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A box contains $n$ white and $n$ black pebbles. A fair sided die with $n$ sides(numbers from $1$ to $n$) is thrown and the number you get, let's call it $r$, is how many pebbles are taken out randomly.

One pebble is chosen randomly from the box, what is the probability that the chosen pebble is black?

(preferably using as simple probabilty as possible, no binomial distribution and stuff like that, it's sadly the basic stuff that's bothering me.)

My attempt:

So I assume we need $\sum_{r=1}^{n} P(r)\cdot \sum_{k=0}^{r} P($k black pebbles are taken out$)\cdot P($black pebble chosen from the remaining$)$

$P(r) = 1/n$

And then, for a fixed $r \in \{1,...,n\}$:

$P($k black pebbles are taken out$) = \frac{\binom{n}{k}\cdot \binom{n}{r-k}}{\binom{2n}{r}}$

$P($black pebble chosen from the remaining $2n - r$ $) = \frac{n-k}{n-r}$

But I'm really not sure how to make anything useful out of this. Thank you in advance!

EDIT: I know the answer is $1/2$, but I want to know how to reach it!

Collapse
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    How could it be more likely to be white than black? How could it be more likely to be black than white? – Angina Seng Aug 27 '18 at 21:13
  • This isn't clear. Say you toss a $d$. Do you remove the $d$ pebbles randomly? If so, then there is symmetry between white and black so the answer would have to be $\frac 12$. If, on the other hand, the $d$ pebbles are not removed randomly we need to know what rule is followed. – lulu Aug 27 '18 at 21:13
  • @lulu yes, sorry, I will edit it. d pebbles are removed randomly indeed.

    I understand the answer is 1/2, but I would really like a literal way to get to 1/2.

     – Collapse Aug 27 '18 at 21:14
  • Symmetry is a perfectly valid method. Whatever answer you get must be the same for black and white and the two must sum to $1$, so... – lulu Aug 27 '18 at 21:15
  • To phrase it (slightly) differently, suppose the pebbles were numbered $1,\cdots , 2n$. Then each pebble has the same probability of being the one chosen (namely, $\frac 1{2n}$), and half of them are black. – lulu Aug 27 '18 at 21:17
  • @lulu Sorry if I'm annoying, but I would really like the bigger picture, how to act generally in problems like these, in case they are similar to this one but let's say not symmetrical, just any hint that would help me compute this more easily.

    But.. you're not choosing $1$ from $2n$. There's more stuff happening xD

     – Collapse Aug 27 '18 at 21:18
  • Well, symmetry is one of the most powerful tools there is. I always look for symmetry, wherever possible. That said, my second method of solution shows you how to handle the case where, initially, there weren't the same number of white and black pebbles. – lulu Aug 27 '18 at 21:20
  • @HirakoShinji You are choosing one from $2n$. The other stuff happening is misdirection. – Angina Seng Aug 27 '18 at 21:21
  • @LordSharktheUnknown I understand it is, but I need to know how to mathematically explain that it's just misdirection, that's what's bothering me. – Collapse Aug 27 '18 at 21:22

1 Answers1

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The summation you need to simplify is $$ \sum_{r=1}^n\frac1n\sum_{k=0}^r\frac{\binom{n}{k}\binom{n}{r-k}}{\binom{2n}{r}}\cdot \frac{n-k}{n-r}=\frac1n\sum_{r=1}^n\frac{1}{\binom{2n}{r}(n-r)}\sum_{k=0}^r\binom{n}{k}\cdot (n-k)\cdot \binom{n}{r-k} $$ Since $\binom{n}{k}(n-k)=\binom{n-1}{k}n$, this reduces to $$ \sum_{r=1}^n\frac{1}{\binom{2n}{r}(n-r)}\sum_{k=0}^r\binom{n-1}{k}\binom{n}{r-k} $$ Using Vandermonde's$^*$ identity, the inner summation is equal to $\binom{2n-1}r$. Combined with the fact that $\binom{2n}{r}(n-r)=\binom{2n-1}{r}2n$, this becomes $$ \sum_{r=1}^n\frac{\binom{2n-1}r}{2n\binom{2n-1}r}=\sum_{r=1}^n\frac1{2n}=\frac12. $$

$^*$ The Wikipedia link gives several proofs of Vandermonde's identity, but here is one with a probabilistic flavor. The probability that a hypergeometric random variable with parameters $2n-1,n-1$ and $r$ is equal to $k$ is $\binom{n-1}k\binom{n}{r-k}/\binom{2n-1}r$. Summing over all possible values of $k$, the result must be one.

Mike Earnest
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