Probabilities in a game with six dice

Question

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:

1. exactly one ace
2. a least one ace
3. 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.

1. 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 $$ P=\frac{5^5}{6^5}\approx0.402.\qquad\checkmark $$
2. 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: $$ P(M)=1-\frac{5^6}{6^6}\approx 0.6651.\qquad \checkmark $$
3. Similarly to item (1) the probability $P$ of getting exactly 2 as is given by: $$ P= =\begin{pmatrix} 6\\2 \end{pmatrix}\left(\frac 16\right)^2\left(\frac 56\right)^4=0.201\not=0.040. $$

Answer 1

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.)