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An alien contra-organization wants to spur the development of Earth to help humans in the near-future in presence of a prime-directive like law. They cannot establish actual contact, so they do the next best thing: covertly deliver some important technology.

Unfortunately they do something not-so-clever here. Assume our secret organization replaces some percentage of batteries P with new, upgraded batteries that do the following:

  • About 90% of the mass of the new battery is antimatter (the technology is very miniaturized, so most of the 10% is the outer shell for point 3).
  • It converts this into energy using the matter in air at near-100% efficiency.
  • It looks exactly the same as the old battery from the outside, and performs the same, with the caveat that:
  • The new battery never seems to run dry...

The replaced batteries are completely random, and could be anywhere. In multi-cell batteries, each cell has its own probability P of being swapped. We also assume the replacement happens at-once. Exactly how that is done is left unanswered and isn't the point of the question.

A problem arises when somehow one of the new batteries is destroyed. The result is a large thermonuclear explosion (minus the fallout). A standard AA battery would produce about a 1.25 MT-equivalent, a 9V battery would produce a 1.87 MT blast. For the P=1 case: a Tesla pack weighs 540 kg. that's about 24 GT.

If we plot the human survival rate R against the probability P, we know two points: at P < 10^-10, R = 1, as it's likely that nothing happens. At P = 1, R ~ 0. It's likely the only survivors are in very isolated places. Say: in space, aboard a submarine, aboard a ship that isn't in the very busy shipping lanes ,rainforest natives, or the antarctic as any battery that explodes sets off a giant chain reaction.

Our organization wants to know, what would be an optimal P? Thus, they want to know: what would the graph look like in-between these two P values?

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closed as off-topic by Mołot, Renan, Ash, Mathaddict, John Dallman Apr 26 at 13:46

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question does not appear to be about worldbuilding, within the scope defined in the help center." – Renan, Ash, Mathaddict, John Dallman
If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ Considering chain reactions between exploding batteries, for these purposes R is never 0 as there will be large enough gaps between human habitations that the chain won't reach all areas. Batteries are not evenly distributed across the planet. Specifically these will be very rural regions that are more independent of "central civilisation" and hence more likely to be able to survive and repopulate. Equally your chain reaction is unlikely to be able to cross large oceans, so you could easily exterminate a large portion of Europe independently of East Asia or Australia. $\endgroup$ – Separatrix Apr 3 at 9:17
  • $\begingroup$ It is true a single chain reaction won't reach everywhere. But consider that it's likely multiple batteries are destroyed before people know about the dangers... I also say R ~0, not R = 0. Basically, a negligible fraction of people survive. $\endgroup$ – aphid Apr 3 at 9:32
  • $\begingroup$ Certainly you'd lose a few cities, but the blast radius isn't actually that large from batteries in household goods, only up to 3miles. The bigger issue would be car batteries, UPSs etc. What would be the blast radius from a Tesla for example. Even then, if London went boom, Scotland could be fine, there's a lot of empty space. $\endgroup$ – Separatrix Apr 3 at 9:39
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    $\begingroup$ Query: many large batteries like laptop batteries are actually collections of smaller lithium ion battery cells. ( 2.bp.blogspot.com/-AnPIzc2j-nc/UZo_wGD2NoI/AAAAAAAAAsM/… ) Do all cells convert or is each battery cell independent. It would increase the number of batteries dramatically but decrease individual blast size. $\endgroup$ – Murphy Apr 3 at 11:09
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    $\begingroup$ I have the impression that is a pure probability problem and that, with a slight rephrasing, might find a better and more formal answer on mathematic.SE $\endgroup$ – L.Dutch Apr 3 at 13:31