# Probabilities in a game with six dice

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: $$P= =\begin{pmatrix} 6\\2 \end{pmatrix}\left(\frac 16\right)^2\left(\frac 56\right)^4=0.201\not=0.040.$$
• You solved this correctly. Maybe the author forgot to divide by 2? Commented Sep 10, 2016 at 19:41

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.)