6 blocks in 4 minutes

[biodieselchris]

For more clarity, I was wondering what the probability of such of an event was.

The time it takes to mine one block is exponentially distributed with λ (mean) = 10 minutes.

This means the total time to mine four blocks is Erlang-distributed with k=4 and λ=10 minutes, and the probability to find 4 blocks within 6 minutes is about 0.336%.

So, roughly speaking (and strongly simplified, probability-wise), once every 100/0.33581 ≈ 298 chunks of 4 blocks, or once every (100/0.33581)*4*10 minutes ≈ 8.3 days, you can expect 4 subsequent blocks to appear within 6 minutes.

thank you Jace, good quant analysis. Also worth noting that you are assuming the difficulty, which could've been set as long as two weeks ago, still accurately matches the hash rate, producing blocks with a mean target of 600s.

Also note that the probability (W) being calculated is (p) mining 6 blocks within 6 minutes (so it would be the probability of 5 subsequent blocks being mined after any particular block, not 4) and also, given p, (q) the probability of no block being mined for an hour, so the probability I was looking for was:

W = p * q

I was guessing originally around 10,000 to 1

[spazzdla]

I wonder what the chances of that are.  I just like seeing rare things lol.

[jonald_fyookball]

Here is something to think about:

Regardless of how long ago the last block was added, the average time to the next block is 10 minutes from now.

Yep, and if starting now you wait 2 hours and no block gets mined, then the expected time for the next block to appear is still 10 minutes after that (rather than 'any moment now!')

Exponential probability distribution vs people's intuition
 

Not sure if this is 100% correct.  I think the longer time that passes, the more likely it is a block will be found sooner.
All the nonces that have to be tried and failed, will be tried (and fail) as time goes on, bringing us
closer to the solution. 

Think of an isolated intersection on the outskirts of town where only one car drives through
every ten minutes.  The longer that passes, the closer you get to the next car.

[spazzdla]

For more clarity, I was wondering what the probability of such of an event was.

The time it takes to mine one block is exponentially distributed with λ (mean) = 10 minutes.

This means the total time to mine four blocks is Erlang-distributed with k=4 and λ=10 minutes, and the probability to find 4 blocks within 6 minutes is about 0.336%.

So, roughly speaking (and strongly simplified, probability-wise), once every 100/0.33581 ≈ 298 chunks of 4 blocks, or once every (100/0.33581)*4*10 minutes ≈ 8.3 days, you can expect 4 subsequent blocks to appear within 6 minutes.

thank you Jace, good quant analysis. Also worth noting that you are assuming the difficulty, which could've been set as long as two weeks ago, still accurately matches the hash rate, producing blocks with a mean target of 600s.

Also note that the probability (W) being calculated is (p) mining 6 blocks within 6 minutes (so it would be the probability of 5 subsequent blocks being mined after any particular block, not 4) and also, given p, (q) the probability of no block being mined for an hour, so the probability I was looking for was:

W = p * q

I was guessing originally around 10,000 to 1

what about the 7th block being an hour?   1 /

[NorrisK]

I wonder what the chances of that are.  I just like seeing rare things lol.

I think it was calculated in the post above you

It's small, but big enough to happen. And it gets bigger as the hash power increases before diff adjustment.

[spazzdla]

I wonder what the chances of that are.  I just like seeing rare things lol.

I think it was calculated in the post above you

It's small, but big enough to happen. And it gets bigger as the hash power increases before diff adjustment.

That was just the 6 blocks in a row though!!


Unless it's 1 / 1 000 000 000 000 000 000 000

I'm not too worried. The situation doesn't sound that rare but rare.

[jonald_fyookball]

For more clarity, I was wondering what the probability of such of an event was.

The time it takes to mine one block is exponentially distributed with λ (mean) = 10 minutes.

This means the total time to mine four blocks is Erlang-distributed with k=4 and λ=10 minutes, and the probability to find 4 blocks within 6 minutes is about 0.336%.

So, roughly speaking (and strongly simplified, probability-wise), once every 100/0.33581 ≈ 298 chunks of 4 blocks, or once every (100/0.33581)*4*10 minutes ≈ 8.3 days, you can expect 4 subsequent blocks to appear within 6 minutes.

thank you Jace, good quant analysis. Also worth noting that you are assuming the difficulty, which could've been set as long as two weeks ago, still accurately matches the hash rate, producing blocks with a mean target of 600s.

Also note that the probability (W) being calculated is (p) mining 6 blocks within 6 minutes (so it would be the probability of 5 subsequent blocks being mined after any particular block, not 4) and also, given p, (q) the probability of no block being mined for an hour, so the probability I was looking for was:

W = p * q

I was guessing originally around 10,000 to 1

what about the 7th block being an hour?   1 /

Uh no...guys its a Poisson distribution.  Different math entirely.
This is what you use when there is "X number of something every Y minutes."

https://en.wikipedia.org/wiki/Poisson_distribution

[Jace]

Uh no...guys its a Poisson distribution.  Different math entirely.

The poisson and exponential distributions are closely related, but be careful not to mistake one for the other:

Block time (a continuous variable, can by anything ≥0) is distributed exponentially, in this case with mean λ=10 minutes.
The number of blocks per time interval (a discrete variable, can be any integer ≥0) is a poisson distribution, in this case with mean λ=one block per 10 minutes.

This is what you use when there is "X number of something every Y minutes."

When calculating the probability of 6 blocks being found within 4 minutes, using Poisson distribution is most likely the wrong approach.

However "the probability of 6 blocks being found within 4 minutes" is somewhat ambiguous. Do we mean:
1. the probability that the number of blocks in any random 4 minute interval is 6 (could ask the same question for e.g. 5 or 13 or 0 blocks)
or
2. the probability that you succeed in mining 6 blocks within 4 minutes (could ask the same question for e.g. 2 or 19.54 or 180 minutes)

If you're observing the total time time it takes to mine 6 blocks, then it's an Erlang distributed variable, because each individual block's mining time is exponentially distributed, and we're observing the sum of 6 of those.

[Jace]

I think the longer time that passes, the more likely it is a block will be found sooner.
All the nonces that have to be tried and failed, will be tried (and fail) as time goes on, bringing us
closer to the solution.

Nope. I haven't actually measured the following statistic myself, but if you were to take all blocks that took more than 37 minutes to mine (since their preceding block), I'm sure you will find that they still took 10 minutes *on average* after that 37 minute mark.
(obviously the 37 minutes is just an arbitrary period, any other will yield the same result).

Think of an isolated intersection on the outskirts of town where only one car drives through
every ten minutes.  The longer that passes, the closer you get to the next car.

Intuitionally this may seem the case, but it's not true. Exponentially distributed variables (or poisson distributed events) are independent of the past.

In a fair casino roulette, one can expect that red and black will roughly come up equally as much.
If you wait patiently until a sequence of straight 23 reds comes up, is the probability of the next roll being black now larger than 50%?

[jonald_fyookball]

I think the longer time that passes, the more likely it is a block will be found sooner.
All the nonces that have to be tried and failed, will be tried (and fail) as time goes on, bringing us
closer to the solution.

Nope. I haven't actually measured the following statistic myself, but if you were to take all blocks that took more than 37 minutes to mine (since their preceding block), I'm sure you will find that they still took 10 minutes *on average* after that 37 minute mark.
(obviously the 37 minutes is just an arbitrary period, any other will yield the same result).

Think of an isolated intersection on the outskirts of town where only one car drives through
every ten minutes.  The longer that passes, the closer you get to the next car.

Intuitionally this may seem the case, but it's not true. Exponentially distributed variables (or poisson distributed events) are independent of the past.

In a fair casino roulette, one can expect that red and black will roughly come up equally as much.
If you wait patiently until a sequence of straight 23 reds comes up, is the probability of the next roll being black now larger than 50%?

I see your point.  Because the inputs keep changing (different timestamp header, etc), there
is a near infinite set of values we could hash, so it more resembles a random function, even
though the hash function is deterministic.

[Jace]

I see your point.  Because the inputs keep changing (different timestamp header, etc), there
is a near infinite set of values we could hash, so it more resembles a random function, even
though the hash function is deterministic.

Exactly, I believe there are two 32-bit nonces in the block header, and several other variables you can change (timestamp, and which transactions to include or exclude, amongst other things) so that gives a huge set of possibilities. Any number of possibilities we can feasibly process through brute force (which is what Bitcoin mining does) is still insignificant, so indeed it resembles a random process.