<p>I would first like to comment that there is widespread confusion about how double-spending works. In particular, the question you've asked here is irrelevant. I urge anyone who is seriously examining this issue to carefully read <a href="https://bitcoil.co.il/Doublespend.pdf" rel="nofollow">Analysis of Hashrate-based Double Spending</a>, which also examines the effect of changing the average time between blocks.</p>

<p>For your question as is - do you mean</p>

<ol>
<li>The probability that in a given hour, the miner will find at least 6 blocks? Or</li>
<li>The probability that the event "the miner finds 6 blocks in a row, without the other miners finding blocks between him" will happen in a given hour?</li>
<li>Or something else?</li>
</ol>

<p>The first option is relatively easy - If the average time between blocks is T, the miner's proportion of the hashrate is q, and you're looking at a time period S, he will find on average (Sp/T) blocks, following the Poisson Distribution. So the probability this will be at least 6 is</p>

<p>1 - \sum_{x=0}^{5} (exp(-Sq/T) * (Sq/T)^x / x!)</p>

<p>For example, if q = 0.5 then that's 8.4% for 1 hour, and for 1 day it's 99.9999999999999999999999%.</p>

<p>I'll need to work out the 2nd option, but the answer to the easier question "What is the chance to find 6 blocks in a row starting right now" is of course q<sup>6</sup>.</p>

CC BY-SA 3.0
2013-09-04T11:16:09.107Z
I will read your article with the thought "What can an attacker do with a q=0.5 and a probability of 8.4%"... I suspect it will cause some chain reorgs and possibly the chain to be built on top of these 6 blocks, possibly not
makerofthings7
CC BY-SA 3.0
2013-09-04T05:55:48.267Z
2013-09-04T08:00:05.643Z
2013-09-04T08:00:05.643Z
Meni Rosenfeld
<p>I'd like to determine the probability of a fairly powerful miner (or pool) generating 6 sequential blocks in an hour. (so they would rebroadcast it later) </p>

<p>For example, given </p>

<ul>
<li>Hash/second: X</li>
<li>A target of: Y </li>
<li>A probability of getting 6 blocks in less than an hour: Z</li>
</ul>

<p>What is the probability that sequential 6 blocks will be created in</p>

<ul>
<li>1 hour</li>
<li>1 Day</li>
<li>1 Week</li>
<li>3 months</li>
</ul>

<p>I'd like to find the formula to determine this so I can graph the exponential difficulty of how much power an attacker will need to thwart the network.</p>

<p>There is discussion that Litecoin is more resistant to this type of attack, but so far it seems unfounded.  I'm looking for thought that would indicate otherwise.</p>

<blockchain><security><hashpower>
What is the probability of generating 6 sequential blocks, less than an hour?
makerofthings7
# What is the probability of generating 6 sequential blocks, less than an hour?

(Asked by: username_0 on 2013-09-03)

I'd like to determine the probability of a fairly powerful miner (or pool) generating 6 sequential blocks in an hour. (so they would rebroadcast it later)

For example, given

* Hash/second: X
* A target of: Y
* A probability of getting 6 blocks in less than an hour: Z

What is the probability that sequential 6 blocks will be created in

* 1 hour
* 1 Day
* 1 Week
* 3 months

I'd like to find the formula to determine this so I can graph the exponential difficulty of how much power an attacker will need to thwart the network.

There is discussion that Litecoin is more resistant to this type of attack, but so far it seems unfounded. I'm looking for thought that would indicate otherwise.

**Tags:** blockchain, security, hashpower


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## 1 answers

### Answer by username_1 on 2013-09-04. Score: 1

I would first like to comment that there is widespread confusion about how double-spending works. In particular, the question you've asked here is irrelevant. I urge anyone who is seriously examining this issue to carefully read [Analysis of Hashrate-based Double Spending](https://bitcoil.co.il/Doublespend.pdf), which also examines the effect of changing the average time between blocks.

For your question as is - do you mean

1. The probability that in a given hour, the miner will find at least 6 blocks? Or
2. The probability that the event "the miner finds 6 blocks in a row, without the other miners finding blocks between him" will happen in a given hour?
3. Or something else?

The first option is relatively easy - If the average time between blocks is T, the miner's proportion of the hashrate is q, and you're looking at a time period S, he will find on average (Sp/T) blocks, following the Poisson Distribution. So the probability this will be at least 6 is

1 - \sum\_{x=0}^{5} (exp(-Sq/T) \* (Sq/T)^x / x!)

For example, if q = 0.5 then that's 8.4% for 1 hour, and for 1 day it's 99.9999999999999999999999%.

I'll need to work out the 2nd option, but the answer to the easier question "What is the chance to find 6 blocks in a row starting right now" is of course q6.

> **Comments:**
> **username_0**: I will read your article with the thought "What can an attacker do with a q=0.5 and a probability of 8.4%"... I suspect it will cause some chain reorgs and possibly the chain to be built on top of these 6 blocks, possibly not  

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