If you were eating a soup from a bowl with 500ml of soup taking 25ml spoonfuls, and the rain replaced the volume that you ate at the same rate as you ate it, how many spoon fulls would it take for the soup to be completely replaced with water? Also, when that happens, would it still be the same soup?
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: 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_N
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.
Again using Wolfram Alpha, the Nth harmonic number is H_N = 52.438
With the formula given we get λ = -log(475/500) = 0.051293
Plugging it all in we get the expected number of spoonfuls:
1022.8
zxqwas@lemmy.world 3 weeks ago
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.
Boddhisatva@lemmy.world 3 weeks ago
I feel seen.
Onomatopoeia@lemmy.cafe 3 weeks ago
You get the upvote for shaving the pubes of a bear
Valmond@lemmy.world 3 weeks ago
At one point you only have 1 molecule of “soup” left, and a 50/50 chance of getting it so statistically you might never.
zxqwas@lemmy.world 2 weeks ago
My soup I’ve got for lunch today is guesstimated at 40-60% water pre-rain. So it could very well be indistinguishable from rain water long before a one molecule scenario.