Has anyone calculated this:

Assuming that 10000 nodes randomly divided into 100 shards. What is the number of malicious nodes, such that the situation where at least one shard having 51% or more malicious nodes occurs with probability p?

For example, it would be important to know the number of malicious nodes such that there is a 50% chance of launching a 1% attack.

In analogy to the birthday problem, the number should be significantly less than 5000.

# On the probability of 1% attack

**daniel-tung**#1

Of course, it was one of the critical things to calculate/evaluate at the beginning of the sharding research.

Roughly speaking, if we assume: 1) a large validator pool with no more than 1/3 of malicious validators and 2) a secure RNG (this is extremely important), than a committee of a few hundred validators (e.g. 350) has an extremely small probability (probably around 2^-40) of having a malicious majority.

For more details, you can check this answer in Sharding FAQs.

**daniel-tung**#3

I think the reply and the link given are pertain to a different problem.

The answer given is on the probability of randomly choosing more than half (> N/2) malicious nodes, given p% of chance to be chosen.

The problem posted above is related to finding the probability of *at least one shard* having more than half malicious nodes.

To illustrate this, letâ€™s adapt the birthday problem to this situation:

Say there are 1095 nodes in total, being divided into 365 shards, 3 nodes in each shard. Then if the malicious node only controls 23 of the nodes (merely 2.1% of the nodes), there is already a 50% chance of at least one shard having 2 malicious nodes. This means in every two rounds of random sampling, there is, on average, one round where the malicious player controls one shard.

The probability is therefore much higher than what the link is given.

Thatâ€™s exactly what my reply pertains to.

I fail to understand how the birthday problem is relevant for analyzing this issue. We are discussing probabilities, and we already have a number of proven formulas/methods to calculate them (the most appropriate one in this case would be Binomial distribution).

This example is not appropriate because the validator set in Ethreum will be much larger. Simply put, thatâ€™s exactly why the probability of a malicios majority on a single shard is negligible.

**drcode1**#5

Ah, now I understand why such great emphasis has been placed recently into keeping staking requirements low.

Yep, and that (negligible probability for electing a malicious committee) is only one of the benefits of this model. The others are:

- Low barriers to entry (we donâ€™t want this to be rich peopleâ€™s game; now even a small group of enthusiasts in a developing country can join forces and run a validator node)
- Extremely high censorship resistance (instead of big pools and farms that adversaries such as governments can easily locate, attack, confiscate etc. you have thousands of anonymous nodes all around the world, each running nothing but a single machine)
- Rich are not getting richer (unlike conventional PoS models, if you want to scale your stake in order to increase your earnings/rewards, you also need to linearly scale a number of machines youâ€™re running; this maxes substantial stake concentration almost impossible).

IMHO, this (by â€śthisâ€ť I mean low-barrier, highly decentralized, equally staked validator set model) is the single most brilliant design choice in Ethereum 2.0.

**daniel-tung**#7

One-shard takeover is about controlling any one shard (i.e. at least one shard).

Your link gives the probabilities of *one given shard* being controlled.

The chance of at least one shard being controlled could be much larger than this, depending on the value of N.

How big is the validator set? If N is too large then it beats the purpose of sharding.

Also, the more reshuffling going on to avoid collusion, the higher expected number of times of one shard having majority of malicious nodes.

â€śThe chance of at least one shard being controlledâ€ť has nothing to do with N in this case, itâ€™s simply â€śthe probability of a single shard takeoverâ€ť * 1000 (number of shards), i.e. roughly 2^-40 * 1000 (this is the worst case, it might be much lower than that).

I can see youâ€™re really interested in this issue, so let me put it into perspective for you. If we assume:

- constant presence of 1/3 malicious validators (basically a permanent attack)
- 2^-40 * 1000 probability
- reshuffling every hour (this is not final yet)

then we can realistically expect that â€śat least one shardâ€ť will be taken over every 125,515 years. And I repeat, the system needs to be under permanent attack for all this time.

AFAIK, the long-term target is 10M ETH staked, i.e. 312,500 validators. Would you mind explaining why do you believe this defeats the purpose of sharding?

Of course, I took this into consideration in the calculation above.

**Enigmatic**#9

Hi Mihailo - Please donâ€™t mind me catching up with the implementation details with regard to the staking amount. Is the restriction on stake scaling per validating node enforced as â€śa maximum stake of 32ETHâ€ť or â€śthe higher the amount stake the lower the returnâ€ť, or etc?

Thanks in advance.

**MihailoBjelic**#10

Sure, no problem.

Itâ€™s the former, a single validator has a fixed stake of 32ETH. So, if you want to stake e.g. 100*32ETH, you need to run 100 machines/validators, which makes stake concentration in the validator pool extremely unlikely.