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 2
64 ( denoted by 2
64 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 K
1, ..., K
n; their balances b
1, ..., 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 (t
1, ..., t
n). In order that our wallet would generate block the following condition must be satisfied t
0 <t
m where t
0 time of block generation by our wallet. After substituting [1] we get x / (A * b
0) <t
m where b
0 is the number of coins in our wallet , transform and obtain the condition for the random variable x
x <t
m * A * b
0taking into account the uniformity of x the probability P is
P = min{1, (t
m * A * b
0) / N} [3]
Here we make assumption that 1 > (t
m * A * b
0) / N then
P = (t
m * A * b
0) / N [3.1]
2) The situation is the same but we have to balance two wallets b
0 / 2 on each of them. Probability that t
01 <t
m by the formula [3] is
p
1 = (t
m * A * b
0 / 2 ) / N = (t
m * A * b
0) / 2N
And substituting the value of [3.1], we obtain
p
1 = P/2
and for the second wallet
p
2 = (t
m * A * b
0 / 2 ) / N = (t
m * A * b
0) / 2N = P/2
the probability that our first wallet will not generate block
q
1 = 1-p
1 = 1 - P/2
and also for the second wallet
q
2 =1-p
2 = 1 - P/2
probability that neither the first nor the second wallet will not generate block is
q = q
1 * q
2 = (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-P
2 ) / 4 = P-P
2 / 4
so the final value
p = P-P
2 / 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
![](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.