A man with a great memory writes down a series of 40 two-digit numbers. He writes 20 even numbers and 20 odd numbers, each at random. After a week, he tries to rewrite the sequence in the correct order. He remembers where there should be an even number and where there should be an odd number. When there is an even number, he remembers what the number should be, but when there is an odd number, he only remembers correctly with probability 0.9.
Find the expectation of number sequences (two numbers in consecutive positions) that he remembers correctly.
I know to use linearity of expectation for this question. Let $X_i$=1 indicate when a pair of numbers starting at position i are guessed correctly.
When trying to find P($X_i$), I figured it's the probability that $pos_i$ is even, times the probability that $pos_{i+1}$ is even plus the probability that $pos_{i+1}$ is odd, and likewise for when $pos_i$ is odd...
$$P(X_i)=\dfrac{20}{40} \dot\ (\dfrac{19}{39} +\dfrac{20}{39}(0.9)) + \dfrac{20}{40} \dot\ (\dfrac{20}{39} +\dfrac{19}{39}(0.9)) = 0.95 $$
However, I know the answer should be $$\dfrac{\binom{20}{2}\ + \binom{20}{2}(0.9)^2 + \binom{20}{1}\binom{20}{1}(0.9)}{\binom{40}{2}} \approx\ 0.9024 $$
What am I missing in my original attempt?