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Problem: Henry flips 10 coins and lays them on the desk. He then chooses one of the coins at random, and if it is tails, he flips it to heads. What is the expected number of heads showing? (source: Mandelbrot)

My solution: We can calculate the expected value separately: (10)(1/2)(1) + (1/2)(1/2)(1) = 21/4 , where the second 1/2 represents the probability of the coin Henry picks to be heads, and the third 1/2 representing if the flip is tails given Henry chooses heads.

Solution manual: 11/2

Can someone explain to me where I went wrong in my reasoning? Thanks

SunlightScreen
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    I don't understand your calculation. It looks like you toss ten coins, to get the expected $5$ heads there, but then I don't understand the second term. As to the problem: The selected coin is clearly $H$ so the only question is what happens with the remaining $9$. As we expect $\frac 92$ of those to be $H$ the answer is $1+\frac 92=\frac {11}2$. – lulu Oct 25 '20 at 19:42
  • but how is the selected coin clearly H – SunlightScreen Oct 25 '20 at 19:48
  • Because whatever it was initially, it is converted to $H$. – lulu Oct 25 '20 at 20:18
  • okay I get it I read the problem wrong -- thought it was "if it is tails, he flips it again" – SunlightScreen Oct 25 '20 at 20:23

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Let $X_i$ be the $i^{th}$ flip. Then $P(X_i=1)=\frac{1}{2}+\frac{1}{2}\frac{1}{10}$ that is half chance that its heads then $\frac{1}{10}\frac{1}{2}$ chance its picked to be turned to heads when it was tails. Note $X_i$ are not independent, however we still have

$E(\sum_{i}^{10}X_i)=\sum_{i}^{10}E(X_i)=\sum_{i}^{10} \big( \frac{1}{2}+\frac{1}{2}\frac{1}{10}\big) = 5+\frac{1}{2}=\frac{11}{2}$.

Monty
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