10. $[7]$ Justine has a coin which will come up the same as the last flip $\frac{2}{3}$ of the time and the other side $\frac{1}{3}$ of the time. She flips it and it comes up heads. She then flips it 2010 more times. What is the probability that the last flip is heads?
Answer: $\frac{3^{2010}+1}{2 \cdot 3^{2010}}$ Let the "value" of a flip be 1 if the flip is different from the previous flip and let it be 0 if the flip is the same as the previous flip. The last flip will be heads if the sum of the values of all 2010 flips is even. The probability that this will happen is $\sum_{i=0}^{1005}\binom{2010}{2 i}\left(\frac{1}{3}\right)^{2 i}\left(\frac{2}{3}\right)^{2010-2 i}$.
We know that
$$
\begin{array}{c}
\sum_{i=0}^{1005}\binom{2010}{2 i}\left(\frac{1}{3}\right)^{2 i}\left(\frac{2}{3}\right)^{2010-2 i}+\binom{2010}{2 i+1}\left(\frac{1}{3}\right)^{2 i+1}\left(\frac{2}{3}\right)^{2010-(2 i+1)}= \\
\sum_{k=0}^{2010}\binom{2010}{k}\left(\frac{1}{3}\right)^{k}\left(\frac{2}{3}\right)^{2010-k}=\left(\frac{1}{3}+\frac{2}{3}\right)^{2010}=1
\end{array}
$$
and
$$
\begin{array}{c}
\sum_{i=0}^{1005}\binom{2010}{2 i}\left(\frac{1}{3}\right)^{2 i}\left(\frac{2}{3}\right)^{2010-2 i}-\binom{2010}{2 i+1}\left(\frac{1}{3}\right)^{2 i+1}\left(\frac{2}{3}\right)^{2010-(2 i+1)}= \\
\sum_{k=0}^{2010}\binom{2010}{k}\left(\frac{-1}{3}\right)^{k}\left(\frac{2}{3}\right)^{2010-k}=\left(\frac{-1}{3}+\frac{2}{3}\right)^{2010}=\left(\frac{1}{3}\right)^{2010}
\end{array}
$$

Summing these two values and dividing by 2 gives the answer $\frac{1+\left(\frac{1}{3}\right)^{2010}}{2}=\frac{3^{2010}+1}{2 \cdot 3^{2010}}$.