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Author Topic: Shouldn't 51% attack really be >50% attack?  (Read 1001 times)
statix
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July 02, 2012, 07:33:57 AM
 #1

Why is it commonly known as 51% attack when, as far as I know, you'll only need to have >50% of hashing power to successfully pull out the attack.

The term has bothered me quite a bit for awhile.
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Gareth Nelson
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July 02, 2012, 07:36:43 AM
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Why is it commonly known as 51% attack when, as far as I know, you'll only need to have >50% of hashing power to successfully pull out the attack.

The term has bothered me quite a bit for awhile.

Because the 50.000001% attack isn't as catchy
the joint
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July 02, 2012, 07:38:37 AM
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Why is it commonly known as 51% attack when, as far as I know, you'll only need to have >50% of hashing power to successfully pull out the attack.

The term has bothered me quite a bit for awhile.

Because the 50.000001% attack isn't as catchy

Yeah, it's kinda like how "we are the 47%" isn't as catchy as "we are the 99%"

But I admit, I have thought the same thing.

Meni Rosenfeld
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July 02, 2012, 07:49:00 AM
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I usually call it the >50% attack.

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ribuck
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July 02, 2012, 08:02:24 AM
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There's not really a hard limit for the percentage.

If an attacker has 47%, they might get lucky and maintain the longest chain for 6 blocks, which is enough for a devastating attack. On the other hand, they might have 53% for a while, and not manage a 6-block attack during that time.

Satoshi did a mathematical analysis of this in section 11 of his paper.

So I think "51% attack" is a good name for this attack, even though it's not a complete explanation.
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July 02, 2012, 08:43:29 AM
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There's not really a hard limit for the percentage.

If an attacker has 47%, they might get lucky and maintain the longest chain for 6 blocks, which is enough for a devastating attack. On the other hand, they might have 53% for a while, and not manage a 6-block attack during that time.

Satoshi did a mathematical analysis of this in section 11 of his paper.

So I think "51% attack" is a good name for this attack, even though it's not a complete explanation.
With 47% the attacker has a chance to succeed, but no certainty. With >50% he is guaranteed to win eventually, no matter how many blocks are waited (though "eventually" can be a very long time, on average inversely proportional to the excess).

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July 02, 2012, 08:58:32 AM
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The right term is majority attack.
Gareth Nelson
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July 02, 2012, 09:11:14 AM
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Why is it commonly known as 51% attack when, as far as I know, you'll only need to have >50% of hashing power to successfully pull out the attack.

The term has bothered me quite a bit for awhile.

Because the 50.000001% attack isn't as catchy

Yeah, it's kinda like how "we are the 47%" isn't as catchy as "we are the 99%"

But I admit, I have thought the same thing.

Nice quote!

53% of the US population pay taxes
DannyHamilton
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July 02, 2012, 03:44:43 PM
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Nice quote!

53% of the US population pay taxes

Actually, I'm pretty sure 100% of the U.S. population pays taxes, but we will avoid getting into a political discussion in this thread.

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February 06, 2018, 03:19:06 PM
 #10

Sorry for galvanizing the corpse, don't want to create a new topic.

Nope, the 51% is true only for two-miners network.

For more complex configuration, supposing the finding bock events are completely independent and the probability is only proportional to the hashrate.

Consider a [relatively simple] minig network where hashrates are divided as 40:3x20 %%.

Then, Probability of at least one of "weak" miners would mine a block is 1 minus product of probabilities of the inverse event:
https://www.khanacademy.org/math/ap-statistics/probability-ap/probability-multiplication-rule/a/general-multiplication-rule

Thus, Pr = 1 - 0.8*0.8*0.8 = 1 - 0.512 = 48.8%

Hence, [for given above configuration] one need only 40% of the network hashrate to gain 51.2% probability to mine a block, that means in long therms his fork will eventually be longer.

If, for example, we consider 7x10% and handful of 1%-hashrate-miners - then we need even less, slightly above 30% of total hashrate to overtake the 50% probability.

BTW, according the formula above, the swarm of micro-miners DOES NOT contribute to decentralization but vice versa, decrease the threshold for this kind of "attack".
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February 06, 2018, 06:20:00 PM
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Sorry for galvanizing the corpse, don't want to create a new topic.

Nope, the 51% is true only for two-miners network.

For more complex configuration, supposing the finding bock events are completely independent and the probability is only proportional to the hashrate.

Consider a [relatively simple] minig network where hashrates are divided as 40:3x20 %%.

Then, Probability of at least one of "weak" miners would mine a block is 1 minus product of probabilities of the inverse event:
https://www.khanacademy.org/math/ap-statistics/probability-ap/probability-multiplication-rule/a/general-multiplication-rule

Thus, Pr = 1 - 0.8*0.8*0.8 = 1 - 0.512 = 48.8%

Hence, [for given above configuration] one need only 40% of the network hashrate to gain 51.2% probability to mine a block, that means in long therms his fork will eventually be longer.

If, for example, we consider 7x10% and handful of 1%-hashrate-miners - then we need even less, slightly above 30% of total hashrate to overtake the 50% probability.

BTW, according the formula above, the swarm of micro-miners DOES NOT contribute to decentralization but vice versa, decrease the threshold for this kind of "attack".

Check your math.  You've clearly got it wrong.  You make multiple conflicting statements, and you've made some false assumptions,


DevilOper
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February 06, 2018, 10:30:48 PM
 #12

Check your math.  You've clearly got it wrong.  You make multiple conflicting statements, and you've made some false assumptions

Enlighten me.
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