Pick White Marble

If one jar contains $a$ white marbles and $b$ black marbles, then the probability of picking a white marble is $f(a,b) \; = \; \tfrac12\Big[\frac{a}{a+b} + \frac{50-a}{100-a-b} \Big]$ provided that $0 < a+b < 100$ . Note that $f(0,0) = f(50,50) = \tfrac14$ (when all the balls are in one jar).

If $1 \le N = a+b \le 99$ then $f(a,b) \; = \; f(a,N-a) \; = \; \tfrac12\Big[\frac{a}{N} + \frac{50-a}{100-N}\Big] \; = \; \frac{25N + (50-N)a}{N(100-N)}$ and since the coefficient of $a$ in the numerator of this fraction is $50-N$ , we see that, for fixed $N$ , $f(a,b)$ is maximized by maximizing $a$ when $1 \le N \le 50$ , while $f(a,b)$ is maximized by minimizing $a$ when $51 \le N \le 99$ . In other words, $\begin{array}{rcl} \displaystyle \mathrm{max}_{a+b=N} f(a,b) & = & \displaystyle \left\{ \begin{array}{lll} f(N,0) \; = \; \frac{75-N}{100-N} & \qquad & 1 \le N \le 50 \\ f(N-50,50) \; = \; \frac{N - 25}{N} & & 51 \le N \le 99 \end{array}\right. \\ & = & \displaystyle \frac{25 + |50-N|}{50 + |50 - N|} \end{array}$ The symmetry about $N=50$ is to be expected; if one jar contains $N$ balls, then the other contains $100-N$ balls. Anyway (since any of these last probabilities is greater than $f(0,0) = f(50,50)$ ): $\mathrm{max}_{0 \le a,b \le 50} f(a,b) \; = \; \mathrm{max}_{1 \le N \le 99} \frac{25 + |50-N|}{50 + |50-N|} \; = \; \boxed{\frac{74}{99}}$ with this largest probability being $f(1,0) = f(49.50)$ (with $N=1$ or $N=99$ respectively), so that one jar contains a single white ball, and the other jar all the other balls.

The maximum value of the probability that the marble picked is white is $\dfrac{74}{99}$ .

The idea of the proof consists of showing that any other distribution of marbles can be modified to increase the desired probability.

Assume that one jar contains $x$ white marbles and $y$ black marbles, and the other one has $50-x$ white marbles and $50-y$ black marbles. Then, the probability of getting a white marble at the end is given by the following function:

$f(x,y) = \dfrac{1}{2}.\dfrac{x}{x+y} + \dfrac{1}{2}.\dfrac{50-x}{100-x-y}$

We distinguish three cases:

• Case 1: If both jars contain the same number of white and black marbles ( $x=y$ ) then the probability is $\dfrac{1}{2}$ , which is less than $\dfrac{74}{99}$ .

• Case 2: If in each jar the number of white marbles is different from the number of black marbles, then one of the jars will have more white marbles than black marbles, i.e., $x>y$ . Calling $u = x-y$ we have (after doing some algebra) that the probability is:

$f(x,x-u) = \dfrac{1}{2} + \dfrac{u}{4}\left[\dfrac{1}{2x-u}-\dfrac{1}{100+u-2x}\right]$

That is a decreasing function in $x$ . Hence the probability can be increased by moving a white marble and a black marble from that jar to the other. If we keep doing that until no black marbles remain in that jar ( $x=u$ ) we will find ourselves in:

• Case 3: If one jar contains only white marbles, then the probability is

$f(x,0) = \dfrac{1}{2} + \dfrac{1}{2 + \dfrac{100}{50 - x}}$ .

That is also a decreasing function in $x$ , so by moving white marbles from that jar to the other one the probability will increase, until we end up with a single white marble in that jar and all the other marbles in the other jar, yielding which we claimed to be the maximum probability: $f(1,0) = \dfrac{74}{99}$ .

That exhausts all possible configurations of marbles and ends the proof.

So, $m+n=\boxed{173}$ .

In order to maximize your probability of picking a white marble you should put only one white marble in a jar and all the other marbles in the other, this is one way to get a probability higher than 50% and in fact is the max probability achievable. So the sought probability is $\frac{1}{2} + \frac{1}{2} \times \frac{49}{99} = \frac{74}{99}$ So the solution is $\boxed{173}$