I thought that when rolling two dices to get only pairs is for a single roll $\frac{1}{6}$ for two $\left(\frac{1}{6}\right)^2$, and for three $\left(\frac{1}{6}\right)^3$.
But since I have 4 rolls I could get the three doubles with 4 possible solutions, i.e. $4\times\left(\frac{1}{6}\right)^3$. Since I can as well obtain a double I have $\left(+ \left(\frac{1}{6}\right)^4\right)$
i.e. $4 \times \left(\frac{1}{6}\right)^3 + \left(\frac{1}{6}\right)^4$
but somehow this is incorrect. Where is the mistake?
Get a "double" in exactly $3$ out of $4$ rolls:
$$\binom43\cdot\left(\frac16\right)^{3}\cdot\left(1-\frac16\right)^{4-3}$$
Get a "double" in at least $3$ out of $4$ rolls:
$$\sum\limits_{n=3}^{4}\binom4n\cdot\left(\frac16\right)^{n}\cdot\left(1-\frac16\right)^{4-n}$$