# Calculating average number of hashes tried before hitting a valid block

I'm toying with the idea of starting to mine, and was wondering how to calculate how many hashes on average must be tried before one will "hit" a valid block? I'm talking about a single user (not a pool) using one rig.

• I rephrased the question to make it general instead of specific to the current point in time, to make it more useful for future visitors.
– D.H.
Aug 27 '12 at 18:43

Well, currently the difficulty is 2,440,643. There will be (difficulty * 2^32) (roughly) hashes per block.

So that means there are (2,440,643 * 2^32) hashes that must occur before one block is solved. That's is across the entire network.

A block is solved in ten minutes (on average) or 600 seconds.
So the network is hashing (2,440,643 * 2^32) / 600 ~= 17 Thash/s. (This calculation is performed and the result is displayed on BitcoinWatch.com)

Let's say your dual 5970 rig gets 1.4 Ghash/s. That is 17,770 Ghash/s / 1.4 Ghash/s = 12,693. That is the number of rigs, if all were exactly like yours, operating today.

So if 12,693 rigs can do the work in 10 minutes, one solo rig will tka 12,693 * 10 minutes, or 126,930 minutes. With 1,440 minutes in a day, that means it will be about 88 days before you will have a 50% chance of solving a block operating solo.

Now that was the long way.

The short way:

For Hash Rate, plunk in 1400 (1.4 Ghash/s) and click Calculate.

So then, how many hashes is that? There are 86,400 seconds in a day and you have to has for 88 days. 86,400 * 88 = 7,603,200 seconds. At 1.4 Ghash/s X 7,603,200 seconds means you hashed 10,644 Thashes.

That is: 10,644,000,000,000,000 hashes.

• Since it takes the network `2^32 * difficulty` hashes to find a block on average, it will also take you `2^32 * difficulty` hashes to find a block on average, roughly. So you can take the current difficulty, multiply by about 4.3 billion, and that's how many hashes you will need, on average, to find one block. (Assuming the difficulty doesn't change on you.) Aug 27 '12 at 10:24