# Intuitive reason why sampling without replacement doesnt change expectation?

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?

• The way I intuit this is to take advantage of the inherent symmetry. Imagine the same problem, but replace all red balls with blue balls and vice versa. Should the outcome change? – Ryan Feb 15 '16 at 22:12

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?

• Great answer. Thank you! – user3000877 Feb 15 '16 at 22:18

There is a clear bijection between the outcomes where the $i^{th}$ ball is red to the outcomes where the first ball is red. This implies that the number of outcomes for each event is the same, and thus the probabilities are the same.

Think of it this way:

We have ten labeled slots, labeled $1$, $2$, $3,\dots$. When we remove a ball on the first turn, we may place it in the first slot. Then on the second turn, when we remove a ball we can place it in the second slot, etc...

Compare this to the scenario where on the first turn when we remove the ball we instead place the ball in the fourth slot, and then on the second turn we place the ball in the first slot, then second slot, then third slot, then fifth slot.

Because the action of pulling the ball on the first turn is not affected at all by which slot we place the ball in, we can conclude that the probability that the first slot is occupied by a red ball is the same as the probability that the fourth slot is occupied by a red ball.

Now, interpreting the slots as the order in which the balls are pulled, the result follows.

You can come up with a more pen-and-paper argument for the same using conditional probability and induction.