One White Marble, Alone in a Jar, Is Worth 74/99

You have $50$ white marbles and $50$ black ones, and two empty jars. Distribute all hundred marbles between the jars however you like, with neither jar left empty. Someone then picks a jar by a fair coin toss and draws one marble from it, uniformly at random. Arrange the marbles to make white as likely as you can.

The colours are perfectly balanced and the jar is chosen by a fair coin, so the answer feels pinned at $1/2$. It is not. The best arrangement reaches

and the arrangement that gets there is grotesquely lopsided: a single white marble alone in one jar, and the other ninety-nine marbles piled into the second.

Why the even split gives exactly one half

The reflex arrangement is $25$ white and $25$ black in each jar. Both jars are then half white, the coin cannot help or hurt, and the law of total probability returns

Equation (2) is correct. That is worth stressing, because the trap here is not an arithmetic slip. The candidate computes an honest $1/2$ and then makes a second, silent claim: that no arrangement can do better. The arithmetic is right and the conclusion drawn from it is wrong.

A much larger set of arrangements gives exactly one half

Before improving on $1/2$, it is worth seeing how much company the even split has. Write $(w_1, n_1)$ for the white count and total in jar one, and $(w_2, n_2)$ for jar two, so that $w_1 + w_2 = 50$ and $n_1 + n_2 = 100$. The quantity you are maximising is

Now suppose the two jars merely have the same size, so $n_1 = n_2 = 50$, and colour them however you please. Equation (3) collapses:

Every equal-size split gives exactly one half, including the wildest ones. Fifty white in one jar and fifty black in the other is worth exactly as much as $25$ and $25$. So the answer is completely insensitive to colour and completely sensitive to size, which is the opposite of where attention naturally goes.

Buy one branch outright

The coin decides which jar you draw from and you cannot touch it. What you can do is decide what each of its two outcomes is worth. One of them can be made worth exactly $1$, and the price of doing that is one white marble.

Put a single white marble alone in jar one. That jar now returns white with certainty. The remaining $49$ white and $50$ black marbles go into jar two, which returns white with probability $49/99$, barely below a half. Average the two branches:

The gain over the even split is

which is almost a quarter of probability created out of nothing but rearrangement. Note also what (5) is not. It is $0.7474\ldots$, and

so the honest headline is nearly three quarters and never three quarters. The gap is tiny and it is real, and equation (13) below explains why it can never be closed.

Why nothing beats it

Finding a good arrangement is easy. Proving that none is better takes an argument, and the argument is short enough to give in an interview.

Label the jars so that $n_1 \le n_2$. If $n_1 = n_2$ then equation (4) already tells us the value is $W/(W+B)$, which for $W = B$ is exactly $1/2$, so assume $n_1 < n_2$ from here.

Suppose the small jar still contains a black marble. Two cases. If the large jar contains a white marble, swap the two. Jar sizes do not change, $w_1$ rises by one and $w_2$ falls by one, so (3) changes by

because $n_1 < n_2$. Strictly better. If instead the large jar holds no white marble at all, then every white marble is already in the small jar; move a black marble across, which raises $w_1/n_1$ and leaves $w_2/n_2 = 0$ alone. Strictly better again.

Each step removes a black marble from the small jar without ever making things worse, so after finitely many steps the small jar is pure white. Say it holds $k$ white marbles. The value is then

for $W = B = 50$, and differentiating the closed form $f(k) = (150-2k)/(2(100-k))$ gives

so $f$ is strictly decreasing and $k = 1$ wins. That closes the proof, and an exhaustive sweep of all $2599$ legal arrangements confirms it: the maximum is $74/99$ and it is attained by exactly one arrangement.

One marble, not a handful

Equation (10) is the piece most people get wrong on their second guess. Having seen that isolating white marbles helps, the instinct is to isolate a few of them, which feels safer. It is worse:

Each extra white marble you move into the certain jar is wasted there, since that jar was already returning white with probability one, and it is missed in the other jar, where it was the only thing fighting the fifty black marbles.

Follow the curve to its right end and something pleasing happens. At $k = 50$ the arrangement is all white in one jar and all black in the other, and $f(50) = 1/2$. The most aggressive colour separation imaginable is worth exactly as much as the even split, which is equation (4) turning up again from the other direction.

The general urn, and the ceiling at three quarters

Redo (9) with $W$ white and $B$ black marbles and $k = 1$:

With equal counts $W = B = m$ this is $(3m-2)/(2(2m-1))$, and the sequence climbs slowly:

Three quarters is the supremum and it is never attained, for a reason you can read off (12): the second jar is always short exactly one white marble, the one you spent. As $m$ grows that single missing marble matters less, and $(m-1)/(2m-1)$ creeps up on $1/2$ from below without arriving. At $m = 1$ the effect vanishes entirely, since spending your only white marble leaves the other jar all black.

One more variant is worth a line, because it changes the answer completely. If the person drawing gets to pick the jar rather than tossing a coin for it, the second jar becomes irrelevant and the whole problem collapses: put one white marble alone, they pick that jar, and the answer is $1$. The coin toss is the entire source of difficulty, and the arrangement in Fig. 1 is best understood as buying one of its two branches outright at the lowest possible price.

Sources and further reading

• The decomposition in equation (3) — Law of total probability
• The standard model for drawing marbles — Urn problem
• Why averaging two fractions is not the same as pooling them — Weighted arithmetic mean, and Simpson's paradox for the same effect used to mislead

Every number above was checked before publication in exact rational arithmetic, by an exhaustive sweep of all $2599$ legal arrangements, and by a seeded simulation of $400{,}000$ draws per arrangement, which measured $0.74794$ for the winner against the exact $0.747475$.