A system will randomly generate 4 numbers, each from 0-9. So there're 10,000 possible combinations.
Next I'll pick 4 numbers, each also from 0-9. What is the probability that at least 3 of my 4 numbers match the system-generated numbers (order doesn't matter)?
For example, if the system-generated combination is 0-0-1-2, eligible combinations are 0-1-0-2 (4 matches), 3-0-1-0 (3 matches), 2-1-0-8 (3 matches)...
The required probability depends on how much there is matching among the picked numbers.
Say, if all picked numbers are different ($XYZT$), to get four matches the generated conbunations can be $XYZT$ or any permutation of this numbers. The probability to get four matches is $$\dfrac{4!}{10^4}=\dfrac{24}{10^4}.$$ To get exactly three matches, one need to chose four ways to choose which numbers will coincide with generated: $XYZ$ or $XYT$ or $YZT$ or $XZT$. Say, it will be $XYZ$. Denote the last generated number by $U\neq T$.
The last number $U$ can either coincide with one of $XYZ$ or not. In the first case there is three ways to choose it and $\binom{4}{2}\cdot 2$ ways to rearrange $XXYZ$. In the second case there is six ways to choose it and $4!$ ways to rearrange $XYZU$.
Finally, the probability to get exactly three matches when the previously picked numbers are different one from another is $$ \dfrac{4\cdot\left(3\cdot\binom{4}{2}\cdot 2+6\cdot 4!\right)}{10^4}=\dfrac{720}{10^4}. $$ And the total probability to get at least 3 of 4 different numbers match the system-generated numbers is $$\dfrac{24+720}{10^4}=\dfrac{744}{10^4}.$$
Similarly, you can find probability when two of four picked numbers are the same, then three, then four. The last case is simplest one: if picked numbers are $XXXX$, the probability of four matches is $1/10^4$ and the probability of three matches is $9\cdot 4/10^4$, totally $$\dfrac{1+9\cdot 4}{10^4}=\dfrac{37}{10^4}.$$