Extra Credit Random-Ade Recipe

Once again I’m preparing random-ade, but this time I have three 12-ounce glasses.

I fill the first glass with a random amount of lemon juice, the second glass with a random amount of lime juice, and the third glass with a random amount of water. As before, each amount is chosen uniformly between 0 and 12 ounces, and all amounts are independent. 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.

Then I refill that now-empty glass with yet another random amount of the same liquid it previously contained. 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?

OK, so in the extra credit version we can let $E, I, W \sim U(0,12)$ represent the random volume in ounces of l"E"mon juice, l"I"me juice and "W"ater, respectively, that are initially poured into the glasses. Let $V_1, V_2 \sim U(0,12)$ be the additional random volumes that are poured into the then empty glasses. Here again we have $E, I, W, V_1,$ and $V_2$ independent. We have $$\tilde{R}(E,I,W,V_1,V_2) = \begin{cases} \min \{E,I,W\} + V_1 + V_2, & \text{if $V_1 + V_2 \leq \text{med} \{E, I, W \} - \min \{E, I, W \}$} \\ \text{med} \{E, I, W\}, &\text{if $V_1 \leq \text{med} \{E, I, W \} - \min \{E, I, W\} \leq V_1 + V_2$}\\ \text{med} \{E, I, W \} + \min \{ V_1, V_2 \}, &\text{if $\text{med}\{E, I, W\} - \min \{E, I, W\} \leq V_1$} \\ & \,\,\,\,\,\, \text{ and $\min \{V_1, V_2 \} \leq \max \{E, I, W \} - \text{med} \{E, I, W\}$}\\ \max \{E, I, W \}, &\text{if $\text{med} \{E, I, W\} - \min \{ E, I, W\} \leq V_1$}\\ & \,\,\,\,\,\, \text{ and $\min \{V_1, V_2\} \geq \max \{E, I, W\} - \text{med} \{E, I, W\}.$} \end{cases}$$

As we can see there are many possible cases to go through in order to get the theoretical solution; however, it is relatively easy to set up a Monte Carlo simulation. Doing so with $N= 100,000$ simulations gives an average $\tilde{R}$ value of $\hat{R} = 7.400516$ with standard error of $0.0081.$ So we can confidently say that the expected volume of the Extra Credit Ranom-Ade is about 7.40 ounces.