Posted on May 8, 2026
I'm preparing a mixture of "random-ade" using a large, empty pitcher and two 12-ounce glasses.
First, I fill one glass with some amount of lemon juice chosen randomly and uniformly between 0 and 12 ounces. I fill the other glass with some amount of water, also chosen randomly and uniformly between 0 and 12 ounces. Next, I pour an equal amount from each glass into the pitcher until one of the glasses is empty.
At this point, I refill that empty glass with yet another random amount of the same liquid it previously contained. Once again, I pour an equal amount from each glass into the pitcher until one of the glasses is empty.
On average, how much random-ade can I expect to prepare? Note that all three random amounts are chosen independently.
Let the initial amounts be \(X\) and \(Y\), and let the refill amount be \(Z\). Each is independently uniform on \([0,12]\).
During the first pour, the process stops when the smaller of \(X\) and \(Y\) runs out. So the pitcher receives
\[ 2\min(X,Y) \]
ounces, and the amount left in the nonempty glass is
\[ |X-Y|. \]
After refilling the empty glass with \(Z\), the second pour contributes
\[ 2\min(|X-Y|,Z). \]
So the total amount prepared is
\[ T=2\min(X,Y)+2\min(|X-Y|,Z). \]
Now scale by 12. Write
\[ x=\frac{X}{12}, \qquad y=\frac{Y}{12}, \qquad z=\frac{Z}{12}, \]
so \(x,y,z\) are independent and uniform on \([0,1]\). Then
\[ T=24\left(\min(x,y)+\min(|x-y|,z)\right). \]
We therefore need two expected values.
First, for two independent uniform random variables on \([0,1]\),
\[ \mathbb{E}[\min(x,y)]=\frac{1}{3}. \]
Next, let \(D=|x-y|\). For a fixed value \(D=d\),
\[ \mathbb{E}[\min(d,z)] =\int_0^d z\,dz+\int_d^1 d\,dz =d-\frac{d^2}{2}. \]
So
\[ \mathbb{E}[\min(D,z)] =\mathbb{E}[D]-\frac{1}{2}\mathbb{E}[D^2]. \]
For \(D=|x-y|\) with \(x\) and \(y\) independent uniform on \([0,1]\), the standard values are
\[ \mathbb{E}[D]=\frac{1}{3}, \qquad \mathbb{E}[D^2]=\frac{1}{6}. \]
Therefore
\[ \mathbb{E}[\min(D,z)] =\frac{1}{3}-\frac{1}{2}\cdot\frac{1}{6} =\frac{1}{4}. \]
Putting both pieces together,
\[ \mathbb{E}[T] =24\left(\frac{1}{3}+\frac{1}{4}\right) =24\cdot\frac{7}{12} =14. \]
Answer: \( \boxed{14 \text{ ounces}} \).
The same two-glass setup has a clean extension. Suppose we keep going for \(n\) total rounds: after each round, we refill the empty glass with a fresh independent uniform amount and continue.
Again work on the scaled interval \([0,1]\). Let \(F_r(w)\) be the expected sum of the round minima over the next \(r\) rounds if one glass currently contains a fixed amount \(w\) and the other is freshly refilled with a uniform random amount \(U\).
For one round,
\[ F_1(w)=\mathbb{E}[\min(w,U)]=w-\frac{w^2}{2}. \]
For later use, the two-round case is
\[ F_2(w) =\mathbb{E}\bigl[\min(w,U)+\min(|w-U|,V)\bigr] =\frac{w}{2}+\frac{1}{3}, \]
where \(V\) is another independent uniform random variable on \([0,1]\).
Now assume \(r \ge 2\) and \(F_r(w)=\dfrac{w}{2}+c_r\) for some constant \(c_r\). After one more round, the contribution is \(\min(w,U)\), and the leftover amount is \(|w-U|\). So
\[ F_{r+1}(w)=\mathbb{E}[\min(w,U)]+\mathbb{E}[F_r(|w-U|)]. \]
Using the inductive form,
\[ F_{r+1}(w) =\left(w-\frac{w^2}{2}\right)+\frac{1}{2}\mathbb{E}[|w-U|]+c_r. \]
But
\[ \mathbb{E}[|w-U|] =\int_0^w (w-u)\,du+\int_w^1 (u-w)\,du =w^2-w+\frac{1}{2}. \]
Therefore
\[ F_{r+1}(w) =\left(w-\frac{w^2}{2}\right)+\frac{1}{2}\left(w^2-w+\frac{1}{2}\right)+c_r =\frac{w}{2}+c_r+\frac{1}{4}. \]
So once we reach \(F_2\), every additional round just adds another \(1/4\) to the expected scaled total. Since \(F_2(w)=\dfrac{w}{2}+\dfrac{1}{3}\), induction gives
\[ F_r(w)=\frac{w}{2}+\frac{1}{3}+\frac{r-2}{4} \qquad (r \ge 2). \]
Now return to the original process. Let \(S_n\) be the expected scaled sum of the \(n\) round minima when the first two glass amounts are independent uniforms \(X\) and \(Y\). The first round contributes \(\mathbb{E}[\min(X,Y)]=\dfrac{1}{3}\), and the leftover amount is \(D=|X-Y|\), with \(\mathbb{E}[D]=\dfrac{1}{3}\). For \(n \ge 2\),
\[ S_n =\frac{1}{3}+\mathbb{E}[F_{n-1}(D)]. \]
For \(n=2\), this reproduces \(S_2=\dfrac{7}{12}\). For \(n \ge 3\), plugging in the formula for \(F_{n-1}\) gives
\[ S_n =\frac{1}{3}+\frac{1}{2}\mathbb{E}[D]+\frac{1}{3}+\frac{n-3}{4} =\frac{1}{3}+\frac{1}{6}+\frac{1}{3}+\frac{n-3}{4} =\frac{3n+1}{12}. \]
This same formula also matches \(n=1\) and \(n=2\). Undoing the scaling, the expected amount of random-ade after \(n\) rounds is
\[ 24S_n =24 \cdot \frac{3n+1}{12} =6n+2. \]
General formula for \(n\) rounds: \( \boxed{6n+2 \text{ ounces}} \).
This time there are three 12-ounce glasses.
I fill them with random amounts of lemon juice, lime juice, and water, each chosen independently and uniformly between 0 and 12 ounces. I pour equally from all three glasses until one is empty, refill that empty glass with a new random amount of the same liquid, and repeat. Then I refill the newly emptied glass again and do the same thing a third time.
On average, how much random-ade do I prepare now?
Again scale everything to \([0,1]\). Let the three initial fills be \(a\), \(b\), and \(c\), and sort them as
\[ x_1 \le x_2 \le x_3. \]
The first pour contributes
\[ 3x_1. \]
After that pour, the two nonempty glasses contain \(x_2-x_1\) and \(x_3-x_1\). So if we define
\[ u=x_2-x_1, \qquad v=x_3-x_1, \]
then the remaining problem is this: the next two rounds start from amounts \(u\), \(v\), and a fresh uniform random refill \(z\), where \(0 \le u \le v \le 1\).
Let \(g(u,v)\) be the expected sum of the round minima from those last two rounds. Since the next minimum is \(\min(u,z)\), there are three cases:
Averaging over \(z \sim \text{Uniform}(0,1)\) gives
\[ g(u,v) =\int_0^u \left(u-\frac{(u-z)^2}{2}\right)\,dz +\int_u^v \left(z-\frac{(z-u)^2}{2}\right)\,dz +\int_v^1 \left(v-\frac{(v-u)^2}{2}\right)\,dz. \]
Simplifying,
\[ g(u,v)=v-v^2+uv-\frac{uv^2}{2}+\frac{v^3}{3}. \]
Now we average over the distribution of \(u\) and \(v\). For three independent uniform variables on \([0,1]\), the joint density of
\[ u=x_2-x_1, \qquad v=x_3-x_1 \]
is
\[ f(u,v)=6(1-v), \qquad 0 \le u \le v \le 1. \]
So the expected contribution from the last two rounds is
\[ \mathbb{E}[g(u,v)] =\int_0^1 \int_0^v g(u,v)\,6(1-v)\,du\,dv =\frac{11}{30}. \]
Also, for three independent uniform random variables, the expected minimum is
\[ \mathbb{E}[x_1]=\frac{1}{4}. \]
Therefore the expected total sum of the three round minima is
\[ \frac{1}{4}+\frac{11}{30}=\frac{37}{60}. \]
Each round minimum is multiplied by 3 glasses and then by 12 ounces, so the expected amount of random-ade is
\[ 36 \cdot \frac{37}{60} =\frac{111}{5} =22.2. \]
Extra credit answer: \( \boxed{\frac{111}{5} \text{ ounces}} = \boxed{22.2 \text{ ounces}} \).
There is also a clean \(m\)-glass version of the extra-credit setup. Start with \(m\) glasses, each initially filled to an independent uniform level between 0 and 12 ounces. Pour equally from all \(m\) glasses until one empties, refill that glass with a fresh independent uniform amount of the same liquid, and continue for a total of \(m\) rounds.
After scaling by 12 again, each round contributes \(m\) times its minimum glass level. So if \(\tau_m\) is the total scaled time through the first \(m\) emptying events, the expected amount prepared is
\[ 12m\,\mathbb{E}[\tau_m]. \]
To compute \(\mathbb{E}[\tau_m]\), view each glass as its own renewal process. For glass \(i\), let
\[ U_{i,1}, U_{i,2}, U_{i,3}, \dots \]
be its successive independent uniform fills on \([0,1]\). Then its emptying times are
\[ S_{i,1}=U_{i,1}, \qquad S_{i,2}=U_{i,1}+U_{i,2}, \qquad S_{i,3}=U_{i,1}+U_{i,2}+U_{i,3}, \dots \]
The process ends its \(m\)-th round at the \(m\)-th emptying event overall, so \(\tau_m\) is the \(m\)-th smallest value among all these \(S_{i,r}\).
Now fix \(0 \le t \le 1\). Let \(N_i(t)\) be the number of times glass \(i\) has emptied by time \(t\). Since \(t \le 1\), the event \(N_i(t) \ge r\) is exactly
\[ U_{i,1}+\cdots+U_{i,r} \le t, \]
which has probability \(t^r/r!\). Therefore
\[ \Pr(N_i(t)=r)=\frac{t^r}{r!}-\frac{t^{r+1}}{(r+1)!}, \qquad r=0,1,2,\dots \]
and the probability generating function of \(N_i(t)\) is
\[ G_t(s) =\sum_{r \ge 0} \Pr(N_i(t)=r)s^r =e^{st}-\frac{e^{st}-1}{s}. \]
Because the glasses are independent, the total number of emptyings by time \(t\),
\[ N(t)=N_1(t)+\cdots+N_m(t), \]
has generating function \(G_t(s)^m\). The event \(\tau_m > t\) means that fewer than \(m\) emptyings have occurred by time \(t\), so
\[ \Pr(\tau_m > t) =\Pr(N(t) < m) =\sum_{r=0}^{m-1} [s^r]\,G_t(s)^m. \]
Integrating this survival probability from 0 to 1 gives
\[ \mathbb{E}[\tau_m] =\int_0^1 \Pr(\tau_m > t)\,dt =H_{2m}-H_m =\sum_{k=1}^m \frac{1}{m+k}, \]
where \(H_n=1+\frac12+\cdots+\frac1n\) is the \(n\)-th harmonic number. Equivalently, the expected minimum on round \(k\) is
\[ \frac{1}{m+k}, \qquad k=1,2,\dots,m. \]
Undoing the scaling, the expected amount of random-ade in the \(m\)-glass, \(m\)-round problem is
\[ 12m\,(H_{2m}-H_m). \]
\(m\)-glass formula: \( \boxed{12m\,(H_{2m}-H_m)\text{ ounces}} \).
This matches the earlier cases:
\[ m=2:\ 24\!\left(\frac13+\frac14\right)=14, \qquad m=3:\ 36\!\left(\frac14+\frac15+\frac16\right)=\frac{111}{5}. \]