I thought hashing power is an expression of the probability of being the one to solve the cryptographic puzzle and placing a new block on the network.

If that's true, owning 51% of the hashing power should just give a slightly higher chance of mining a new block than 49% hashing power. However when I read articles explaining Bitcoin and mining, owning 51% of the hashing power in the network is presented as a guarantee to be able to outpace the rest of the network, to the point where an alternate chain can be created and funds double spent.

Given what I assume (could be wrong) is a marginally higher probability of solving the puzzle first at 51%, how does that make sense? Couldn't someone who got a little lucky do that with 40% of the hashing power? How does 51% of the hashing power guarantee that one will be able to carry out an attack on the network?

2 Answers 2


With any probability distribution, a larger sample size will converge onto the expected probabilistic outcomes.

So if you had 51% of mining power, there is a chance the other 49% of miners will outcompete you on a short time scale, but over longer time scales we’d expect you to win.

Really, the threshold for this is just >50%, but a ‘51% attack’ has become the colloquial term for this.

Note that a 1-2% of an advantage may not sound like much, but that’s the house favour on many casino games, and casinos generally seem to have no problem staying in business (despite sometimes paying out big to lucky players). Over long enough time scales, we reliably see a regression to the mean.

  • But couldn't it take a large sample size to converge onto the true probability distribution with reasonable confidence? In other words, it seems to me the slightly >50% party (let's say 50.5%) could mine 100 blocks and easily not have outpaced the rest of the network if he gets even a little bit unlucky. Is the assumption that he would just keep going for as long as it takes, eventually being able to fork his entire chain and cancel the now irrelevant chain mined by the rest of the network?
    – bitsearch
    Apr 28, 2020 at 1:08
  • @bitsearch yes, and likewise, an attacker with <50% of the hashpower could pull off an attack, though they would fail on average. With >50% of the hashpower, they would succeed on average, with a continuously growing likelihood of averaging success of as time increases. I don't doubt that articles written for laymen may not communicate this effectively, but you can check out section 11 of the original whitepaper for some of Satoshi's calculations regarding this: bitcoin.org/bitcoin.pdf
    – chytrik
    Apr 28, 2020 at 1:19
  • Thank you. I was reading a lot of articles but I'm not sure the writers understand statistics and it seems they're all copying from the same article. I guess going directly to the technical source makes more sense.
    – bitsearch
    Apr 28, 2020 at 1:21

If you do not bound the time the attacker has, if he has 51% of the hash power (and is able to hold that majority during the entire attack time), he will eventually always overtake the rest of the network.

More formally: the probability of the attacker overtaking the honest network will converge to 100% as the attack time goes to infinity. There is no guarantee that this happens quickly at all, and that probability will never reach 100%, but it will come arbitrarily close.

Someone with 51% of the hash rate can simply produce blocks faster than the rest of the network combined, so this should not be a surprise. Sure, he may be unlucky for the first block, or the second, or even the first 100 blocks, but in the long term, the rest of the network simply cannot keep up with the attacker.

  • 1
    That makes sense, thank you. I understand the convergence properties of probability distributions as the number of iterations of a process goes to infinity. I was confused because the layman articles I was reading (my mistake) seemed to implicitly define a 51% attack as the instantaneous downfall of the entire network.
    – bitsearch
    Apr 28, 2020 at 1:25

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