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Topic: solo mining in lottery terms - page 2. (Read 2978 times)

legendary
Activity: 1344
Merit: 1023
Mine at Jonny's Pool
June 17, 2015, 10:01:51 AM
#7
as you can see by my bitcoinwsdom  method   
Yeah, but you sneakily edited your post - it only said how much 1Th/s would earn in 1 day before Wink

this is the actual formula..
difficulty * 2^32 / hash rate / 60 / 60 / 24
Well, it's the 'close enough' formula Smiley
(2^256 / (((2^224 - 2^208) / Difficulty) * Hashrate)) / 60 / 60 / 24
Yes, it's the close enough formula that most everyone uses.  I'm certainly guilty of using the easy-mode version Smiley.

WBF1, I try not to express mining chances in lottery terms (although I certainly have in the past).  With a lottery, you have a very specific set of numbers in play, which define a closed system that allows you to provide accurate chances at a jackpot win.  For example, to guarantee a jackpot win in the powerball, you'd have to purchase 175,223,510 tickets.  To guarantee that you just win *something* you need purchase only 35 tickets.

You can't make that same analysis with BTC.  Yes, you expect that if you produce 49,692,386,354 shares you will solve a block.  However, that's not guaranteed.  It might take 1 share, or it might take 100,000,000,000 shares or theoretically that miner may never ever produce a valid block solve.

Because each share is wholly independent and has a chance at solving a block, we can't truly state that "miner X has Y chance" like we can in lottery terms.  Furthermore, most every lottery has some kind of prize tiering.  Using the powerball again, as long as I match the powerball number, I will win $4.  That's why it only takes 35 tickets to guarantee a win.  There's no such concept in BTC.  You either solve a block or you don't.
hero member
Activity: 686
Merit: 500
FUN > ROI
June 16, 2015, 11:28:00 PM
#6
as you can see by my bitcoinwsdom  method   
Yeah, but you sneakily edited your post - it only said how much 1Th/s would earn in 1 day before Wink

this is the actual formula..
difficulty * 2^32 / hash rate / 60 / 60 / 24
Well, it's the 'close enough' formula Smiley
(2^256 / (((2^224 - 2^208) / Difficulty) * Hashrate)) / 60 / 60 / 24
legendary
Activity: 4116
Merit: 7849
'The right to privacy matters'
June 16, 2015, 11:04:29 PM
#5
So in a lottery  you have a 1 in xxx,xxx,xxx chance of winning.  [...] how do you state it according to mining hardware speed? Is it as simple as [hashrate of miner]/[global hashrate] ? Or does difficulty or target come into play?
That's roughly it.  You can express it better as a function of the current difficulty (it's the same for everyone, so a straightforward calculation - just not as straightforward as the emphasised bit) rather than the current estimated network hash rate - but if the network hash rate figure is based on the current difficulty rather than block intervals anyway, then it doesn't matter.

as you can see by my bitcoinwsdom  method    



  I get close to the same as this =    1th/347ph   this is 347,000 to one for any given block.
legendary
Activity: 1736
Merit: 1006
June 16, 2015, 11:03:59 PM
#4
this is the actual formula..
difficulty * 2^32 / hash rate / 60 / 60 / 24
just put it in google for the answer.

so this would be for 800th/s
(((49692386354 * 2^32) / 800000000000000) / 60 / 60 / 24)



hero member
Activity: 686
Merit: 500
FUN > ROI
June 16, 2015, 11:00:16 PM
#3
So in a lottery  you have a 1 in xxx,xxx,xxx chance of winning.  [...] how do you state it according to mining hardware speed? Is it as simple as [hashrate of miner]/[global hashrate] ? Or does difficulty or target come into play?
That's roughly it.  You can express it better as a function of the current difficulty (it's the same for everyone, so a straightforward calculation - just not as straightforward as the emphasised bit) rather than the current estimated network hash rate - but if the network hash rate figure is based on the current difficulty rather than block intervals anyway, then it doesn't matter.
legendary
Activity: 4116
Merit: 7849
'The right to privacy matters'
June 16, 2015, 10:56:41 PM
#2
So in a lottery  you have a 1 in xxx,xxx,xxx chance of winning.

Is it meaningful to state something similar about solo mining? And if so, how do you state it according to mining hardware speed? Is it as simple as [hashrate of miner]/[global hashrate] ? Or does difficulty or target come into play?

Note that I'm aware of calculators that show expected length of time to produce a block based on hashrate.  That's not what I'm talking about here. What I mean is if we think of each block found as a winning lottery ticket, is there a meaningful way to say any given miner has a 1 in xxx chance of winning that lottery?

https://bitcoinwisdom.com/bitcoin/difficulty

1th would earn 0.0102 btc a day under current difficulty in 1 day.

so 25/0.0102 =  2450.98 to one in finding a block in a day  since 144 is 'norm' for blocks in a day mutiply  2450.98 by 144 and that is your chance for any given block under current diff


so 352,941.12 to 1 for any one block
sr. member
Activity: 419
Merit: 250
June 16, 2015, 10:20:09 PM
#1
So in a lottery  you have a 1 in xxx,xxx,xxx chance of winning.

Is it meaningful to state something similar about solo mining? And if so, how do you state it according to mining hardware speed? Is it as simple as [hashrate of miner]/[global hashrate] ? Or does difficulty or target come into play?

Note that I'm aware of calculators that show expected length of time to produce a block based on hashrate.  That's not what I'm talking about here. What I mean is if we think of each block found as a winning lottery ticket, is there a meaningful way to say any given miner has a 1 in xxx chance of winning that lottery?
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