<p>As MeshCollider's comment mentions, it is simply the difference between the total number of bits against the target bits.</p>

<p>As to why it is this way is just a function of probability. If you have a range of 1-100, and your target is <code>&lt;= 20</code>, there are 20 numbers that satisfy your requirement. Thus, <code>20 * 1/100</code>, or 20 tries out of 100 should result in a valid  number against your target.</p>

<p>It works the same way, although on a much larger scale here. Your target and attempt are both 256 bits numbers, which means they range from 0 to <code>2^256-1</code>. If your target is <code>2^179</code>, it means that <code>256-179</code>, or <code>2^67</code> tries will, on average, produce a valid number against the target.</p>

CC BY-SA 4.0
2018-10-04T05:35:54.233Z
2018-10-04T05:35:54.233Z
Raghav Sood
<p>First, I'm sorry if I'm asking a stupid question. I'm just trying to understand bitcoin. I'm reading the <a href="https://github.com/ethereum/wiki/wiki/White-Paper#mining" rel="nofollow noreferrer">Ethereum white paper</a>, and I couldn't understand this sentence about proof of work:</p>

<blockquote>
  <p>At the current target of ~2^187, the network must make an average of
  ~2^69 tries before a valid block is found</p>
</blockquote>

<p>My question: how the 2^69 number of tries is computed?</p>

<hr>

<h2>Updated:</h2>

<p>My question: why 2^69 instead of 2^70 or 2^169?</p>

CC BY-SA 4.0
2018-10-04T01:36:37.583Z
Are you asking how to determine the number of tries required to solve a block given a difficulty target? Or are you asking what physical operations take place in order for a miner to "try" to solve a block?
Mason Ticehurst
CC BY-SA 4.0
2018-10-04T04:51:13.673Z
The 69 is just 256-187 because the hashes are 256 bits long
meshcollider
<proof-of-work>
Proof of work: average number of tries to create a block
sean
# Proof of work: average number of tries to create a block

(Asked by: username_0 on 2018-10-04)

First, I'm sorry if I'm asking a stupid question. I'm just trying to understand bitcoin. I'm reading the [Ethereum white paper](https://github.com/ethereum/wiki/wiki/White-Paper#mining), and I couldn't understand this sentence about proof of work:

> At the current target of ~2^187, the network must make an average of
> ~2^69 tries before a valid block is found

My question: how the 2^69 number of tries is computed?

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## Updated:

My question: why 2^69 instead of 2^70 or 2^169?

**Tags:** proof-of-work

> **Comments:**
> **username_1**: Are you asking how to determine the number of tries required to solve a block given a difficulty target? Or are you asking what physical operations take place in order for a miner to "try" to solve a block?  
> **username_2**: The 69 is just 256-187 because the hashes are 256 bits long  

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## 1 answers

### Answer by username_3 on 2018-10-04. Score: 3 | Accepted answer

As MeshCollider's comment mentions, it is simply the difference between the total number of bits against the target bits.

As to why it is this way is just a function of probability. If you have a range of 1-100, and your target is `<= 20`, there are 20 numbers that satisfy your requirement. Thus, `20 * 1/100`, or 20 tries out of 100 should result in a valid number against your target.

It works the same way, although on a much larger scale here. Your target and attempt are both 256 bits numbers, which means they range from 0 to `2^256-1`. If your target is `2^179`, it means that `256-179`, or `2^67` tries will, on average, produce a valid number against the target.


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