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.

[–] [email protected] 11 points 7 months ago* (last edited 7 months ago) (2 children)

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?

[–] kameecoding 5 points 7 months ago

Yeah lets try, hasnt been done before

https://youtu.be/5LM3uh6PyHU?si=sP9-t4OCmA52Nmwn

[–] ChickenLadyLovesLife 5 points 7 months ago

Should we try?

Absolutely. You can feed all the unsuccessful attempts to the Shakespeare-typing monkeys.

[–] Donjuanme 4 points 7 months ago

Except Kerbal popping is rate limited by energy input, there's not an instant of energy flow, there's 150 seconds of energy input, each second increasing the energy, popped kernals absorb less energy allowing the unpopped ones to absorb the incoming energy to each the same state.

If you wanted them to all pop at once you'd need to put that amount of energy in all at once. Not impossible, but not going to happen with your home microwave oven

[–] visnae 3 points 7 months ago

I can't imagine it would be equally distributed? Probably normal distribution applies over the span, most of the kernels would probably pop within say 20s of each other, and none in the beginning.