<p>One difficulty-1 share takes on average 2^32 = 4.3 billion hashes to find.</p>

<p>If your hashrate is X GH/s, then in one day the number of difficulty-d shares you'll find is</p>

<p>(X * 1000000000 * 86400) / (2^32 * d)</p>

<p>Or simply</p>

<p>20,000 * X / d</p>

CC BY-SA 3.0
2013-03-25T09:32:21.880Z
2013-03-25T09:32:21.880Z
Meni Rosenfeld
<p>Many pools offer a difficulty of 1 per share.  </p>

<p>Since variable difficulty is a feature of a couple of pools, I'd like to understand how many shares would be created at varying levels of difficulty.</p>

<ul>
<li>What is the formula to determine the average number of shares when the hashrate and difficulty are known?</li>
</ul>

<p>This is a variation of the following question:</p>

<p><a href="https://bitcoin.stackexchange.com/a/8075/1878">Given the probability of finding a block, and time taken, can I infer hashrate?</a></p>

<mining-pools><hash><difficulty>
How many shares will be found for a given hashrate?
makerofthings7
# How many shares will be found for a given hashrate?

(Asked by: username_0 on 2013-03-24)

Many pools offer a difficulty of 1 per share.

Since variable difficulty is a feature of a couple of pools, I'd like to understand how many shares would be created at varying levels of difficulty.

* What is the formula to determine the average number of shares when the hashrate and difficulty are known?

This is a variation of the following question:

[Given the probability of finding a block, and time taken, can I infer hashrate?](https://bitcoin.stackexchange.com/a/8075/1878)

**Tags:** mining-pools, hash, difficulty


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## 1 answers

### Answer by username_1 on 2013-03-25. Score: 1 | Accepted answer

One difficulty-1 share takes on average 2^32 = 4.3 billion hashes to find.

If your hashrate is X GH/s, then in one day the number of difficulty-d shares you'll find is

(X \* 1000000000 \* 86400) / (2^32 \* d)

Or simply

20,000 \* X / d


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