Bitcoin Forum

I am trying to follow through the steps in the link below to find the average generation time.
https://en.bitcoin.it/wiki/Generation_Calculator
 

• Get the current target.
• Divide by 115792089237316195423570985008687907853269984665640564039457584007913129639935, which is the maximum value of a 256-bit number. You now have the probability of a single hash solving a block.
• Take the reciprocal of the probability to get the average number of hashes to solve a block.
• Divide the average number of hashes by your hash/s (not khash/s) to get the average number of seconds required to solve a block.

So the current target is given here, http://blockexplorer.com/q/hextarget, in hexadecimal format. Converted to base ten it is

38110290672195532365762668664552282566878756832852091863040


If you use this number in the calculation above it comes out to be right, if you use the difficulty you get a number that is hugely off.

Why is this not the same value are the one reported in mining software, such as CGMiner, it reports 707M. All websites also list it as 707M and not the number above.

Is there a way to go from 707408283 (Bitcoin difficulty) to 38110290672195532365762668664552282566878756832852091863040 (Bitcoin target)? Are those the proper names for the two?

difficulty * 2^32 = average number of hashes required to solve a block.

Both values express the same concept in different form.

difficulty = 2^256 / (target * 2^32)
 

And how do you combine this with your speed, block reward to get the avg reward you'd earn per day?

Because you need to divide the maximum target by the current target.

So 00000000FFFF0000000000000000000000000000000000000000000000000000 / 0000000000000006124200000000000000000000000000000000000000000000 = 707 408 283.

https://en.bitcoin.it/wiki/Target
https://en.bitcoin.it/wiki/Difficulty

(blockReward/difficulty)*((mhs*1000000*86400)/(POWER(2,48)/65535)) = btc / day based on pps

Oh wow, thanks! If you can post the one for proportional too, you'd be golden!

Why not 2^31? Base Bitcoin difficulty is 2^32, so 2^(32-1) should be required in average.

If the odds of rolling a six are 1 in 6 then how many rolls on average will it take to roll a six.
a) Six
b) Thee
c) One because I am lucky. Come on lucky six.
d) I don't see the connection

b)

That is incorrect, thank you for trying but it does indicate the reason for your incorrect claim above.

Hint: the reason why I picked dice in the example is you can test it yourself.  Just be sure to roll a lot of dice.

Ah, right. I thought that if I rolled "3" then I got a die with "1", "2", "4", "5" and "6" (without "3"). With mining this doesn't work.

Actually, the odds of rolling a 6 with 6 rolls is 66.5%