Let's do the math! If you assume there are 300 kernels, the popcorn will be finished within two minutes, and all kernels popping within 100 ms of each other is sufficient for a big bada boom...

There are 2×60×10 epochs where the bang could occur. Each of the 300 kernels needs to pop in the same epoch, so 1/(2×60×10) is the probability of the second kernel popping in the same epoch as the first kernel. The probability of all 299 popping in the same epoch as the first kernel is (1/(2×60×10))^299 = (2×60×10)^(-299).

Crunching the numbers in the Google search calculator... the probability is zero. That was anticlimactic.

[–] Malgas 21 points 7 months ago (1 children)

That math assumes a flat distribution of popping times, which I suspect is incorrect.

Listening to a bag of microwave popcorn, it starts off slow, gets more rapid, and then tapers off again, implying that kernels are more likely to pop near the average time, which makes it somewhat more likely for two kernels to pop simultaneously.

But yeah, whole bag at once is probably still basically zero. Unless you use one of these, of course.

[–] 0x0@lemmy.dbzer0.com 6 points 7 months ago (1 children)

Agreed! I admit I made a few sweeping simplifications to shoehorn this into a discrete math problem.

[–] Toribor@corndog.social 1 points 7 months ago

Assume a frictionless spherical microwave.

[–] Hjalamanger@feddit.nu 6 points 7 months ago* (last edited 7 months ago)

The exact probability is something more like 2*10^-921. Given that it would take around 9 gogol (9*10^926) years of constantly popping popcorns until that happens. Should we try?