This is awesome, but just a small comment about "Next block expected in 7 minutes". The next expected block is always due in 10 minutes, no matter when the previous block was generated. Mining is a memoryless process.
Not if the current network speed is higher than the difficulty level due to more mining power having come online since the last difficulty change.
You could estimate this by examining the block times over the last 120 (or some N) blocks, estimating the difficulty for that solve rate, and using the ratio of current network difficulty over the the estimated difficulty times 10 minutes.
I think the best solution is to remove the expected next block time counter altogether. One of the reasons UX wise I went for such an expected point in the future was that after reading it you are aware a new block will come again (just like when you first load the project) and that might keep you waiting for it.
Just like when you flip 10 coins they likely won't result in: head - tail - head - tail - (...). Though on average you end up with 5 tails and 5 heads.
That's why it says expected! To be 100% correct you could plot the cumulative density function as time progresses, but I think just printing the expected time is fine.
Testing a block header is just a binary (yes/no) test with a given probability (target/max_target). Finding a block is repeating this test again and again, i.e. a binomial distribution, right?
Finding a block averages 10 minutes, which means that, after 10 minutes trying hashes randomly, the expectation of finding a block is high, i.e. the CDF of the binomial distribution approaches 1.
The result will deviate from the expected result? Of course! After all, it's a distribution... but this doesn't contradict the fact that as you try again a again, the probability of AT LEAST ONE hitting the target gets higher.
So, yes, it doesn't change the individual probability of each outcome, but it sure does mean that more hashes have been tried, i.e. there's a greater probability of a hash being found simply because we've tried more guesses as time passes, i.e. we're measuring the CDF with a high N.
To put it another way: the more coins you flip, the higher the expectation of AT LEAST ONE yielding tails (even if the previous attempts don't change the outcome probability).
Think about it: even if each coin outcome is always 1/2, we're not assessing whether it's going to be tails or not in flip N, but whether after N flips we'll see at least one flipping tails, whose probability is (1 - (1/2)^N).
Flipping a coin 7 times yields a 99% probability of flipping tails (or heads) at least once, regardless of previous outcomes.
And I'll stop here, I think I'm repeating myself :P
Of course it does not describe time. It describes attempts (which grow with time because you're doing N attempts per second, i.e. the network hashrate). My point still stands.
I see your point, but I think we're unknowingly discussing a semantic issue: the expected time is not the time when you're expected to find a block, it is the time where the probability of at least one block being found is high (including the previous attempts). It's only gambler's fallacy if you're discarding the previous attempts and only measuring the probability of hitting a block in a single attempt, which of course is constant throughout all attempts.
> The inter-arrival time -- the time between valid blocks -- is a random variable with an Exponential distribution.
Sure, but if you pick 10 minutes, you'll be right more often than not (not counting hashrate changes, of course) because it averages 10 minutes...
One block is found in 1 minute and the next in 19? No problem. That's why it says "expected" and not "sure".
----
How many times do you have to flip a coin to have a 99% confidence of flipping tails at least once? Seven. You might need less than seven (the first attempt is already 50%) or more than seven (because 99% != 100% and we never reach 100%) but 99 out of 100 times you flip 7 coins you'll hit tails at least once.
I think that's what the "expected" means. You expect to flip tails within 7 flips (because I've arbitrarily set the bar at 99%) so we might be arguing what "expected" means.
If you flip it only 3 additional times you get only a 7/8=88% probability of getting at least one tail.
You need to flip it 7 additional times you get 127/128=99.2% probability of getting at least one tail.
The coin doesn’t remember that the last 4 times it landed head.