Plan ₿: BlockIntervalTime

BlockIntervalTime

Due to the random nature, the specific realized block interval time can be beyond Average Joe's imagination. Denote:

p the probability of not finding a block in one second
k the random variable of number of seconds before finding a block successfully

Then k is of negative binomial distribution with r=1. The mean is p/(1-p), therefore the seconds to find a block is 1+p/(1-p). For example of bitcoin, it is designed with adjustment of finding difficulty such that the p is 599/600 and therefore finding a block every 600 seconds on average. Denote the cdf of Beta distribution of alpha and beta by $I_x( \alpha , \beta )$ , the cdf of negative binomial distribution of r is $1-I_p( k+1 , r )$ .

The cdf of a single block time is $1-p^{k+1}$ and finding the second block after a block is a likely event. People may blame bitcoin miners are not doing a good job because miners find a second block in 6 seconds but statistically the probability for such event is 0.011608.

The daily block numbers and therefore the daily estimated global hash rate is also addressed. The number of blocks of a day is equal or less than n is the equivalent event of $k_1+..+k_n+n\geq 86400$

$Pr(k_1+..+k_n+n\geq 86400)=1-Pr(k_1+..+k_n \leq 85399-n)$ $=1-(1-I_p(85399-n+1,n))=I_p(86400-n,n)$

The designate number of daily blocks for bitcoin is 144 which is 86400/600. People may blame miners are gaming the hash power to cause a 20% swing of daily average block time or estimated global hash power but statistically the probability of the event that number of daily block is less than 115 or greater than 173 is 0.015799.

From the microscope of view of a miner, the block interval time is in fact the 1-order statistics of a population of miners. Mining being a non-repeat trial-and-error process about hash, the block-found time of a miner is a uniform distribution. For example, 10000 hash to be explored and the hashing speed is 10 hash per second, then the block will be found surely in 1000 seconds by any specific miner. Denote the cdf and pdf by F and f. Let the uniform distribution be distributed in interval 0 to A seconds and there are N miners. Then

F(t)=t/A, f(t)=1/A
the 1-order statistics cdf Pr(T<t) is $1-(1-F(t))^N$ whose expectation is $\frac{A}{N+1}$ which shall be equal to 600. Therefore $A=(N+1)600$ and $Pr(T<t)=1-(1- \frac{t}{600(N+1)} )^N \approx 1-e^{-t/600}$

probability of orphan

The probability of a miner finding a block at time $T_2$ while another miner has found it before at $T_1$ with $T_2<T_1+t$ due to internet propagation can be calculated by the joint pdf of 1-2-order statistics and it is

$\int_0^ \infty \int_{T_1}^{T_1+t} N(N-1)(1-F(T_2))^{N-2} f(T_2) f(T_1) dT_2 dT_1$ $=1-N \int_0^ \infty (1-F(T_1+t))^{N-1}f(T_1)dT_1 =1-(1-t/A)^N$

Again, this approaches to $1- e^{- t/600 }$ and $Pr(T_2>T_1+t)=e^{-t/600}$ as probability of no orphan.

stable block time

People often blame the notorious uncertainty of block time. Note that the sigma is almost the same as the expectation of the 1-order statistics. Knowing that $\frac{T_k}{A}$ the k-order statistics is beta distribution with $\alpha =k$ and $\beta =N-k+1$ , its mean is $\frac{Ak}{N+1}$ and its sigma is $\frac{A}{N+1} \sqrt{ \frac{k(N-k+1)}{N+2} }$ , therefore the sigma/mean is $\sqrt{ \frac{N-k+1}{k(N+2)} } \approx \frac{1}{ \sqrt{k} }$ . For 1-order statistics, it is $\sqrt{ \frac{N}{N+2} }$ . For 9-order statistics, it is almost 1/3. If the bitcoin protocol were designed as the 9-order statistics as the valid block, the sigma/block time ratio would be 33%.