Dervish (OP)
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January 17, 2014, 07:33:14 AM |
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Imagine we have 5 people only. Each of them owns 200M. Total forging power is 200+200+200+200+200. Agree?
What do you mean when saying "Total forging power"? Why don't we imaging 3 people with each of them owns 333,333333M? I have already solve that task. And simple example to illustrate what we talking about: 1) We have wallet with 2/3 of billion NXT and other person have wallet with 1/3 of the billion 2) We have two wallets with 1/3 of billion NXT and other person have wallet with 1/3 of the billion 2) is obviously we have probabilty of succes 2/3 but in case 1) we have probabilty of succes 3/4 ![](https://ip.bitcointalk.org/?u=http%3A%2F%2Fs12.postimg.org%2F4vxmgb2jh%2Fimage.png&t=665&c=j2ypJiGIxlH9NQ) Here on horisontal axis value of opponent hit function and on vertical value of our hit function. Red space - we win. Blue space - opponent win. P.S You may notice that for our last example [4] is not correct. The reason that P is very big and our assumption that 1 > (t m * A * b 0) / N for each possible t m is not true. It's became true for P<1/2. Yes, you are correct in this. You mistake is, as CFB has pointed out, assuming P(200M) = 0.2. This is wrong. Make it simpler, suppose there's only one account, P(1000M) != 1. It should be around 0.5, so that the average interval is around 1 min. If the probability of hit target in 1 min is 1, then the average interval is definitely smaller than 1 min (maybe only 0.5 min depends on the distribution). Ok P(200M) = 0.2 is incorrect but it was not my main point. In my main math post I have not such assumption. And even if I have mistaken in 2 times (do not considered retargeting), the idea is still correct.
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Dervish (OP)
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January 17, 2014, 07:34:29 AM |
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Alice owns N, Bob owns N/2 + N/2 (==N).
Let's assess Bob's chance to hit target assuming that base target is such that Alice hits it within 1 minute (say, 0.2 probability):
Every of the accounts has 0.0005 chance to hit the target (0.2 * N/2 / N) Chance that none of the accounts does it == (1 - 0.1) * (1 - 0.1) = 0.9 * 0.9 = 0.81 Chance that any of the accounts does it == 1 - 0.81 = 0.19
Thus the combined stake does have advantage which proves Dervish point of view. Let's calculate this advantage:
Alice / Bob = 0.2 / 0.19 = 1.0526
Thus Alice quotient of advantage = 1.0526 - 1 = 0.0526 = 5.26%
I marked incorrect assumption with red. Isn't the part highlighted in blue also incorrect? "My first account forges" and "my second account forges" are not independent events. In my post I study another events: Event A - time that wallet #1 need to generate block less that time other wallets except ours need to generate block. Event B - time that wallet #2 need to generate block less that time other wallets except ours need to generate block. And this events are not mutual exclusive.
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Come-from-Beyond
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January 17, 2014, 07:35:22 AM |
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Imagine we have 5 people only. Each of them owns 200M. Total forging power is 200+200+200+200+200. Agree?
Why don't we imaging 3 people with each of them owns 333,333333M? I have already solve that task. And simple example to illustrate what we talking about: 1) We have wallet with 2/3 of billion NXT and other person have wallet with 1/3 of the billion 2) We have two wallets with 1/3 of billion NXT and other person have wallet with 1/3 of the billion 2) is obviously we have probabilty of succes 2/3 but in case 1) we have probabilty of succes 3/4 ![](https://ip.bitcointalk.org/?u=http%3A%2F%2Fs12.postimg.org%2F4vxmgb2jh%2Fimage.png&t=665&c=j2ypJiGIxlH9NQ) Here on horisontal axis value of opponent hit function and on vertical value of our hit function. Red space - we win. Blue space - opponent win. P.S You may notice that for our last example [4] is not correct. The reason that P is very big and our assumption that 1 > (t m * A * b 0) / N for each possible t m is not true. It's became true for P<1/2. I don't get why u rely on this graph. U don't take into account that base target will be adjusted. This is what I call "Poisson process". Every account can forge is block, this is a Poisson process. But there is also another Poisson process that determines changes of forging difficulty (a process of a higher level). U don't take it into account, that's why u get wrong numbers.
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BitThink
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January 17, 2014, 07:37:41 AM |
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Anyway, Could we reach a conclusion that "The large holder does have an advantage in forging. If some large account be divided into many small accounts, they lose this advantage and will give the rest large holders a little bit more forging advantage. Nonetheless, the difference is not so high so it will not jeopardize the system."
An extreme case:
suppose there's an account with 500M Nxt, and other 5000 users with 100K Nxt. To be fair, the large account should forge 50% of the coins, but actually at least one user in the 5000 hits the target in 1 min is 1 - (1 - p/5000)^5000. If p = 0.2, then 1 - (1 - p/5000) = 0.181. Which means the large holder actually can mine around 52.5% of the coins. The sums of small accounts can mine around 47.5%, but when they all combine to one account together, they can mine 50% (not 52.5%) and the advantage of the large holder goes away too. (The exact value of p here does not matter, it could be 0.3 or 0.18. The conclusion is that the advantage exists but not so large).
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Come-from-Beyond
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January 17, 2014, 07:38:25 AM |
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Therefore, this is a WRONG argument: "If I have p percent of the total Nxt, I have p percent of chance to hit the target in 1 min." But this is true "If A account has twice Nxt as B, A have twice as much chance to hit the target in 1 min."
In other words, If we add up all the probability of hit the target in 1 min of all accounts, it is NOT 1.
U should understand that forging process is unfair from point of view of pure math, but it is fair from point of view of statistics, coz advantage of larger stake lays within the dispersion.
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Dervish (OP)
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January 17, 2014, 07:38:31 AM |
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Imagine we have 5 people only. Each of them owns 200M. Total forging power is 200+200+200+200+200. Agree?
What do you mean when saying "Total forging power"? Each coin is like a mining rig. So we could use "forging power" to describe possibility to forge a block. But "forging power" of coin in different size wallet is different. How can sum them?
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Come-from-Beyond
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January 17, 2014, 07:41:02 AM |
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And even if I have mistaken in 2 times (do not considered retargeting), the idea is still correct.
It's correct. But the numbers r not. Already answered in that russian thread and here: U should understand that forging process is unfair from point of view of pure math, but it is fair from point of view of statistics, coz advantage of larger stake lays within the dispersion.
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BitThink
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January 17, 2014, 07:42:19 AM |
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Alice owns N, Bob owns N/2 + N/2 (==N).
Let's assess Bob's chance to hit target assuming that base target is such that Alice hits it within 1 minute (say, 0.2 probability):
Every of the accounts has 0.0005 chance to hit the target (0.2 * N/2 / N) Chance that none of the accounts does it == (1 - 0.1) * (1 - 0.1) = 0.9 * 0.9 = 0.81 Chance that any of the accounts does it == 1 - 0.81 = 0.19
Thus the combined stake does have advantage which proves Dervish point of view. Let's calculate this advantage:
Alice / Bob = 0.2 / 0.19 = 1.0526
Thus Alice quotient of advantage = 1.0526 - 1 = 0.0526 = 5.26%
I marked incorrect assumption with red. Isn't the part highlighted in blue also incorrect? "My first account forges" and "my second account forges" are not independent events. Why? I think the account does not know who the owner is. There's no communication between your two accounts. Therefore, they are just like any two accounts in the world.
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Dervish (OP)
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January 17, 2014, 07:43:03 AM |
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Imagine we have 5 people only. Each of them owns 200M. Total forging power is 200+200+200+200+200. Agree?
Why don't we imaging 3 people with each of them owns 333,333333M? I have already solve that task. And simple example to illustrate what we talking about: 1) We have wallet with 2/3 of billion NXT and other person have wallet with 1/3 of the billion 2) We have two wallets with 1/3 of billion NXT and other person have wallet with 1/3 of the billion 2) is obviously we have probabilty of succes 2/3 but in case 1) we have probabilty of succes 3/4 ![](https://ip.bitcointalk.org/?u=http%3A%2F%2Fs12.postimg.org%2F4vxmgb2jh%2Fimage.png&t=665&c=j2ypJiGIxlH9NQ) Here on horisontal axis value of opponent hit function and on vertical value of our hit function. Red space - we win. Blue space - opponent win. P.S You may notice that for our last example [4] is not correct. The reason that P is very big and our assumption that 1 > (t m * A * b 0) / N for each possible t m is not true. It's became true for P<1/2. I don't get why u rely on this graph. U don't take into account that base target will be adjusted. This is what I call "Poisson process". Every account can forge is block, this is a Poisson process. But there is also another Poisson process that determines changes of forging difficulty (a process of a higher level). U don't take it into account, that's why u get wrong numbers. Ok. Just answer the question "We have wallet with 2/3 of billion NXT and other person have wallet with 1/3 of the billion what probability that we find a block?" Poisson process is about different time moments, not about probability in current time moment.
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notsoshifty
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January 17, 2014, 07:45:32 AM |
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In my post I study another events: Event A - time that wallet #1 need to generate block less that time other wallets except ours need to generate block. Event B - time that wallet #2 need to generate block less that time other wallets except ours need to generate block. And this events are not mutual exclusive.
But are they independent? Isn't it simpler if you ignore time and just consider who forges blocks N, (N+1), (N+2), etc? Suppose: - I have 400,000nxt in account A1 and it is actively forging - You have 400,000nxt which is split into 2 accounts A2 and A3 (200,000 per account), both of which are actively forging - The other 200,000nxt is either held by other people, but is constantly being involved in transactions and so never actually forging, or is destroyed nxt sent to the genesis block My assumption is: - If account A has X nxt, and a total of Y nxt is forging across the whole network, then P(account A forges next block)=X/Y. Is that assumption correct? For block N, chance of each account A1,A2,A3 forging: P(A1 forges block N)=0.5 P(A2 forges block N)=0.25 P(A3 forges block N)=0.25 i.e.: P(I forge block N) = P(you forge block N) = 0.5 Difficulty can go up and down, and time taken to forge a block may vary, but the calculations above are independent of this. Or where did I go wrong?
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Come-from-Beyond
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January 17, 2014, 07:45:56 AM |
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But "forging power" of coin in different size wallet is different. How can sum them?
We can sum them coz forging power of 1M == forging power of 0.5M + 0.5M + epsilon, where epsilon is small if compared to 1M. Even more, due to orphaned blocks 1 single pool will lose more than get, we already saw this in early days when were DDoSed.
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BitThink
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January 17, 2014, 07:47:55 AM |
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In my post I study another events: Event A - time that wallet #1 need to generate block less that time other wallets except ours need to generate block. Event B - time that wallet #2 need to generate block less that time other wallets except ours need to generate block. And this events are not mutual exclusive.
But are they independent? Isn't it simpler if you ignore time and just consider who forges blocks N, (N+1), (N+2), etc? Suppose: - I have 400,000nxt in account A1 and it is actively forging - You have 400,000nxt which is split into 2 accounts A2 and A3 (200,000 per account), both of which are actively forging - The other 200,000nxt is either held by other people, but is constantly being involved in transactions and so never actually forging, or is destroyed nxt sent to the genesis block My assumption is: - If account A has X nxt, and a total of Y nxt is forging across the whole network, then P(account A forges next block)=X/Y.Is that assumption correct? For block N, chance of each account A1,A2,A3 forging: P(A1 forges block N)=0.5 P(A2 forges block N)=0.25 P(A3 forges block N)=0.25 i.e.: P(I forge block N) = P(you forge block N) = 0.5Difficulty can go up and down, and time taken to forge a block may vary, but the calculations above are independent of this. Or where did I go wrong? No, the probability cannot be simply added together, so that assumption is also not correct. Moreover, we can only use probability to forge one block in 1 minute. P(A1 forges block N)=0.25 P(A2 forges block N)=0.125 P(A3 forges block N)=0.125 P(you forge block N) = 1 - (0.875) * (0.875) = 0.234375. So before splitting, the probability to forge one block in 1 minute is 0.5, but after splitting 0.484375, the difficulty may be adjusted in following rounds.
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Come-from-Beyond
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January 17, 2014, 07:48:48 AM |
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Ok. Just answer the question "We have wallet with 2/3 of billion NXT and other person have wallet with 1/3 of the billion what probability that we find a block?" Poisson process is about different time moments, not about probability in current time moment.
U'll find ~ 2001 blocks of 3000 ones.
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Dervish (OP)
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January 17, 2014, 07:50:43 AM |
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Ok. Just answer the question "We have wallet with 2/3 of billion NXT and other person have wallet with 1/3 of the billion what probability that we find a block?" Poisson process is about different time moments, not about probability in current time moment.
U'll find ~ 2001 blocks of 3000 ones. Can you prove it?
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Come-from-Beyond
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January 17, 2014, 07:51:29 AM |
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Ok. Just answer the question "We have wallet with 2/3 of billion NXT and other person have wallet with 1/3 of the billion what probability that we find a block?" Poisson process is about different time moments, not about probability in current time moment.
U'll find ~ 2001 blocks of 3000 ones. Can you prove it? Already proved upthread.
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Dervish (OP)
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January 17, 2014, 07:57:59 AM |
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Ok. Just answer the question "We have wallet with 2/3 of billion NXT and other person have wallet with 1/3 of the billion what probability that we find a block?" Poisson process is about different time moments, not about probability in current time moment.
U'll find ~ 2001 blocks of 3000 ones. Can you prove it? Already proved upthread. This is not a proof. You prove that epsilon is small making assumption that epsilon is small. Some situation with ddos is also is non a prof.
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Dervish (OP)
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January 17, 2014, 07:58:58 AM |
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As I say Poisson process is about different time moments, not about probability in the current time moment.
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Come-from-Beyond
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January 17, 2014, 08:01:28 AM |
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As I say Poisson process is about different time moments, not about probability in the current time moment.
There are TWO Poisson processes. I'm talking about higher-level one. Ur math doesn't take into account BASE TARGET. This is why ur numbers r INCORRECT.
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BitThink
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January 17, 2014, 08:02:30 AM |
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Dervish, why not to discuss with CFB a getwork like of pool. Your pool will be more attractive if people don't need to send Nxt to you. I guess it's doable by distributing work to multiple clients if there's some support from the client code?
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January 17, 2014, 08:06:11 AM |
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Dervish, why not to discuss with CFB a getwork like of pool. Your pool will be more attractive if people don't need to send Nxt to you. I guess it's doable by distributing work to multiple clients if there's some support from the client code?
When we implement Account Control users will be able to lease forging power instead of transfering real coins. No need in getwork. Actually we should be getting used to use pools, they r necessary in later stages of TF. Their official name is "hubs" though (time to reserve good domains in advance ![Wink](https://bitcointalk.org/Smileys/default/wink.gif) ).
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