3

If many parties are mining to try to establish a block, and one succeeds, does that mean the efforts of the other miners (whose blocks have at least some common transactions) are lost; the time and energy spent to create their blocks must be discarded?

[I have read some of the links, and some of the links of the links, and am still a bit confused.]

Here is an example using a simplified coin. Imagine there are only 1000 possible choices to complete the block. And only one miner, Alice. Since she has to guess which one of the 1000 choices will complete the block, she will make, on average, 500 tries to complete the block. It could be more or less, but 500 would be the average.

Now add another miner, Bill. Alice and Bill start mining at the same time and at the same rate. Since there are now two miners, and only one needs to succeed to complete the block, together, they need to each try only about 293 times before, on average, the block will be complete. That is quite a bit faster, because only 293 tries-times have elapsed, but the total number of tries to complete the block is larger, 586 instead of 500. The 86 tries seem wasted because a single miner would only need, on average, 500 tries.

UPDATE:

I now realize that each miner (or at least each mining group) is working on their own, unique block. The block is unique if only because of the miner designation (who gets the bitcoin reward) that is part of the block. So, in general, no two miners, outside of a mining group, are ever working on the same block.

2 Answers 2

5

Mining is a 'memory-less process', which means that each and every attempt at solving the block is unrelated to how many past attempts you've taken. It is like rolling dice: no matter what you rolled previously, your chance of rolling a 6 is still just 1 in 6.

The network difficulty automatically adjusts according to the recent rate of block production, and as the difficulty increases, so too does the average number of hash attempts needed in order to find a new block. So each previous attempt is not 'wasted', it is simply expected that the vast majority of attempts will not be valid. And when considering all of the miners en masse, the sheer number of guesses required to find each new block is what allows the bitcoin network to be considered secure.

9
  • 1
    @Jiminion All the players roll the die until one of them gets a 6. When Carlos gets a 6 first (on his 3rd try), does that mean Alice and Bob wasted their rolls?
    – user253751
    Apr 29, 2022 at 14:45
  • 2
    @user253751 Nice analogy, that's exactly it: no, the unsuccessful rolls/hashes aren't wasted because they could have been successful. Apr 29, 2022 at 15:08
  • 1
    @user253751 Yes, if Alice and Bob both rolled a 3 then one of them duplicated work unnecessarily.
    – Andy
    Apr 29, 2022 at 15:32
  • 3
    @Jiminion: Miners never repeat work done by other miners because they all work on unique block templates: bitcoin.stackexchange.com/a/36194/5406
    – Murch
    Apr 29, 2022 at 18:33
  • 1
    @Jiminion That scenario is actually lost effort indeed, and it is the primary reason why the inter block time is that long: giving (honest) miners time to learn about each others blocks (see bitcoin.stackexchange.com/a/113506/208). But that's a different case than what you are asking about, I think? Hash attempts that just never beat the target value are not wasted - they're just a byproduct of the process of creating hashes that do beat the target, at a fixed (expected) ratio. Apr 30, 2022 at 15:17
3

All miners create and discard billions of blocks every second. This work, including the discarded work, acts to benefit the entire Bitcoin community by securing the network. This is the work that is an essential basis of the "Proof of Work" on which Bitcoin is based.

In the long term, miners are rewarded fairly for their share of all work done.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.