Place b blue balls and r red balls in a box. Prove that the probability that the kth ball taken from the box is blue is the same for 1 ≤ k ≤ b.

Question

For some fixed b, r > 0, place b blue balls and r red balls in a box. Select balls from the box, one after the other, without replacement. Prove that the probability that the kth ball taken from the box is blue is the same for 1 ≤ k ≤ b.

Does this just come from the fact that if you do not know what you are pulling out, the probability remains the same? Like if I had a deck of cards face down in a line, the probability that any kth card is an Ace is 4/52. How do you create a proof on this?

Answer 1

We have $b$ blue balls and $r$ red balls, making for $b+r$ total balls in the box. Let $B_k$ be the event the $k^{th}$ ball drawn is blue.

Clearly, $P(B_1) = \frac{b}{b+r}$, because of the total number of balls in the box, we have $b$ blue balls to chose from.

What is $P(B_2)$? This requires a little more thought. Since we are drawing as second ball, this means we have already drawn a ball from the box, meaning we are now drawing from a box of $b+r - 1$ balls. Also, we must consider the cases where the already drawn ball is either blue or red.

In the first case, there are $b-1$ balls to choose from out of the $b+r-1$ balls. This case happens with $\frac{b}{b+r}$ probability. In the other case, we drew red and there are still $b$ balls to choose from. This case happens with $\frac{r}{b+r}$ probability. So, we have:

$P(B_2) = \frac{b}{b+r}\cdot\frac{b-1}{b+r-1} + \frac{r}{b+r}\frac{b}{b+r-1}$ $=\frac{b(b+r-1)}{(b+r)(b+r-1)} = \frac{b}{b+r}$

Using this information, can you figure out a way we can frame the expression for the general case of $P(B_k)$? (Hint: conditional probabilities, like we used for the case of $B_2$).

Answer 2

For all $k\in [b+r]$ the probability that the $k^{\rm th}$ ball is blue is the same, namely ${b\over b+r}$.

Imagine that the blue balls are secretly numbered from $1$ to $b$ and the red balls from $1$ to $r$. The devil arranges the $b+r$ now distinguishable balls in a random linear order in one of $(b+r)!$ ways. Since each of the balls is equally likely to land on place $k$ of this order, the probability that the ball at place $k$ is blue amounts to ${b\over b+r}$.