<s>[INST] Probability that atleast $3$ from $5$ dice show the same face I think I have solved this problem correctly, but my solution does not seem to match the one in the textbook - Problem IV.2 from Feller's introduction to probability theory & applications page 101. Could you help with where I've possible gone wrong?
Note. The author expects the student to use the result for the realisation of at least $m$ among $N$ events.
$$P_m = S_m - {m \choose m-1}S_{m+1} + {m + 1 \choose m-1}S_{m+2}-{m+2 \choose m-1}S_{m+3} + \ldots \pm {N-1 \choose m-1}S_N$$

Five dice are thrown. Find the probability that at least three of them show the same face. (Verify by the methods of chapter II, section 5).

Solution.
Define $A_i := \text{the event that dice i shows the common face}$.
$P(\text{3 selected dice show the same face})=P(A_i A_j A_k)=\frac{6}{6^3}$
$P(\text{4 selected dice show the same face})=P(A_i A_j A_k A_l)=\frac{6}{6^4}$
$P(\text{5 selected dice show the same face})=P(A_i A_j A_k A_l A_m)=\frac{6}{6^5}$
$S_3 = \sum_{1 \le i < j < k \le 5} p_{ijk}={5 \choose 3}\frac{6}{6^3}$
$S_4 = \sum_{1 \le i < j < k < l \le 5} p_{ijkl}={5 \choose 4}\frac{6}{6^4}$
$S_5 = \sum_{1 \le i < j < k < l < m \le 5} p_{ijklm}={5 \choose 5}\frac{6}{6^5}$
\begin{align*}
    P_3 &= S_3 - {3 \choose 2}S_4 + {4 \choose 2}S_5\\
    &= \frac{15}{36}-\frac{15}{216} + \frac{1}{216}\\
    &= \frac{76}{216} \\
    &= \frac{19}{54} \approx 0.35
\end{align*}
However, the answer in the book is $0.21$.
 [/INST] In case you are interested in alternative approaches, here is a slightly different solution, also using the formula for the realization of at least $m$ among $N$ events.
Let's see if we can solve a simpler problem: What is the probability that at least $3$ of $5$ dice roll a $1$? Here $A_i$ is the event that die $i$ rolls a $1$, and
$$S_k = \binom{5}{k} \left( \frac{1}{6} \right)^k$$
for $1 \le k \le 5$.
Applying the formula for the probability of at least $m$ of $N$ events with $m=3$ and $N=5$,
$$P_3 = S_3 - \binom{3}{2} S_4 + \binom{4}{2} S_5 = \frac{23}{648}$$
So $P_3$ is the probability that at least $3$  of $5$ dice will roll a $1$.  The same probability holds for $3$ of $5$ dice rolling a $2,3,4,5$ or $6$.  So the probability that at least $3$ of $5$ dice roll the same number is
$$6 \cdot P_3 = \frac{23}{108}$$
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