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datadata | 2 years ago

Your argument (I think) is that bitcoin mining machines that don't solve a block directly are a complete waste, and only bitcoin mining machines that that solve a block do anything useful. Since 99.7% of mining machines don't solve a block, bitcoin is 333 times more wasteful than in needs to be. Is that right?

This misses the point that mining is probabilistic. Mining pools pay out based on the number of hashes tried, not the number of blocks solved. Your misunderstanding seems to be that the utility of something can be based on a probabilistic outcome. It is akin to saying that airbags are 99% (or whatever) wasteful because most of them aren't actually involved in a crash, and therefore 100x inefficient. You pay extra for a car with an airbag because it has value in the rare event you do have a crash it has immense value. Similarly, a mining pool would pay a single bitcoin mining rig to mine for years without it solving a single block, because the mining pool would realize the entire value if it did find a block, and so it pays out just under the expected value of this reward times the probability of the event happening.

Second, your argument ignores the fact that the unit of a "mining machine" is completely arbitrary. If mining machines were on average 2x larger and more powerful (but there were half as many of them), then by your argument bitcoin mining would be 2x less wasteful. That makes zero sense.

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arcticbull|2 years ago

> Since 99.7% of mining machines don't solve a block, bitcoin is 333 times more wasteful than in needs to be. Is that right?

Absolutely not, it's millions of times more wasteful than that. You don't need to allocate a single miner to a single block and then throw it out. It's just so many order of magnitude more wasteful than it needs to be that it's hard to reason about. You should be able to run the entire Bitcoin network on a single Raspberry Pi. It's literally 2tps, each a few bytes.

> Your misunderstanding seems to be that the utility of something can be based on a probabilistic outcome.

There's no misunderstanding. The kWh and kT of e-waste don't need to happen, it's just a poorly designed proof of concept that ran amok.

> If mining machines were on average 2x larger and more powerful (but there were half as many of them), then by your argument bitcoin mining would be 2x less wasteful.

No, I'm measuring waste by weight, in kilotons per year. In addition to energy consumption. So if a machine was twice as large it would contribute the same amount of waste in both columns.

datadata|2 years ago

OK, let's make this a multiple choice question. Assume bitcoin mining machines magically each become twice as powerful, AND twice as heavy, but there are half as many of them, so the total energy and resources used is exactly the same. According to your argument is this change:

a) good, because now twice as many miners actually solve a block, so instead of 99.7% waste it is 99.4% waste.

b) doesn't matter, because I just changed the unit of measurement-- each machine is just doing twice as much work.

c) doesn't matter, because bitcoin is already infinitely wasteful-- in which case pointing out that 99.7% of machines are somehow more wasteful than the other 0.3% does not make sense.

d) bad, for some other reason.