Can You Drink the “Random-ade”?

Question: 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.)

Extra Credit: 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?

(Fiddler on the Proof)

Solution

say we draw $N$ random initial volumes. the probability that the smallest volume is greater than $v$ is equal to the probability that none of the $N$ volumes occupy the space between $0$ and $v,$ which is $(1-v)^N.$ using the fact that $\langle v_1\rangle = \int_0^1 \text{d}v\, P(\text{space}>v)$ we get that the average minimum of the random volumes, and therefore the volume that each cup contributes on the first pour is $1/(N+1)$ for a total pour of $N/(N+1)$ from all $N$ cups.

<–! ${a_1, b_1, c_1, \ldots n_1}$ with $a_1 < b_1 < c_1< \ldots < n_1. $ –>

after the first pour, the volumes will be $b_1-a_1, c_1-a_1,$ and so on, in the same order.

now, the probability that $b_1-a_1$ is bigger than $v$ is equal to the probability that none of the $N$ volumes entered the space of length $v$ between $a_1$ and $b_1$ times the probability that the new random volume is not less than $b_1-a_1,$ or $P(\text{space} > v) = (1-v)^N\times(1-v) = (1-v)^{N+1}.$ again using the identity above, we get each cup contributing an average volume $\langle v_2\rangle = 1/(N+2).$

it keeps going on like this until we hit the cup $N+1.$ at this point, it is not possible for the first volume $a_1$ to have had a chance to land within the interval. so, from round $N$ onwards, the minimum volume is the product of $2N$ random volumes avoiding landing in the interval.

for the $N$ round prep of random-ade, the expected volume is therefore