jmc

This problem requires a little bit of casework. There are four ways in which they can both get the same number: if they both get 1's, both get 2's, both get 3's or both get 4's. The probability of getting a 1 is $\frac{1}{10}$, so the probability that they will both spin a 1 is ${\left(\frac{1}{10}\right)}^2=\frac{1}{100}$. Similarly, the probability of getting a 2 is $\frac{2}{10}$, so the probability that they will both spin a 2 is ${\left(\frac{2}{10}\right)}^2=\frac{4}{100}$, the probability of getting a 3 is $\frac{3}{10}$, so the probability that they will both get a 3 is ${\left(\frac{3}{10}\right)}^2=\frac{9}{100}$ and the probability of getting a 4 is $\frac{4}{10}$, so the probability that they will both get a 4 is ${\left(\frac{4}{10}\right)}^2=\frac{16}{100}$. So our answer is $\frac{1}{100}+\frac{4}{100}+\frac{9}{100}+\frac{16}{100}=\frac{30}{100}=\boxed{\frac{3}{10}}$.