Bitcoin Forum
April 25, 2024, 08:45:52 AM *
News: Latest Bitcoin Core release: 27.0 [Torrent]
 
   Home   Help Search Login Register More  
Pages: [1]
  Print  
Author Topic: Shouldn't 51% attack really be >50% attack?  (Read 1116 times)
statix (OP)
Newbie
*
Offline Offline

Activity: 12
Merit: 0


View Profile
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.
1714034752
Hero Member
*
Offline Offline

Posts: 1714034752

View Profile Personal Message (Offline)

Ignore
1714034752
Reply with quote  #2

1714034752
Report to moderator
1714034752
Hero Member
*
Offline Offline

Posts: 1714034752

View Profile Personal Message (Offline)

Ignore
1714034752
Reply with quote  #2

1714034752
Report to moderator
1714034752
Hero Member
*
Offline Offline

Posts: 1714034752

View Profile Personal Message (Offline)

Ignore
1714034752
Reply with quote  #2

1714034752
Report to moderator
Even in the event that an attacker gains more than 50% of the network's computational power, only transactions sent by the attacker could be reversed or double-spent. The network would not be destroyed.
Advertised sites are not endorsed by the Bitcoin Forum. They may be unsafe, untrustworthy, or illegal in your jurisdiction.
1714034752
Hero Member
*
Offline Offline

Posts: 1714034752

View Profile Personal Message (Offline)

Ignore
1714034752
Reply with quote  #2

1714034752
Report to moderator
1714034752
Hero Member
*
Offline Offline

Posts: 1714034752

View Profile Personal Message (Offline)

Ignore
1714034752
Reply with quote  #2

1714034752
Report to moderator
1714034752
Hero Member
*
Offline Offline

Posts: 1714034752

View Profile Personal Message (Offline)

Ignore
1714034752
Reply with quote  #2

1714034752
Report to moderator
Gareth Nelson
Hero Member
*****
Offline Offline

Activity: 721
Merit: 503


View Profile
July 02, 2012, 07:36:43 AM
 #2

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
Legendary
*
Offline Offline

Activity: 1834
Merit: 1020



View Profile
July 02, 2012, 07:38:37 AM
 #3

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
Donator
Legendary
*
Offline Offline

Activity: 2058
Merit: 1054



View Profile WWW
July 02, 2012, 07:49:00 AM
 #4

I usually call it the >50% attack.

1EofoZNBhWQ3kxfKnvWkhtMns4AivZArhr   |   Who am I?   |   bitcoin-otc WoT
Bitcoil - Exchange bitcoins for ILS (thread)   |   Israel Bitcoin community homepage (thread)
Analysis of Bitcoin Pooled Mining Reward Systems (thread, summary)  |   PureMining - Infinite-term, deterministic mining bond
ribuck
Donator
Hero Member
*
Offline Offline

Activity: 826
Merit: 1039


View Profile
July 02, 2012, 08:02:24 AM
 #5

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.
Meni Rosenfeld
Donator
Legendary
*
Offline Offline

Activity: 2058
Merit: 1054



View Profile WWW
July 02, 2012, 08:43:29 AM
 #6

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).

1EofoZNBhWQ3kxfKnvWkhtMns4AivZArhr   |   Who am I?   |   bitcoin-otc WoT
Bitcoil - Exchange bitcoins for ILS (thread)   |   Israel Bitcoin community homepage (thread)
Analysis of Bitcoin Pooled Mining Reward Systems (thread, summary)  |   PureMining - Infinite-term, deterministic mining bond
mrb
Legendary
*
Offline Offline

Activity: 1512
Merit: 1027


View Profile WWW
July 02, 2012, 08:58:32 AM
 #7

The right term is majority attack.
Gareth Nelson
Hero Member
*****
Offline Offline

Activity: 721
Merit: 503


View Profile
July 02, 2012, 09:11:14 AM
 #8

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
Legendary
*
Offline Offline

Activity: 3374
Merit: 4606



View Profile
July 02, 2012, 03:44:43 PM
 #9


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.
DevilOper
Member
**
Offline Offline

Activity: 280
Merit: 26


View Profile
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".
DannyHamilton
Legendary
*
Offline Offline

Activity: 3374
Merit: 4606



View Profile
February 06, 2018, 06:20:00 PM
 #11

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
Member
**
Offline Offline

Activity: 280
Merit: 26


View Profile
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.
Pages: [1]
  Print  
 
Jump to:  

Powered by MySQL Powered by PHP Powered by SMF 1.1.19 | SMF © 2006-2009, Simple Machines Valid XHTML 1.0! Valid CSS!