[NXT] nxtpool.com - first forging NXT pool [161 KNXT] - closed - page 3.

Come-from-Beyond

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Quote from: Dervish on January 17, 2014, 03:38:31 AM

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.

notsoshifty

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Quote from: Dervish on January 17, 2014, 03:34:29 AM

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?

Dervish

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Quote from: Come-from-Beyond on January 17, 2014, 03:35:22 AM

Quote from: Dervish on January 17, 2014, 03:10:45 AM

Quote from: Come-from-Beyond on January 17, 2014, 02:29:14 AM

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.

Quote from: Dervish on January 16, 2014, 10:08:55 AM

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

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 * b₀) / 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.

BitThink

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Quote from: notsoshifty on January 17, 2014, 03:28:37 AM

Quote from: Come-from-Beyond on January 17, 2014, 02:30:07 AM

Quote from: BitThink on January 17, 2014, 02:28:05 AM

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.

Come-from-Beyond

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Quote from: Dervish on January 17, 2014, 03:33:14 AM

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:

Quote from: Come-from-Beyond on January 17, 2014, 03:38:25 AM

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.

Dervish

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Quote from: Come-from-Beyond on January 17, 2014, 03:30:39 AM

Quote from: Dervish on January 17, 2014, 03:10:45 AM

Quote from: Come-from-Beyond on January 17, 2014, 02:29:14 AM

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?

Come-from-Beyond

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Quote from: BitThink on January 17, 2014, 03:12:13 AM

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.

BitThink

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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).

Come-from-Beyond

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Quote from: Dervish on January 17, 2014, 03:10:45 AM

Quote from: Come-from-Beyond on January 17, 2014, 02:29:14 AM

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.

Quote from: Dervish on January 16, 2014, 10:08:55 AM

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

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 * b₀) / 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.

Dervish

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Activity: 713

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Quote from: notsoshifty on January 17, 2014, 03:28:37 AM

Quote from: Come-from-Beyond on January 17, 2014, 02:30:07 AM

Quote from: BitThink on January 17, 2014, 02:28:05 AM

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.

Dervish

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Activity: 713

Merit: 508

Quote from: BitThink on January 17, 2014, 03:17:59 AM

Quote from: Dervish on January 17, 2014, 03:10:45 AM

Quote from: Come-from-Beyond on January 17, 2014, 02:29:14 AM

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.

Quote from: Dervish on January 16, 2014, 10:08:55 AM

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

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 * b₀) / 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.

Come-from-Beyond

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Quote from: Dervish on January 17, 2014, 03:10:45 AM

Quote from: Come-from-Beyond on January 17, 2014, 02:29:14 AM

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.

notsoshifty

sr. member

Activity: 378

Merit: 250

Quote from: Come-from-Beyond on January 17, 2014, 02:30:07 AM

Quote from: BitThink on January 17, 2014, 02:28:05 AM

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.

BitThink

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Activity: 882

Merit: 1000

Quote from: Dervish on January 17, 2014, 03:10:45 AM

Quote from: Come-from-Beyond on January 17, 2014, 02:29:14 AM

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.

Quote from: Dervish on January 16, 2014, 10:08:55 AM

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

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 * b₀) / 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).

BitThink

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Quote from: Come-from-Beyond on January 17, 2014, 03:01:39 AM

Quote from: BitThink on January 17, 2014, 02:57:38 AM

Quote from: Come-from-Beyond on January 17, 2014, 02:30:07 AM

Quote from: BitThink on January 17, 2014, 02:28:05 AM

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.

Why?

The probability is 0.2 if he has 20% of Nxt, right?

No, the point is that if base target == 100%, then guy with 200M can hit the target within 1 minute only with 0.1 probability. If someone starts playing with probabilities the system will just raise difficulty lowering advantage of combined stake.

Ok, I got a little bit more.

In your example, 5 people have equally 200M, does the system adjust the difficulty so that the expected interval is 1 min. Therefore the probability of none of the 5 people to hit the target (P0) needs to be around 50%? (What's the value in your algorithm?). P0 = (1 - P) ^ 5. If P = 0.2, then P0 = 0.32, a little bit too high. If P = 0.1, then P0 = 0.59, almost. Seems P should be around 0.125.

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.

Dervish

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Activity: 713

Merit: 508

Quote from: Come-from-Beyond on January 17, 2014, 02:29:14 AM

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.

Quote from: Dervish on January 16, 2014, 10:08:55 AM

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

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 * b₀) / N for each possible t_m is not true. It's became true for P<1/2.

Come-from-Beyond

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Quote from: BitThink on January 17, 2014, 03:00:40 AM

Quote from: Come-from-Beyond on January 17, 2014, 02:29:14 AM

Quote from: Dervish on January 17, 2014, 02:23:04 AM

Can you please repeat your calculations with 20% but not 0,1% or just say that there is no lie in my words.

Ok. Let's do it step by step.

Imagine we have 5 people only. Each of them owns 200M. Total forging power is 200+200+200+200+200. Agree?

Edit: Maybe u don't lie but just confused. I don't state that u spread incorrect information intentionally.

Now I think, if one of them split it to (100, 100), all other four actually may have a little bit more probability due to that one's loss. That only makes the difference larger.

I believe that actually the large holder does have a considerable advantage in forging, although it may be not that important anyway. As I always think the forging profit is negligible.

Let's discuss situation when 2 guys combine 200 and 200 into 400. What total forging power will be in this case?

Come-from-Beyond

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Quote from: BitThink on January 17, 2014, 02:57:38 AM

Quote from: Come-from-Beyond on January 17, 2014, 02:30:07 AM

Quote from: BitThink on January 17, 2014, 02:28:05 AM

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.

Why?

The probability is 0.2 if he has 20% of Nxt, right?

No, the point is that if base target == 100%, then guy with 200M can hit the target within 1 minute only with 0.1 probability. If someone starts playing with probabilities the system will just raise difficulty lowering advantage of combined stake.

BitThink

legendary

Activity: 882

Merit: 1000

Quote from: Come-from-Beyond on January 17, 2014, 02:29:14 AM

Quote from: Dervish on January 17, 2014, 02:23:04 AM

Can you please repeat your calculations with 20% but not 0,1% or just say that there is no lie in my words.

Ok. Let's do it step by step.

Imagine we have 5 people only. Each of them owns 200M. Total forging power is 200+200+200+200+200. Agree?

Edit: Maybe u don't lie but just confused. I don't state that u spread incorrect information intentionally.

Now I think, if one of them split it to (100, 100), all other four actually may have a little bit more probability due to that one's loss. That only makes the difference larger.

I believe that actually the large holder does have a considerable advantage in forging, although it may be not that important anyway. As I always think the forging profit is negligible.

BitThink

legendary

Activity: 882

Merit: 1000

Quote from: Come-from-Beyond on January 17, 2014, 02:30:07 AM

Quote from: BitThink on January 17, 2014, 02:28:05 AM

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.

Why?

The probability is 0.2 if he has 20% of Nxt, right?

Topic: [NXT] nxtpool.com - first forging NXT pool [161 KNXT] - closed - page 3. (Read 7425 times)