Topic: [NXT] nxtpool.com - first forging NXT pool [161 KNXT] - closed - page 4. (Read 7474 times)

Thank you for your attention to my math. I agree that my words may be confusing, but they correct. And I also agree that in small amounts phenomenon is negligible.

And I want to remember that when I first point out that paradox I even has no idea to make a pool. I just study NXT forging algorithm. And then you answer:


Do you think it is necessary to do a pool? you said.
And I think. Hm... He is right! And start to make pool. Spent about ten days for it on winter holidays. And now make final steps with codding.

I marked incorrect assumption with red.

Ok. Let's do it step by step.

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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?

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

I will try for CFB


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 also find out the reason why split reduces the chance. It's because these two accounts do not exchange information so they cannot avoid duplicate work.

Sorry mistake. It should be read

for 0,1% (1 million) d = 4/3,99 = 1,0025 difference 0,025% CFB have rounded it to 0,03%
for 5% (50 millions) d = 4/3,95 = 1,01265 difference 1,265% I have rounded it to 1,27%
for 20% (200 millions) d = 4/3,8 = 1,05263 difference 5,26%

Here I wrote correct numbers:

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

Do you assume the total coins join forging is 100 millions?

Dervish overlooked that forging is a Poisson process. That's why he got insane numbers.

We need another wallets with non zero balance to consider the way I does and get this equation.

Ok, Now I understand better.
After split to two equal parts, p = 1 - (1 - P/2) * (1 - P/2) = 1 - (1 - P + P^2/4) = P - (P^2) / 4

So the equation Dervish used was correct.

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, 0.001 probability):

Every of the accounts has 0.0005 chance to hit the target (0.001 * N/2 / N)
Chance that none of the accounts does it == (1 - 0.0005) * (1 - 0.0005) = 0.9995 * 0.9995 = 0.99900025
Chance that any of the accounts does it == 1 - 0.99900025 = 0.00099975

Thus the combined stake does have advantage which proves Dervish point of view. Let's calculate this advantage:

Alice / Bob = 0.001 / 0.00099975 = 1.000250062515629

Thus Alice quotient of advantage = 1.000250062515629 - 1 = 0.0003 = 0.03%

Max value of advantage quotient can be reached if Bob splits all his coins among very big number of accounts. It will be close to 0.05% in this case.

PS: 0.05% advantage is compensated by dispersion and can work only in very long run which doesn't make sense coz currency where stakeholders don't spend coins will die much earlier.

Do you have any reference of p = P-P²/4?

Say if I have all the Nxts, my probability is certainly 1. Then I divide all my coins to two accounts, the probability of one account is (1 - 1/4) / 2 = 3/8 and the sum of my two accounts becomes 3/4? Does that mean there's 1/4 chance no forge in a round?

If I further divide them to 4 accounts, then the forge probability further decreases? So there're more empty rounds? Seems quite hard for me to understand.

EDIT:

What I said was not true.

The equation for difference is
d=P/p=P/(P-P²/4)=4/(4-P)

for 1% (1 million) d = 4/3,99 = 1,0025 difference 0,025% CFB have rounded it to 0,03%
for 5% (5 millions) d = 4/3,95 = 1,01265 difference 1,265% I have rounded it to 1,27%
for 20% (20 millions) d = 4/3,8 = 1,05263 difference 5,26%

I said that my coefficient was incorrect, but not whole idea.


It means:

Yes I have lost coefficient calculation.
Benefit somewhat less than I expected.
For 1 million in a single splitting the difference is 0.03% (and here I quoted Come-from-Beyond)
For 50 millions in a single splitting the difference is 1.27%

And now I say that for 200 millions (20%) in a single splitting the difference is 5.26% (20%/19%)

Where is the lie?

I sure that it can not be fixed easily.

We generate two more blocks but they was empty
http://87.230.14.1/nxt/nxt.cgi?action=1000&blk=15008136651107453394
http://87.230.14.1/nxt/nxt.cgi?action=1000&blk=2778462017552493231

Lets add some math:

We will consider the algorithm for generating units in the system NXT.
The algorithm is as follows :
For each  wallet hash function hit is calculated with value range from 0 to 2⁶⁴ ( denoted by 2⁶⁴ of N). By the cryptographic properties of hit its value can be regarded as a random variable x with a uniform distribution on the interval [0 , N]. Value of the function is recalculated when a new block appears.
On the function hit value time t is calculated in seconds by the formula:
t = x / (A * b) [1]
where x - is a random variable whose value hit, A - base target ratio is recalculated each unit in order to maintain the required rate of generation units ( performs the same function as the complexity of the system Bitcoin ), b - the balance of the wallet (coins). Wallet for which the value of t is the minimal generates the next block.
Consider two situations : 1) All of our available funds belong to the one wallet 2) Funds split into 2 wallet with equals sum.
Lets find out in which case the probability to generate a new block is higher.

1 ) Let wallets except our involved in the generation unit K₁, ..., K_n; their balances b₁, ..., b_n. Here we make assumption that each b_i > 0. Then the time to generate a block for each of them
t_k = x_k / (A * b_k); [2]
Denote t_m = min (t₁, ..., t_n). In order that our wallet would generate block the following condition must be satisfied t₀ m where t₀ time of block generation by our wallet. After substituting [1] we get x / (A * b₀) m where b₀ is the number of coins in our wallet , transform and obtain the condition for the random variable x
x m * A * b₀
taking into account the uniformity of x the probability P is
P = min{1, (t_m * A * b₀) / N} [3]
Here we make assumption that 1 > (t_m * A * b₀) / N then
P = (t_m * A * b₀) / N [3.1]

2) The situation is the same but we have to balance two wallets b₀ / 2 on each of them. Probability that t₀₁ m by the formula [3] is
p₁ = (t_m * A * b₀ / 2 ) / N = (t_m * A * b₀) / 2N
And substituting the value of [3.1], we obtain
p₁ = P/2
and for the second wallet
p₂ = (t_m * A * b₀ / 2 ) / N = (t_m * A * b₀) / 2N = P/2
the probability that our first wallet will not generate block
q₁ = 1-p₁ = 1 - P/2
and also for the second wallet
q₂ =1-p₂ = 1 - P/2
probability that neither the first nor the second wallet will not generate block is
q = q₁ * q₂ = (2-P) ² / 4
the probability that either the first or the second wallet block is generate
p = 1-q = 1 - (2-P) ² / 4 = (4-(2-P) ² ) / 4 = (4P-P² ) / 4 = P-P² / 4
so the final value
p = P-P² / 4 [4]
Ie probability to generate a block in the event of a less than half of funds in the case of storage facilities on one wallet to the square of this probability.

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 but for each transaction you will lost @minimal 1 NxT .... for transaction.

So for 1 Nxt sent , you use 1..... pool for NxT by grouping all account doesn't worth .... and too risky , if your wallet get hacked ... you will refund all users back ?