Nice try, Samuel Hahnemann!!!
this post was submitted on 02 Aug 2025
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I just got a flashback to an open air concert I was at. It was raining like mad. At some point my beer tasted only like rain water. Oh and the second thing is, after I returned home, not a single thing I carried along was dry. Clothes wet, underwear wet, even everything in my wallet was wet. Still, the beer was worst.
But it was an amazing concert.
If you assume it gets thoroghly mixed between spoonfuls:
(All maths done while in the bathroom, should be checked before used for soup science) At spoonful 14 you'll have less than 50% soup in your rainwater.
At 45 you'll be down to a 10% soup contamination of your delicious bowl of rainwater.
At just over 100 spoonfuls you'll be down to less than 0.5%.
Just like when you're shaving the pubes of a bear you'll have to draw the line somewhere.
Just like when you’re shaving the pubes of a bear you’ll have to draw the line somewhere.
I feel seen.
You get the upvote for shaving the pubes of a bear
Why are you eating rainy soup? I'm not eating any soup that's getting rain in it.
The answer is about 1144.
Ok, let's do the probability math properly. Others have mentioned how it's a matter of probability how long until the last molecule of soup is taken out.
Suppositions:
There are N molecules in every ml of soup and every ml of water.
All soup molecules are the same.
Every spoonful takes out exactly 25N molecules out of the bowl selected at random, and they are immediately replaced by 25N molecules of water.
At the start, there are 500N molecules of soup in the bowl.
The question is:
How many spoonfuls is it expected to take until all soup molecules are removed?
For every spoonful, each molecule of soup in the bowl has a 25/500 chance of being removed from the bowl.
For ease of calculation, I will assume that each molecule being removed is independent of all others. This is technically wrong, because this implies that there is a very very tiny chance that all soup molecules are replaced in the very first spoonful. However, for the large number of molecules we are going to be working with, this shouldn't affect the final result in any meaningful way.
Number all soup molecules in the bowl: 1, 2, ..., 500N.
Define X_i to be the number of iterations it took until molecule i was removed. All X_i are I.I.D.:
P(X_i = 1) = 25/500 P(X_i = 2) = (475/500) * 25/500 P(X_i = 3) = (475/500)² * 25/500 ... P(X_i = n) = (475/500)^(n-1) * 25/500 ...
This is a geometric distribution with p = 25/500.
Now what we're interested in if the maximum value between all X_i
That is: max_i { X_i }
Specifically we want the "Expected Value" (basically the average) of it: E[ max_i { X_i } ]
This is exactly the question asked here: https://math.stackexchange.com/q/26167
According to the answer there, there is no closed-form exact answer but a very good approximation for the solution is:
1/2 + (1/λ) H_500N
Where λ = -log(1-p) and H_n is the nth harmonic number.
Now it's just a matter of plugging in the numbers.
According to Wolfram Alpha, there are N = 3.333*10^22 molecules in 1mL of water, or 1.666*10^25 in 500mL.
Again using Wolfram Alpha, the Nth harmonic number is H_500N = 58.652
With the formula given we get λ = -log(475/500) = 0.051293
Plugging it all in we get the expected number of spoonfuls:
0.5 + (1/0.051293)(52.438) = 1143.97 spoonfuls on average.
With enough coercion we can also force Wolfram Alpha to do the whole calculation in one go: 1/2 + 1/(-log(1-25/500)) * harmonic number (number of molecules in 500mL of water/molecule) giving 1143.9743.
Edit: initially used N instead of 500N and got the wrong answer of 1022.
Wow, someone actually bothered to do it properly! I just wrote some horrible R code and ended up with 1146 spoons to get to 50% probability of having either 1 or 0 soup molecules. So good to see that the answers were so close.
:)
What I would like to do is give a margin of error, e.g. "there is a 95% change that it will be between spoonful 1000 and spoonful 1300" or something like that. But I don't have the time to figure that out now, sounds like it would be harder to figure out than the expected value.
Ok, I couldn't resist. Here are the results of running that simulation 6000 times.
Min. 1st Qu. Median Mean 3rd Qu. Max. NA's
1085 1126 1140 1144 1157 1292 0
This means that about half the time it's somewhere between 1126 and 1157 spoons.
FYI: /u/protist@mander.xyz, /u/Sabin10@lemmy.world, /u/neo2478@sh.itjust.works
By the way, how did you actually stimulate it? Surely you didn't keep 10^25 variables in memory...
I thought of making a vector with a length of about 1.671398e+25, but then I remembered what one time when when I tried to make a linear model with hundreds of dimensions. So yeah... We have gigabytes of RAM, and it's still not enough. Not really a problem, as long as you don't try to do anything completely ridiculous.
Instead, I just made a variable that simply contains the number of soup molecules and another one for the number of water molecules. Far simpler that way.
Here's where the magic happens:
# Number of soup molecules drawn
soup_molecules_replaced <- rbinom(1, replacement_count, prob_soup)
The rbinom function is used to generate random numbers from a binomial distribution. It's a discrete probability distribution that models the number of successes, i.e. scooping out a soup molecule. Rest of the codes is just basic infrastructure like variables, loops, etc.
BTW the variable names look ugly, because I couldn't be bothered to tidy everything up. I really prefer camelCase, whereas Mistral seems to prefer underscores. That's what you get for vibing.
Side note: If you do this kind of stuff for private purposes, you have to rely on your own hardware. If you plan to publish your discoveries, universities and publicly funded supercomputers might be an option. If there exists a Journal Of Recreational Mathematics And Useless Simulations (JORMAUS), I could totally publish this stuff and maybe even run my code on a supercomputer.
You rock! Thank you :)
If I find myself in the right mood I might try to work out the actual distribution. If I do, your simulation will be a very handy sanity check!
Thanks!
If you want, I can increase the sample size. Just need to figure out how to add a timer to the code and set it to run for a few hours, maybe even overnight. A histogram with 10^6 bins should look pretty smooth.
Should also upgrade the visualization to ggplot2. Frivolous computations like this deserve no less.
I’m tempted to run a full simulation that really picks those molecules randomly.
This reminds me of this video I saw (YouTube I believe) about a soup restaurant in.... I believe it's Thailand or Indonesia, where they have a ginormous bowl of soup. Every day they serve customers and at the end about half is left and they just add more ingredients and water for the next day.
They've been doing that for decades now and apparently it's the best tasting soup anyone can try in their lives.
This used to be a pretty common practice. It's called a forever soup
Replace rain with “more stew ingredients” and you’re describing Perpetual Stew - a soup that is forever cooking, forever being eaten, forever being added to.
Critically, this does not work with iced coffee. It does work with whiskey though.
Why would you cook your whiskey?
I just meant blending the last of one batch/bottle with the beginning of the next like in a decanter.
Because that's the secret to a truly great gravy.
if the rain is broth too.
Sounds like homeopathic soup.
Since excessive consumption of soup probably causes obesity, joint pain, cardiovascular disease, type 2 diabetes, depression and a long list of other conditions, we can unsafely say that the homeopathic soup should cure all of those.
Let's find out. One... two-hoo... three. CRUNCH. Three.
That’s some chunky soup
Who(heh) doesn’t love a nice rich bowl of tootsie pop soup?
Given that the soup and rainwater would mix together, the question of how long it would take to get that last molecule of soup out is one of probability. I'm not qualified to give you a calculated answer, but I can tell you the most likely outcome is that it's going to take a lot of spoonfuls. The soup will begin tasting watered down very quickly and will basically be a bowl of >95% water for a long time before you get those last molecules
I get 1132 spoonfuls as a lower bound.
Big assumptions:
- Soup is considered identical to water for the sake of molecule count and density.
- Soup and water are evenly mixed before each spoonful.
- Each spoonful is guaranteed to contain the correct fraction of soup to water from the mix.
There are ~1.666×10^25 molecules of water in 500 ml (source: WolframAlpha). We seek what power of (500-25)/500 [= 19/20] is small enough to counter this number in order to get to the level of single molecules. This is about 1132.
But like you point out, it's going to be tasting watery a long, long time before that happens. It's 50% rainwater after about 14 spoonfuls (Sanity check: That would be 10 if the container was big enough and no spoonfuls were being removed.). ~90% at 45 spoons and ~99% at 90 spoons.
I got 1022 as the expected value, see my top level comment.
Edit: oops, made a big mistake. Will fix it!
Edit: after correcting I got 1144, much closer to your result 1132.
If you ignore the fact that soup consists of discrete molecules, the answer is infinite.
In real life though, you have to get probabilities involved. Haven’t done the math yet, but my intuition tells me that it’s going to take a lot of spoons. Quick LLM solution suggests it’s only 14 spoons, but I’m not convinced. Need to do it properly later today.
Edit: That first intuition was very wrong. Also, the LLM was wrong too. It was just counting milliliters, not molecules.
Well 14 spoonfuls wouldn't finish the soup even if it wasn't raining so that's definitely wrong.
Ok, Now I've got some sort of estimate. Still didn't do it "the proper way", because writing a simulation was more fun than reading a few Wikipedia articles about mathematics, which would have taken.... probably only a fraction of the time I spent on writing some horrible R code that produces suspicious results.
My simulation is based on keeping track of different kinds of molecules. First, I calculated how many water and soup molecules there are. I assumed that they both have the same molar mass. I also assumed that 500 ml = 500 g, which is close enough IRL. The number of each molecule type doesn't have to be a whole number, so fractions are allowed. When the soup molecule count drops to 0.5, it means that there's a 50% chance of 1 soup molecule being present. I'm not entirely satisfied with this implementation, but it felt reasonable at the time. Anyway, I set the threshold of my while loop to 0.5 soup molecules.
Anyway, here are the results!
It took only 1146 spoons to scoop out the final molecule with 50% certainty. If you used a smaller 5 ml spoon, it would take 5848 spoons, which is still way smaller than I expected. I really thought it would be something totally absurd like the the number of atoms in the observable universe. I feel kinda skeptical about my code until I see a proper mathematical proof about this.
Close enough, somebody mathed it out to 1144 in the comments
In this case I believe there is a difference between “contain no original molecule of soup” and “so fucking close to water you might as well be having sex in a canoe”
this is like a wet sourdough.
this was a thing in inns and such pretty much from Rome through the middle ages, possibly longer but archeology only knows so much, and they tended to consider it the same soup.
Would it still be soup once there's no soup? No, it'd just be water then.
No the water remembers the soup! You get all the goodness of the soup, but in water.
/s