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Author Topic: solo mining in lottery terms  (Read 2978 times)
WBF1 (OP)
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June 17, 2015, 02:20:09 AM
 #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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June 17, 2015, 02:56:41 AM
 #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

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TheRealSteve
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June 17, 2015, 03:00:16 AM
 #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.

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June 17, 2015, 03:03:59 AM
 #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)



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June 17, 2015, 03:04:29 AM
 #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.

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TheRealSteve
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June 17, 2015, 03:28:00 AM
 #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

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June 17, 2015, 02:01:51 PM
 #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.

Jonny's Pool - Mine with us and help us grow!  Support a pool that supports Bitcoin, not a hardware manufacturer's pockets!  No SPV cheats.  No empty blocks.
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June 17, 2015, 02:31:43 PM
 #8

It's a reasonable comparison to a lottery with no rollover and no guaranteed winner, I think.

As far as tickets goes - pool mining is somewhat analogous to that?  But yes, that's taking concepts far away from OP's Q Smiley

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June 17, 2015, 03:04:37 PM
 #9

It's a reasonable comparison to a lottery with no rollover and no guaranteed winner, I think.

As far as tickets goes - pool mining is somewhat analogous to that?  But yes, that's taking concepts far away from OP's Q Smiley
I guess if I had to compare it to a lottery, I would state that it's like going to the store and continuously buying quick pick tickets, and you could determine whether or not that ticket was a winner immediately.  Even if you bought out the number of tickets to guarantee a jackpot win, you wouldn't necessarily have one simply because the quick pick might give you the same set of losing numbers multiple times and never give you the winning combination.  Or, you could get lucky and the first ticket you bought won you the jackpot.

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WBF1 (OP)
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June 17, 2015, 04:19:55 PM
 #10

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

Steve and Philip methods both give a similar number.

Steve what do all the fixed numbers correspond to in your calculation?
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June 17, 2015, 04:34:43 PM
 #11

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

Steve and Philip methods both give a similar number.

Steve what do all the fixed numbers correspond to in your calculation?
They give similar numbers because they're both based on the same calculation, as described by the formula Steve posted.  Based on expectations (the formula) given a certain hash rate at a certain difficulty, you expect to find a block in a certain time.  Let's take a real example and see how it plays out.  A single Antminer S5 hashes at an advertised 1155GH/s.  Using the formula, I can calculate how long that should take to find a block:
Code:
(2^256 / (((2^224 - 2^208) / 49692386354) * 1155000000000)) / 60 / 60 / 24 = 2138.75
According to expectations, that S5 should take 2138.75 days to find a block, giving me a 1 in 2138.75 chance of finding a block today.

Phil's calculation method uses the above formula as well, but he buries the numbers... he calculates the expected earnings per day, then just takes 25 and divides it by that number.  Using the output from above, I can see that it's 2138.75 days to find a block.  The current block reward is 25BTC.  So, I just take 25 and divide by the number of days to find a block and get my expected earnings per day:
Code:
25 / 2138.75 = 0.01168905

Since that S5 expects to earn 0.01168905BTC per day, which is what the calculator at bitcoinwisdom.com shows as well (they round it off to 0.1169), I just take 25 and divide by it:
Code:
25 / 0.01168905 = 2138.75

Gets me right back to the same 1 in 2138.75 chance.

Jonny's Pool - Mine with us and help us grow!  Support a pool that supports Bitcoin, not a hardware manufacturer's pockets!  No SPV cheats.  No empty blocks.
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June 17, 2015, 04:40:00 PM
 #12

Steve what do all the fixed numbers correspond to in your calculation?
I think there was a better link, but: https://en.bitcoin.it/wiki/Difficulty#How_is_difficulty_calculated.3F_What_is_the_difference_between_bdiff_and_pdiff.3F
https://en.bitcoin.it/wiki/Target#What_is_the_maximum_target.3F
That 0x00000000FFFF0000000000000000000000000000000000000000000000000000 is equal to 2^224 - 2^208

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June 17, 2015, 08:04:18 PM
 #13

What happens after all 21 million bitcoins have been mined? Does this mean no more possible reward for solo mining or will it just be the transaction fees for finding a block?
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June 17, 2015, 08:11:35 PM
 #14

What happens after all 21 million bitcoins have been mined? Does this mean no more possible reward for solo mining or will it just be the transaction fees for finding a block?
https://www.google.com/webhp#q=bitcoin+%22what+happens+when+all%22
tl;dr: yes, transaction fees - no, all Bitcoins mined is not any time soon - yes, the halving curve means that the block reward will be fairly low long before then.

I'm sure there's a BitcoinTalk thread somewhere in those Google results, best pick it up over there Smiley

WBF1 (OP)
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June 17, 2015, 08:29:51 PM
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What happens after all 21 million bitcoins have been mined? Does this mean no more possible reward for solo mining or will it just be the transaction fees for finding a block?
https://www.google.com/webhp#q=bitcoin+%22what+happens+when+all%22
tl;dr: yes, transaction fees - no, all Bitcoins mined is not any time soon - yes, the halving curve means that the block reward will be fairly low long before then.

I'm sure there's a BitcoinTalk thread somewhere in those Google results, best pick it up over there Smiley

Yeah but what I am after is a 1 in X chance of hitting a block, not 1 in X chance of receiving Y amt of bitcoin.
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June 17, 2015, 09:38:16 PM
 #16

I explained, with example, just how to get that 1 in X chance.  The problem you face is that it's only an expected chance, because each submitted hash has just as much of a chance at solving the block as the one before it.  There is no concrete way to say that you have exactly 1 in X, for the very reasons I stated above.  You get an expected chance given hash rate and difficulty, not an absolute.

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June 17, 2015, 09:45:37 PM
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What happens after all 21 million bitcoins have been mined? Does this mean no more possible reward for solo mining or will it just be the transaction fees for finding a block?
https://www.google.com/webhp#q=bitcoin+%22what+happens+when+all%22
tl;dr: yes, transaction fees - no, all Bitcoins mined is not any time soon - yes, the halving curve means that the block reward will be fairly low long before then.

I'm sure there's a BitcoinTalk thread somewhere in those Google results, best pick it up over there Smiley

Yeah but what I am after is a 1 in X chance of hitting a block, not 1 in X chance of receiving Y amt of bitcoin.
its impossible. its like saying you want a 1 in X chance of flipping a quarter 5 times in a row and each time come up heads.

while it could happen, you cant ever know how many times you have to flip the 5 quarters to make them all come up heads.

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June 17, 2015, 09:55:41 PM
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Not sure exactly what the odds are using lottery terms but it will definitely depend on your hashrate.

For example, an S5 (1.15TH) according to coinwarz will take 2148.02 days ( 5.8 years ) to find a block at current difficulty:
http://www.coinwarz.com/calculators/bitcoin-mining-calculator/?h=1150&p=600&pc=0.10&pf=0.00&d=49692386354.89380000&r=25.00000000&er=250.82000000&hc=0.00

Whereas 1PH ( 1000TH ) should only take 2.47 days to find a block: http://www.coinwarz.com/calculators/bitcoin-mining-calculator/?h=1000000.00&p=600.00&pc=0.10&pf=0.00&d=49692386354.89380000&r=25.00000000&er=250.82000000&hc=0.00

Ignore the watts and power cost its irrelevant Wink.
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June 17, 2015, 10:28:32 PM
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What happens after all 21 million bitcoins have been mined? Does this mean no more possible reward for solo mining or will it just be the transaction fees for finding a block?
https://www.google.com/webhp#q=bitcoin+%22what+happens+when+all%22
tl;dr: yes, transaction fees - no, all Bitcoins mined is not any time soon - yes, the halving curve means that the block reward will be fairly low long before then.

I'm sure there's a BitcoinTalk thread somewhere in those Google results, best pick it up over there Smiley

Yeah but what I am after is a 1 in X chance of hitting a block, not 1 in X chance of receiving Y amt of bitcoin.
its impossible. its like saying you want a 1 in X chance of flipping a quarter 5 times in a row and each time come up heads.

while it could happen, you cant ever know how many times you have to flip the 5 quarters to make them all come up heads.



I think that's still a fair comparison though. Even with, for example, powerball... Ignore all but jackpot. The chance of powerball jackpot is 1 in 175,223,510.

So if each block is a drawing, there is a chance of winning each block drawing. But it's not a guarantee. Same with the powerball. The only guarantee is if you buy all 175,223,510 combinations. Even buying 175,223,509 doesn't guarantee anything except the set odds. In my mind this is akin to having 100% of network hashrate. Odds never guarantee anything.
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June 17, 2015, 10:35:35 PM
 #20

Not sure exactly what the odds are using lottery terms but it will definitely depend on your hashrate.

For example, an S5 (1.15TH) according to coinwarz will take 2148.02 days ( 5.8 years ) to find a block at current difficulty:
http://www.coinwarz.com/calculators/bitcoin-mining-calculator/?h=1150&p=600&pc=0.10&pf=0.00&d=49692386354.89380000&r=25.00000000&er=250.82000000&hc=0.00

Whereas 1PH ( 1000TH ) should only take 2.47 days to find a block: http://www.coinwarz.com/calculators/bitcoin-mining-calculator/?h=1000000.00&p=600.00&pc=0.10&pf=0.00&d=49692386354.89380000&r=25.00000000&er=250.82000000&hc=0.00

Ignore the watts and power cost its irrelevant Wink.


Right so if we take the S5 figure, for instance, and extend it with the average 144 blocks per day, I think it's 1 in 309,314.88. I also think all of these calculations are all restatements of what everyone in this thread said earlier. The lotto-type odds rely on difficulty and local hashrate. Difficulty depends on target. Target is reset every 2016 blocks based on global hashrate.
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