Suppose we have $10$ red balls and $10$ blue balls. We pick $5$ balls. What is the expected number of red balls we have?
Let $X_i$ be the event that the $i$th ball is red. Then the answer is $E[\sum _1 ^5 X_i]= \sum _1 ^5 E[x_i]$. Buy why are all the $E[X_i]=\frac12$? Obviously I can verify this for $X_1$ and the computation works out for $X_2$, but it is not intuitive to me at all why this is true for all $X_i$, especially when it seems to me like $X_i$ is dependent on all the $X_j$s before it- what is the intuitive reason for why this computation works out so nicely?
Answer
Consider $20$ cards $10$ red in front and $10$ blue in front. The back of the cards are all black. Now, shuffle the cards and lay $5$ of them face down (or $6$ , or $7$, or all of them if you prefer). Now point to the first one and ask yourself what is the probability that it is blue? What about red?
Now point to the second, or third or any other and ask yourself
What is the probability that this card is blue? Red?
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