TITLE: Probabilities in a game with six dice
QUESTION [3 upvotes]: I have tried to solve this exercise from Fundamentals of statistical and thermal physics-Frederick Reif,  but the answer of item (3) gives me 0.201, and according to the author the solution is 0.040.

Consider a game in which six true dice are rolled. Find the probability of obtaining:

exactly one ace
a least one ace 
exactly two aces


I tried this:
The probability that a given obtained an ace is 1/6, then the probability that an ace will not get is 5/6.

Thus, the probability $P_s$ that a configuration with exactly one ace on six dice has been obtained is:
$$
P_s=\left(\frac 16\right)\left(\frac 56\right)^5,
$$
If exist $ m $ configurations with equal probability of occurrence, which satisfy the proposition, then the event will have a probability $ P $ given by:
$$
P=m\left(\frac 16\right)\left(\frac 56\right)^5,
$$
The number $ m $ of configurations is given by the binomial distribution as follows:
$$
m=\begin{pmatrix}
6\\1
\end{pmatrix}= \frac{6!}{1!5!}=6,
$$
therefore
$$
P=\frac{5^5}{6^5}\approx0.402.\qquad\checkmark
$$
Let M be the event that at least one ace is obtained, let $ M ^ c $ its complement, ie the event that no ace is not obtained; then it must be met:
$$
P(M)=1-P(M^c),
$$
but $\displaystyle P(M^c)=\begin{pmatrix}
6\\6
\end{pmatrix}\frac{5^6}{6^6}$, therefore:
$$
P(M)=1-\frac{5^6}{6^6}\approx 0.6651.\qquad \checkmark
$$
Similarly to item (1) the probability $P$ of getting exactly 2 as is given by:
$$
P=
=\begin{pmatrix}
6\\2
\end{pmatrix}\left(\frac 16\right)^2\left(\frac 56\right)^4=0.201\not=0.040.
$$

REPLY [1 votes]: Comment: Here is a simulation of a million repetitions of the
experiment, verifying that the probability is $0.201 \pm 0.0008.$ 
m = 10^6;  n = 6
die = c(1, 0,0,0,0,0) # '1' stands for Ace, 0's for rest
faces = sample(die, m*n, rep=T)
DTA = matrix(faces, nrow=m)  # 6 rolls in each roll
nr.aces = rowSums(DTA)       # Aces in each row
mean(nr.aces == 2)           # aprx P(exactly 2 Aces in 6 rolls)
## 0.200768                  # 95% margin of simulation error
2*sd(nr.aces == 2)/sqrt(m)
## 0.0008011501

The histogram below shows the simulated distribution of the number of Aces
in six rolls. (Approximate answers to Parts 1 and 2 can also be read from
the histogram.)
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