \left[1 - \left(\frac 5 6\right)^{12}\right]^6 \\[5 mm] & \approx 0.490 \end{align}

Answer

The solution by my professor gives the required probability as $$1 - \left[\binom 6 1 \left(\frac 5 6\right)^{12} - \binom 6 2 \left(\frac 4 6\right)^{12} + \binom 6 3 \left(\frac 3 6\right)^{12} - \binom 6 4 \left(\frac 2 6\right)^{12} + \binom 6 5 \left(\frac 1 6\right)^{12}\right] \approx 0.438.$$

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The solution does come with an explanation, but I am very confused by it, so I am hoping to find more intuitive suggestions on why my working is wrong as well as how to go about approaching the problem.

Answer 1

You are assuming that the random variables $X_i$'s are independent but this is not the case.

Let $A_i$ be the event that $i$ occur, we want to compute $$P\left(\bigcap_{i=1}^6 A_i\right)=1-P\left( \bigcup_{i=1}^6 A_i^c\right)$$

We then compute $P\left( \bigcup_{i=1}^6 A_i^c\right)$ by inclusion-exclusion principle.

Notice that we have

$$P\left(\bigcap_{i=1}^m A_i^c\right)=\left( \frac{6-m}{6}\right)^{12}$$

By symmetry:

$$P\left( \bigcup_{i=1}^6 A_i^c\right)=\sum_{m=1}^6 (-1)^{m+1}\binom{6}{m}P\left(\bigcap_{i=1}^m A_i^c\right)$$

Answer 2

The issue is that $X_1, X_2, X_3, X_4, X_5, X_6$ as you defined are NOT independent random variables, due to the constraint that $$\sum_{i=1}^6 X_i = 12.$$ Therefore, the probability that each of the six possible outcomes (1, 2, 3, 4, 5 and 6) come up at least once, i.e. $P(X_1 \geq 1, X_2 \geq 1, X_3 \geq 1, X_4 \geq 1, X_5 \geq 1, X_6 \geq 1)$, cannot be factorized as the product of $P(X_i \geq 1), i = 1,\dots,6$.