Probabilistic puzzle

Question

There are $n+1$ boxes and every box contains $n$ balls. For every $k\in\left\{ 0,1,\ldots,n\right\} $ there is exactly $1$ box containing $k$ white balls and $n-k$ black balls. A box is picked out and $m$ balls are taken out. Here $m<n$ so the box is not empty yet. All balls that are taken out appear to be white balls. Now another ball is taken out of that same box. What is the probability that it is a white one?

I allready have an answer of myself, so am not in the need of one. I just like this 'puzzle', and maybe so do you. Next to that I am interested in answers that are 'nicer' than mine. It is beyond doubt that they exist.

Answer 1

I liked the puzzle, and your elegant solution of it. One possibly more direct, or at least somewhat informative, approach is to look at the puzzle's continuous counterpart. Thus, corresponding to the set of $n+1$ urns let ${\bf P}$ be a uniform $[0,1]$ random variable one realization ${\bf P}=p$ of which is used to perform $m$ independent Bernoulli trials, each with success probability $p$. Let ${\bf S}_m$ be the number of successes in these $m$ trials. Then ${\sf P}({\bf S}_m=m) \; = \; \int_{0}^{1}\, p^m \, dp \, = \, \frac{1}{1+m}$, exactly as in the discrete urn case. Also, using Bayes' Theorem the posterior density of ${\bf P}$ given that ${\bf S}_m=m$ is then directly seen to be ${\sf P}({\bf P} = p | {\bf S}_m=m) \, = \, f_{P}(p) \, = \, (1+m) \, p^m$. This in turn implies that given ${\bf S}_m=m$ the $(m+1)-$th Bernoulli trial results in a further success with probability $\int_{0}^{1}\, p \, f_{P}(p) dp \, = \, \frac{m+1}{m+2}$, again exactly corresponding to the discrete urn case. Of course, one may still search for an even ``more direct'' approach.

Answer 2

We can go one step further than your own second answer and derive not just the intermediate result $P(W_m)=\frac1{m+1}$ but also the final result $P(W_{m+1}\mid W_m)=\frac{m+1}{m+2}$ purely by symmetry without calculation.

As in that answer, consider a box with $n+1$ balls in a row where the first ball drawn determines the border between white and black balls among the remaining $n$ balls. As you say, the probability for drawing $m$ white balls is the probability for that border ball to be the last of these $m+1$ balls in the row, and thus by symmetry $\frac1{m+1}$. But if we then draw another ball, it is equally likely to be inserted into the order of these $m+1$ balls at any of the $m+2$ possible places, and only one of these places it beyond the border ball and thus among the black balls, so the probability for this is $\frac1{m+2}$.

Answer 3

I have decided to publish here my own answer. This also because of the comments that were given to my 'question'. Also I have decided not to do this anymore. I do not want to do things here that other people rather see not happen.

Denote the event that the box picked out contains exactly $k$ white balls by $D_{k}$.

Denote the event that the $m$ balls taken out are all white by $W_{m}$.

Then:

$$P\left(W_{m}\right)=\sum_{k=m}^{n}P\left(W_{m}|D_{k}\right)P\left(D_{k}\right)=\frac{1}{n+1}\sum_{k=m}^{n}P\left(W_{m}|D_{k}\right)=$$$$\frac{1}{n+1}\sum_{k=m}^{n}\left({k\atop m}\right)\left({n-k\atop 0}\right)\left({n\atop m}\right)^{-1}=\frac{1}{n+1}\left({n\atop m}\right)^{-1}\sum_{k=m}^{n}\left({k\atop m}\right)$$

With induction we find easily that: $$\sum_{k=m}^{n}\left({k\atop m}\right)=\left({n+1\atop m+1}\right)$$ So:

$$P\left(W_{m}\right)=\frac{1}{n+1}\left({n\atop m}\right)^{-1}\left({n+1\atop m+1}\right)=\frac{1}{m+1}$$

Then: $$P\left(W_{m+1}|W_{m}\right)=P\left(W_{m+1}\cap W_{m}\right)/P\left(W_{m}\right)=P\left(W_{m+1}\right)/P\left(W_{m}\right)=\frac{m+1}{m+2}$$

So the answer is:

$$P\left(W_{m+1}|W_{m}\right)=\frac{m+1}{m+2}$$

Especially the 'nice' equality $P\left(W_{m}\right)=\frac{1}{m+1}$ makes me think that a more direct route to that result exists. If there is one then please let me know, and thanks for that in advance.

Answer 4

I ended my original answer with:

Especially the 'nice' equality $P\left(W_{m}\right)=\frac{1}{m+1}$ makes me think that a more direct route to that result exists. If there is one then please let me know, and thanks for that in advance.

Inspired by this I finally (more than $6$ years later) found an answer for that.

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I use a new setup that actually is the same as the original one.

Instead of choosing one box out of $n+1$ boxes we use one box and then choose uniformly how many of the balls in it are white.

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Let there be a box containing $n+1$ balls.

For convenience we label the balls with the numbers $1,2,\dots,n+1$.

We take out randomly $m+1$ balls one by one and denote the drawn numbers by $X_{0},X_{1},\dots,X_{m}$.

After drawing the first ball, the balls with a number less than $X_{0}$ are painted white and all other remaining balls are painted black.

Then we go on with drawing the other balls.

If $Y$ denotes the number of white balls among the balls with labels $X_{1},\dots,X_{m}$ then by symmetry it is evident that $Y$ has uniform distribution on $\left\{ 0,1,\dots,m\right\} $.

Then: $$P\left(Y=m\right)=\frac{1}{m+1}$$ and the event $\left\{ Y=m\right\} $ corresponds with the event $W_{m}$ mentioned in my original answer.