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Toss a coin 10 times. Find the expected number of heads in the first 5 tosses, given 6 heads in the 10 tosses

The given solution:

Think of a box containing 6 heads and 4 tails. Draw 5. The expected number of heads is $\frac{6}{10}$ * 5 = 3

I reasoned that we already know one of the draws will be heads so the answer is 1 + $\frac{5}{9}$*4 = $\frac{29}{9}$

What is my mistake?

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    $\begingroup$ What would your answer be when the numbers in the question are changed to 3 tosses of a coin, you know there were exactly two heads, and you are asked for the expected number of heads in the first two tosses? Now check your answer by enumerating all the possible outcomes. $\endgroup$ – whuber Jan 25 at 21:44
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There is an even simpler answer. If we know that there are $6$ heads in total, we'd expect $3$ heads in the first five tosses and $3$ heads in the last five tosses. However, according to your answer, expected number of heads in the last 5 tosses is $6-{29\over9}={25\over 9}$. So, you would expect different number of heads in the first and last five tosses. There is no reasonable argument to break the symmetry.

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