Solution to Fiddler’s puzzle, Can You Drink the “Random-ade”? (May 08, 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 in this problem are chosen independently of each other.)

Nothing fancy here. We can simulate the two part “pouring” a bunch of times to compute the average.

simulate_randomade <- function(n = 1e5) {
  L <- runif(n, 0, 12) # lemon
  W <- runif(n, 0, 12) # water
  R <- runif(n, 0, 12) # do refill

  # part 1: pour min(L,W). Our pitcher gets 2x
  poured1 <- pmin(L, W)
  total <- 2 * poured1

  remainL <- L - poured1
  remainW <- W - poured1

  # part 2: empty glass gets refilled with R
  poured2 <- ifelse(L <= W,
    pmin(R, remainW), # new lemon vs remaining water
    pmin(remainL, R) # remaining lemon vs new water
  )
  total <- total + 2 * poured2 # we have to pumtiply this by two again
  total
}

set.seed(123)
results <- simulate_randomade(1e6)

Answer

The average is 14 ounces.

mean(results)
## [1] 13.99866

Bonus:

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? 

simulate_randomade_3 <- function(n = 1e5) {
  glasses <- matrix(runif(n * 3, 0, 12), nrow = n, ncol = 3)
  total <- numeric(n)

  for (round in 1:3) {
    poured <- apply(glasses, 1, min)
    total <- total + 3 * poured
    glasses <- glasses - poured # subtract from all

    # now, the empty glass is the one that was the minimum
    empty_idx <- apply(glasses, 1, which.min)

    # refil with a new random amount
    for (i in 1:n) {
      glasses[i, empty_idx[i]] <- runif(1, 0, 12)
    }
  }

  total
}

results <- simulate_randomade_3(1e6)

mean(results)
## [1] 22.20187