Mixing
POW with ROW who has the higher chance to mine a block?
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A group of $N$ miners trying to mine a block, I am one of them. In one scenario they are using a POW(proof of work) algorithm and in another, they use a ROW(race of work) algorithm.
The mining process is to execute the function $m()$. It returns a random hash number between $0 - 2^32$. In case of POW the first miner the find a hash smaller than some value $L$ wins. In case of ROW, the miner with the smallest hash after $T$ minutes wins.
In both scenarios I have 10% of the mining power, I manage to execute $m()$ 10% of the total time it is executed by all the miners until someone wins.
In what scenario do I have more chances to win the block, POW or ROW?
Also does the answer is changing if instead of 10% I have $p$ percent?
probability probability-theory
asked Jul 23 at 23:42
Ilya Gazman
456927
add a comment |Â
up vote
1
down vote
favorite
A group of $N$ miners trying to mine a block, I am one of them. In one scenario they are using a POW(proof of work) algorithm and in another, they use a ROW(race of work) algorithm.
The mining process is to execute the function $m()$. It returns a random hash number between $0 - 2^32$. In case of POW the first miner the find a hash smaller than some value $L$ wins. In case of ROW, the miner with the smallest hash after $T$ minutes wins.
In both scenarios I have 10% of the mining power, I manage to execute $m()$ 10% of the total time it is executed by all the miners until someone wins.
In what scenario do I have more chances to win the block, POW or ROW?
Also does the answer is changing if instead of 10% I have $p$ percent?
probability probability-theory
asked Jul 23 at 23:42
Ilya Gazman
456927
Assuming ties are impossible, I would have thought they both gave you a $10%$ or $p%$ chance of winning
– Henry
Jul 23 at 23:45
@Henry yeah, this is also what I think, but I would like somehow to prove it mathematically.
– Ilya Gazman
Jul 23 at 23:46
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
A group of $N$ miners trying to mine a block, I am one of them. In one scenario they are using a POW(proof of work) algorithm and in another, they use a ROW(race of work) algorithm.
The mining process is to execute the function $m()$. It returns a random hash number between $0 - 2^32$. In case of POW the first miner the find a hash smaller than some value $L$ wins. In case of ROW, the miner with the smallest hash after $T$ minutes wins.
In both scenarios I have 10% of the mining power, I manage to execute $m()$ 10% of the total time it is executed by all the miners until someone wins.
In what scenario do I have more chances to win the block, POW or ROW?
Also does the answer is changing if instead of 10% I have $p$ percent?
probability probability-theory
asked Jul 23 at 23:42
Ilya Gazman
456927
A group of $N$ miners trying to mine a block, I am one of them. In one scenario they are using a POW(proof of work) algorithm and in another, they use a ROW(race of work) algorithm.
The mining process is to execute the function $m()$. It returns a random hash number between $0 - 2^32$. In case of POW the first miner the find a hash smaller than some value $L$ wins. In case of ROW, the miner with the smallest hash after $T$ minutes wins.
In both scenarios I have 10% of the mining power, I manage to execute $m()$ 10% of the total time it is executed by all the miners until someone wins.
In what scenario do I have more chances to win the block, POW or ROW?
Also does the answer is changing if instead of 10% I have $p$ percent?
probability probability-theory
asked Jul 23 at 23:42
Ilya Gazman
456927
asked Jul 23 at 23:42
Ilya Gazman
456927
asked Jul 23 at 23:42
Ilya Gazman
456927
asked Jul 23 at 23:42
Ilya Gazman
456927
456927
Assuming ties are impossible, I would have thought they both gave you a $10%$ or $p%$ chance of winning
– Henry
Jul 23 at 23:45
@Henry yeah, this is also what I think, but I would like somehow to prove it mathematically.
– Ilya Gazman
Jul 23 at 23:46
add a comment |Â
Assuming ties are impossible, I would have thought they both gave you a $10%$ or $p%$ chance of winning
– Henry
Jul 23 at 23:45
@Henry yeah, this is also what I think, but I would like somehow to prove it mathematically.
– Ilya Gazman
Jul 23 at 23:46
Assuming ties are impossible, I would have thought they both gave you a $10%$ or $p%$ chance of winning
– Henry
Jul 23 at 23:45
Assuming ties are impossible, I would have thought they both gave you a $10%$ or $p%$ chance of winning
– Henry
Jul 23 at 23:45
@Henry yeah, this is also what I think, but I would like somehow to prove it mathematically.
– Ilya Gazman
Jul 23 at 23:46
@Henry yeah, this is also what I think, but I would like somehow to prove it mathematically.
– Ilya Gazman
Jul 23 at 23:46
add a comment |Â
2 Answers
2
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up vote
0
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Assume for simplicity no ties are possible, neither in values hashed, nor in timing (i.e. two miners cannot produce a hash "at the same time", so it's always obvious who won POW). Note that in both cases, the winning miner is the one with the lowest hash: this is obvious in ROW, but also in POW since the first hash below L is less than every hash that came before it. Then, if at the time the winner is declared you computed a fraction $f$ of all the hashes, for symmetry reasons the probability that the lowest hash is among those you computed is $f$.
answered Jul 24 at 0:20
Anonymous
4,8033940
Can you please clearly state your conclusion, I think I get it but just to be more clear.
– Ilya Gazman
Jul 24 at 0:31
@IlyaGazman I have a hard time making it clearer. If you throw a ball (the winning hash) into a random bin, and I own 10% of the bins, there is a 10% probability that the ball will end in a bin of mine. More formally, consider $n$ bins and thus $n$ equiprobable, mutually exclusive events, with event $i$ being "the ball falls into bin $i$ and obviously happening with probability $1/n$. The probability for me to win is the sum of the probabilities of all events favourable to me ($(n/10)cdot (1/n)=1/10$) over the sum of the probabilities of all events ($ncdot(1/n)=1$), and thus it equals 1/10.
– Anonymous
Jul 24 at 0:43
add a comment |Â
up vote
0
down vote
Suppose you split your resources into 10 separate mining nodes and your opponent splits his resources into 90 separate nodes. You then have 100 identically powered nodes between you, and you run the competition between them, and look for who owns the winner. It should then be clear that you have a 10% chance of winning.
However, notice that "split your power into 90 nodes and let them run independently" for both modes of the game means that you're actually doing the same thing as "mine with all of your power at once". It's implicit in your description of the rules that each hashing attempt is independent of all the others, so whether you choose to artificially distribute the attempts you have the power to make among 90 imagined sub-miners or not does not change whether one of the attempts is the eventually winning one.
(In practice, $2^32$ is probably not a large enough search space to justify the assumption of independence. Actual coin mining uses a vastly larger space to approximate that assumption much better).
answered Jul 24 at 0:49
Henning Makholm
225k16290516
As you hint, this argument relies on not having two miners repeat the same calculation. If I have $10%$ of the capacity and can make sure I never try the same value twice while there are $99$ people each with $1%$ of the capacity who do not communicate I will win (very slightly) more than $10%$ of the time because (a tiny bit) of their capacity is wasted by repetition. As you say, if the search space is large enough this will be tiny. It needs to be greater than the square of the number of calculations that will be done as shown in the birthday paradox.
– Ross Millikan
Jul 24 at 1:10
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Assume for simplicity no ties are possible, neither in values hashed, nor in timing (i.e. two miners cannot produce a hash "at the same time", so it's always obvious who won POW). Note that in both cases, the winning miner is the one with the lowest hash: this is obvious in ROW, but also in POW since the first hash below L is less than every hash that came before it. Then, if at the time the winner is declared you computed a fraction $f$ of all the hashes, for symmetry reasons the probability that the lowest hash is among those you computed is $f$.
answered Jul 24 at 0:20
Anonymous
4,8033940
Can you please clearly state your conclusion, I think I get it but just to be more clear.
– Ilya Gazman
Jul 24 at 0:31
@IlyaGazman I have a hard time making it clearer. If you throw a ball (the winning hash) into a random bin, and I own 10% of the bins, there is a 10% probability that the ball will end in a bin of mine. More formally, consider $n$ bins and thus $n$ equiprobable, mutually exclusive events, with event $i$ being "the ball falls into bin $i$ and obviously happening with probability $1/n$. The probability for me to win is the sum of the probabilities of all events favourable to me ($(n/10)cdot (1/n)=1/10$) over the sum of the probabilities of all events ($ncdot(1/n)=1$), and thus it equals 1/10.
– Anonymous
Jul 24 at 0:43
add a comment |Â
up vote
0
down vote
Assume for simplicity no ties are possible, neither in values hashed, nor in timing (i.e. two miners cannot produce a hash "at the same time", so it's always obvious who won POW). Note that in both cases, the winning miner is the one with the lowest hash: this is obvious in ROW, but also in POW since the first hash below L is less than every hash that came before it. Then, if at the time the winner is declared you computed a fraction $f$ of all the hashes, for symmetry reasons the probability that the lowest hash is among those you computed is $f$.
answered Jul 24 at 0:20
Anonymous
4,8033940
Can you please clearly state your conclusion, I think I get it but just to be more clear.
– Ilya Gazman
Jul 24 at 0:31
@IlyaGazman I have a hard time making it clearer. If you throw a ball (the winning hash) into a random bin, and I own 10% of the bins, there is a 10% probability that the ball will end in a bin of mine. More formally, consider $n$ bins and thus $n$ equiprobable, mutually exclusive events, with event $i$ being "the ball falls into bin $i$ and obviously happening with probability $1/n$. The probability for me to win is the sum of the probabilities of all events favourable to me ($(n/10)cdot (1/n)=1/10$) over the sum of the probabilities of all events ($ncdot(1/n)=1$), and thus it equals 1/10.
– Anonymous
Jul 24 at 0:43
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Assume for simplicity no ties are possible, neither in values hashed, nor in timing (i.e. two miners cannot produce a hash "at the same time", so it's always obvious who won POW). Note that in both cases, the winning miner is the one with the lowest hash: this is obvious in ROW, but also in POW since the first hash below L is less than every hash that came before it. Then, if at the time the winner is declared you computed a fraction $f$ of all the hashes, for symmetry reasons the probability that the lowest hash is among those you computed is $f$.
answered Jul 24 at 0:20
Anonymous
4,8033940
Assume for simplicity no ties are possible, neither in values hashed, nor in timing (i.e. two miners cannot produce a hash "at the same time", so it's always obvious who won POW). Note that in both cases, the winning miner is the one with the lowest hash: this is obvious in ROW, but also in POW since the first hash below L is less than every hash that came before it. Then, if at the time the winner is declared you computed a fraction $f$ of all the hashes, for symmetry reasons the probability that the lowest hash is among those you computed is $f$.
answered Jul 24 at 0:20
Anonymous
4,8033940
edited Jul 24 at 0:26
answered Jul 24 at 0:20
Anonymous
4,8033940
answered Jul 24 at 0:20
Anonymous
4,8033940
answered Jul 24 at 0:20
Anonymous
4,8033940
4,8033940
Can you please clearly state your conclusion, I think I get it but just to be more clear.
– Ilya Gazman
Jul 24 at 0:31
@IlyaGazman I have a hard time making it clearer. If you throw a ball (the winning hash) into a random bin, and I own 10% of the bins, there is a 10% probability that the ball will end in a bin of mine. More formally, consider $n$ bins and thus $n$ equiprobable, mutually exclusive events, with event $i$ being "the ball falls into bin $i$ and obviously happening with probability $1/n$. The probability for me to win is the sum of the probabilities of all events favourable to me ($(n/10)cdot (1/n)=1/10$) over the sum of the probabilities of all events ($ncdot(1/n)=1$), and thus it equals 1/10.
– Anonymous
Jul 24 at 0:43
add a comment |Â
Can you please clearly state your conclusion, I think I get it but just to be more clear.
– Ilya Gazman
Jul 24 at 0:31
@IlyaGazman I have a hard time making it clearer. If you throw a ball (the winning hash) into a random bin, and I own 10% of the bins, there is a 10% probability that the ball will end in a bin of mine. More formally, consider $n$ bins and thus $n$ equiprobable, mutually exclusive events, with event $i$ being "the ball falls into bin $i$ and obviously happening with probability $1/n$. The probability for me to win is the sum of the probabilities of all events favourable to me ($(n/10)cdot (1/n)=1/10$) over the sum of the probabilities of all events ($ncdot(1/n)=1$), and thus it equals 1/10.
– Anonymous
Jul 24 at 0:43
Can you please clearly state your conclusion, I think I get it but just to be more clear.
– Ilya Gazman
Jul 24 at 0:31
Can you please clearly state your conclusion, I think I get it but just to be more clear.
– Ilya Gazman
Jul 24 at 0:31
@IlyaGazman I have a hard time making it clearer. If you throw a ball (the winning hash) into a random bin, and I own 10% of the bins, there is a 10% probability that the ball will end in a bin of mine. More formally, consider $n$ bins and thus $n$ equiprobable, mutually exclusive events, with event $i$ being "the ball falls into bin $i$ and obviously happening with probability $1/n$. The probability for me to win is the sum of the probabilities of all events favourable to me ($(n/10)cdot (1/n)=1/10$) over the sum of the probabilities of all events ($ncdot(1/n)=1$), and thus it equals 1/10.
– Anonymous
Jul 24 at 0:43
@IlyaGazman I have a hard time making it clearer. If you throw a ball (the winning hash) into a random bin, and I own 10% of the bins, there is a 10% probability that the ball will end in a bin of mine. More formally, consider $n$ bins and thus $n$ equiprobable, mutually exclusive events, with event $i$ being "the ball falls into bin $i$ and obviously happening with probability $1/n$. The probability for me to win is the sum of the probabilities of all events favourable to me ($(n/10)cdot (1/n)=1/10$) over the sum of the probabilities of all events ($ncdot(1/n)=1$), and thus it equals 1/10.
– Anonymous
Jul 24 at 0:43
add a comment |Â
up vote
0
down vote
Suppose you split your resources into 10 separate mining nodes and your opponent splits his resources into 90 separate nodes. You then have 100 identically powered nodes between you, and you run the competition between them, and look for who owns the winner. It should then be clear that you have a 10% chance of winning.
However, notice that "split your power into 90 nodes and let them run independently" for both modes of the game means that you're actually doing the same thing as "mine with all of your power at once". It's implicit in your description of the rules that each hashing attempt is independent of all the others, so whether you choose to artificially distribute the attempts you have the power to make among 90 imagined sub-miners or not does not change whether one of the attempts is the eventually winning one.
(In practice, $2^32$ is probably not a large enough search space to justify the assumption of independence. Actual coin mining uses a vastly larger space to approximate that assumption much better).
answered Jul 24 at 0:49
Henning Makholm
225k16290516
As you hint, this argument relies on not having two miners repeat the same calculation. If I have $10%$ of the capacity and can make sure I never try the same value twice while there are $99$ people each with $1%$ of the capacity who do not communicate I will win (very slightly) more than $10%$ of the time because (a tiny bit) of their capacity is wasted by repetition. As you say, if the search space is large enough this will be tiny. It needs to be greater than the square of the number of calculations that will be done as shown in the birthday paradox.
– Ross Millikan
Jul 24 at 1:10
add a comment |Â
up vote
0
down vote
Suppose you split your resources into 10 separate mining nodes and your opponent splits his resources into 90 separate nodes. You then have 100 identically powered nodes between you, and you run the competition between them, and look for who owns the winner. It should then be clear that you have a 10% chance of winning.
However, notice that "split your power into 90 nodes and let them run independently" for both modes of the game means that you're actually doing the same thing as "mine with all of your power at once". It's implicit in your description of the rules that each hashing attempt is independent of all the others, so whether you choose to artificially distribute the attempts you have the power to make among 90 imagined sub-miners or not does not change whether one of the attempts is the eventually winning one.
(In practice, $2^32$ is probably not a large enough search space to justify the assumption of independence. Actual coin mining uses a vastly larger space to approximate that assumption much better).
answered Jul 24 at 0:49
Henning Makholm
225k16290516
As you hint, this argument relies on not having two miners repeat the same calculation. If I have $10%$ of the capacity and can make sure I never try the same value twice while there are $99$ people each with $1%$ of the capacity who do not communicate I will win (very slightly) more than $10%$ of the time because (a tiny bit) of their capacity is wasted by repetition. As you say, if the search space is large enough this will be tiny. It needs to be greater than the square of the number of calculations that will be done as shown in the birthday paradox.
– Ross Millikan
Jul 24 at 1:10
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Suppose you split your resources into 10 separate mining nodes and your opponent splits his resources into 90 separate nodes. You then have 100 identically powered nodes between you, and you run the competition between them, and look for who owns the winner. It should then be clear that you have a 10% chance of winning.
However, notice that "split your power into 90 nodes and let them run independently" for both modes of the game means that you're actually doing the same thing as "mine with all of your power at once". It's implicit in your description of the rules that each hashing attempt is independent of all the others, so whether you choose to artificially distribute the attempts you have the power to make among 90 imagined sub-miners or not does not change whether one of the attempts is the eventually winning one.
(In practice, $2^32$ is probably not a large enough search space to justify the assumption of independence. Actual coin mining uses a vastly larger space to approximate that assumption much better).
answered Jul 24 at 0:49
Henning Makholm
225k16290516
Suppose you split your resources into 10 separate mining nodes and your opponent splits his resources into 90 separate nodes. You then have 100 identically powered nodes between you, and you run the competition between them, and look for who owns the winner. It should then be clear that you have a 10% chance of winning.
However, notice that "split your power into 90 nodes and let them run independently" for both modes of the game means that you're actually doing the same thing as "mine with all of your power at once". It's implicit in your description of the rules that each hashing attempt is independent of all the others, so whether you choose to artificially distribute the attempts you have the power to make among 90 imagined sub-miners or not does not change whether one of the attempts is the eventually winning one.
(In practice, $2^32$ is probably not a large enough search space to justify the assumption of independence. Actual coin mining uses a vastly larger space to approximate that assumption much better).
answered Jul 24 at 0:49
Henning Makholm
225k16290516
answered Jul 24 at 0:49
Henning Makholm
225k16290516
answered Jul 24 at 0:49
Henning Makholm
225k16290516
answered Jul 24 at 0:49
Henning Makholm
225k16290516
225k16290516
As you hint, this argument relies on not having two miners repeat the same calculation. If I have $10%$ of the capacity and can make sure I never try the same value twice while there are $99$ people each with $1%$ of the capacity who do not communicate I will win (very slightly) more than $10%$ of the time because (a tiny bit) of their capacity is wasted by repetition. As you say, if the search space is large enough this will be tiny. It needs to be greater than the square of the number of calculations that will be done as shown in the birthday paradox.
– Ross Millikan
Jul 24 at 1:10
add a comment |Â
As you hint, this argument relies on not having two miners repeat the same calculation. If I have $10%$ of the capacity and can make sure I never try the same value twice while there are $99$ people each with $1%$ of the capacity who do not communicate I will win (very slightly) more than $10%$ of the time because (a tiny bit) of their capacity is wasted by repetition. As you say, if the search space is large enough this will be tiny. It needs to be greater than the square of the number of calculations that will be done as shown in the birthday paradox.
– Ross Millikan
Jul 24 at 1:10
As you hint, this argument relies on not having two miners repeat the same calculation. If I have $10%$ of the capacity and can make sure I never try the same value twice while there are $99$ people each with $1%$ of the capacity who do not communicate I will win (very slightly) more than $10%$ of the time because (a tiny bit) of their capacity is wasted by repetition. As you say, if the search space is large enough this will be tiny. It needs to be greater than the square of the number of calculations that will be done as shown in the birthday paradox.
– Ross Millikan
Jul 24 at 1:10
As you hint, this argument relies on not having two miners repeat the same calculation. If I have $10%$ of the capacity and can make sure I never try the same value twice while there are $99$ people each with $1%$ of the capacity who do not communicate I will win (very slightly) more than $10%$ of the time because (a tiny bit) of their capacity is wasted by repetition. As you say, if the search space is large enough this will be tiny. It needs to be greater than the square of the number of calculations that will be done as shown in the birthday paradox.
– Ross Millikan
Jul 24 at 1:10
add a comment |Â
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Assuming ties are impossible, I would have thought they both gave you a $10%$ or $p%$ chance of winning
– Henry
Jul 23 at 23:45
@Henry yeah, this is also what I think, but I would like somehow to prove it mathematically.
– Ilya Gazman
Jul 23 at 23:46