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?

  • $\begingroup$ Not sure I follow. Expectation and Variance of what? The number of correct guesses? Well, the even ones are all correct so it's only the odd ones to worry about. Those follow a binomial distribution. $\endgroup$ – lulu Jul 11 '18 at 11:06
  • $\begingroup$ @lulu just updated the question. Have to find the expectation of the number of correct pairs of numbers (consecutive in their position in the series) that he remembers correctly $\endgroup$ – Adam G Jul 11 '18 at 11:11
  • $\begingroup$ Well, the probability that the pair starting at slot $i$ is correct is: $\frac 12\times \frac {19}{39}\times 1+ 2\times \frac 12\times \frac {20}{39}\times .9+\frac 12\times \frac {19}{39}\times .9^2\approx .9024$. I don't understand your first calculation. $\endgroup$ – lulu Jul 11 '18 at 11:23

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