\section*{Problem 6B - 111236B}

50 white and 50 black balls are to be distributed into two indistinguishable urns in such a way that neither urn remains empty and all balls are used.

How should the balls be distributed between the two urns so that the probability of drawing a white ball when one of the two urns is chosen at random and a ball is drawn from it is as high as possible?

\section*{Hints:}

1. In classical probability theory, the probability \(p\) of an event is defined as the quotient of the number \(g\) of favorable cases for this event and the total number \(m\) of possible cases, i.e., \(p=\frac{g}{m}\).

2. Therefore, the probability of drawing a white ball (blindly) from an urn that contains a total of \(u\) balls, of which \(w\) are white, is set to \(p=\frac{w}{u}\).

3. If there are two urns, with the probability of drawing a white ball being \(p_{1}\) and \(p_{2}\) respectively, then the probability of the combined event "selecting one of the two urns and drawing a white ball from the selected urn" is \(p=\frac{1}{2} p_{1}+\frac{1}{2} p_{2}\).
}

Let \(w\) and \(s\) be the numbers of white and black balls, respectively, placed in Urn 1. By assumption, \((w, s) \neq (0,0)\) and \((w, s) \neq (50,50)\). The probability that a ball drawn from Urn 1 or Urn 2 is white is \(p_{1}(w, s) = \frac{w}{w+s}\) and \(p_{2}(w, s) = \frac{50-w}{100-w-s}\), respectively.

The overall probability of drawing a white ball is thus \(p(w, s) = \frac{1}{2} \left(p_{1}(w, s) + p_{2}(w, s)\right)\). We are looking for \(w\) and \(s\) such that \(p(w, s)\) is maximized.

For \(w = s\), we have \(p(w, s) = \frac{1}{2}\). Later, we will see that this is not the desired maximum value. For now, let \(w \neq s\).

Since it is not possible for both urns to have more black balls than white balls, we can assume without loss of generality that Urn 1 has more white balls than black balls, i.e., \(s < w\). This implies \(p_{1}(w, s) > p_{1}(w-1, s-1)\) and \(p_{2}(w-1, s-1) > p_{2}(w, s)\).

Thus, the maximum of \(p(w, s)\) cannot be achieved for \(0 < s < w\).

The maximum of \(p(w, s)\) is therefore achieved for \(s = 0\) (and \(w > 0\)).

We have \(p(w, 0) = \frac{1}{2} \left(1 + \frac{50-w}{100-w}\right) = 1 - \frac{25}{100-w}\). This is maximized for \(w = 1\), and then \(p(1, 0) = \frac{74}{99}\).

Since \(\frac{74}{99} > \frac{1}{2}\), \(\frac{74}{99}\) is the highest possible probability of drawing a white ball. This probability is achieved when one white ball and no black balls (or 49 white and 50 black balls) are placed in Urn 1.