imc

There are $4$ ways that the same number of heads will be obtained; $0$, $1$, $2$, or $3$ heads. The probability of both getting $0$ heads is ${\left(\frac{1}{2}\right)}^3\left(\genfrac{}{}{0ex}{}{3}{0}\right){\left(\frac{1}{2}\right)}^4\left(\genfrac{}{}{0ex}{}{4}{0}\right)=\frac{1}{128}$ The probability of both getting $1$ head is ${\left(\frac{1}{2}\right)}^3\left(\genfrac{}{}{0ex}{}{3}{1}\right){\left(\frac{1}{2}\right)}^4\left(\genfrac{}{}{0ex}{}{4}{1}\right)=\frac{12}{128}$ The probability of both getting $2$ heads is ${\left(\frac{1}{2}\right)}^3\left(\genfrac{}{}{0ex}{}{3}{2}\right){\left(\frac{1}{2}\right)}^4\left(\genfrac{}{}{0ex}{}{4}{2}\right)=\frac{18}{128}$ The probability of both getting $3$ heads is ${\left(\frac{1}{2}\right)}^3\left(\genfrac{}{}{0ex}{}{3}{3}\right){\left(\frac{1}{2}\right)}^4\left(\genfrac{}{}{0ex}{}{4}{3}\right)=\frac{4}{128}$ Therefore, the probabiliy of flipping the same number of heads is: $\frac{1+12+18+4}{128}=\frac{35}{128}\Rightarrow \boxed{\frac{35}{128}}$