Apologies for any notational problems or lack of clarity: I'm a linguist not a mathematician. Anyway, here goes:

There is an urn with $n$ balls divided into $k$ colours, where the number of each colour is $P = \{p_1, p_2 ... p_k\}$.

Define a constant $t < k$ and select $t$ colours with uniform probability from $k$. Remove all balls of those selected colours. What now is the expected probability of selecting any given colour after this operation, given that we don't know which colours will be selected?


  • $\begingroup$ Label your colors as $S = \{1,2,\ldots,k\}$. You remove $t$ of those colors, call the exact set of colors that you removed $T$. Let $X$ be the color that you pick after the operation that you describe. Then $\mathbb{P}(X = x) = \sum_{T} \mathbb{P}(X = x|T) \mathbb{P}(T)$. By this I mean that you sum over all possible sets of size $t$ that we remove and then condition on this occuring and multiply by the probabilty of removing the colors in set $T$. Thankfully, $\mathbb{P}(T) = t/k$, the other probability is much harder since the number of balls per color is not constant. Hope this hint helps. $\endgroup$
    – Ritz
    Commented Dec 10, 2014 at 13:55
  • $\begingroup$ That's a nice reframing, thanks! Two things though: isn't $\mathbb{P}(T)= \frac{1}{k \choose t}$? Secondly, I have no idea how to approach the other probability apart from brute force. Is there any way of getting out a concise expression? Sorry! $\endgroup$
    – Gorga
    Commented Dec 10, 2014 at 14:31

1 Answer 1


I doubt you'll find a nice expression for this. The probability of drawing a ball of colour $j$ is

$$ \frac{p_j}{\binom kt}\sum_{\scriptstyle T\subseteq[1,k]-\{j\}\atop\scriptstyle|T|=t}\frac1{1-\sum_{i\in T}p_i}\;. $$


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