Rolling a die 100 times and probability of product of results Ello, my teacher gave me a simple exercise but I'm stuck.

Rolling a die 100 times, denote the outcome of roll $i$ by $X_i$. Estimate the probability $$ \Pr\{\prod_{i=1}^{100}X_i\leq a^{100}\}$$ for real $1<a<6$

I think that $$ \Pr\{\prod_{i=1}^{100}X_i\leq a^{100}\}=(\Pr\{X_1\leq a\})^{100}$$ by trying to find a solution numerically. Is it correct and why ?
Thanks,
Herosix
 A: The probability satisfies
$$
\mathbb{P}\left( \prod_{i=1}^{100} X_i \leq a^{100}\right) = \mathbb{P}\left( \frac{1}{100}\sum_{i=1}^{100} \ln X_i \leq \ln a\right) = \mathbb{P}\left( \overline{Y}_{100} \leq \ln a\right) ,
$$
where $\overline{Y}_{100}$ denotes the mean of the variables $Y_i = \ln X_i$ over 100 realizations. The variables $Y_i$ are i.i.d. and uniformly distributed with values $\lbrace \ln 1 \dots \ln 6\rbrace$. Therefore, for all $i$, the variable $Y_i$ has expected value
$$
\mathbb{E} Y_1 = \sum_{i=1}^{6} \frac{\ln i}{6} \approx 1.097 \, ,
$$
and variance
$$
\mathbb{V} Y_1 = \left(\sum_{i=1}^{6} \frac{(\ln i)^2}{6}\right) - \left(\mathbb{E} Y_1\right)^2 \approx 0.3659 \, .
$$
According to the central limit theorem, one has the asymptotic result
$$
\lim_{n\rightarrow\infty} \mathbb{P}\left( \sqrt{n}\, \frac{\overline{Y}_n - \mathbb{E}Y_1}{\sqrt{\mathbb{V}Y_1}} \leq t \right) = \Phi(t) \, ,
$$
where $\Phi(t) = \frac{1}{2}\left(1 + \mathrm{erf}(t/\sqrt{2})\right)$ is the cumulative distribution function (CDF) of the standard normal distribution. Considering that $n=100$ is big enough,
$$
\mathbb{P}\left( \overline{Y}_{100} \leq \ln a \right) \simeq \Phi\left(10\, \frac{\ln a - \mathbb{E}Y_1}{\sqrt{\mathbb{V}Y_1}}\right) \approx \Phi\left( \frac{\ln a - 1.097}{0.06049}\right) .
$$
