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If I want to know how many times I need to roll a die in order to have a given chance of seeing at least one six, I can use the formula:

n = (log(1 - s))/(log(1 - p))

where n is the required number of rolls, s is the desired probability of rolling at least one six, and p is the probablity on each roll. It tells me that I must roll the die 26 times if I want a 99% chance of seeing at least one six. As applied to playing cards (where p is 1/52 rather than 1/6), the same formula tells me that I would have to pick a card randomly from 237 packs if I want a 99% chance of seeing at least one Ace of Spades.

My question is: Does this formula apply to mining yields? For instance, if the network hash rate is H and my local hash rate is h, would I have to mine for the time taken for (log(1 - 0.99))/(log(1-(h/H))) blocks if I want a 99% chance of receiving the reward for at least one of them?

2 Answers2

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Yes, kind of. Your model is an idealized representation of mining, and works with one exception. The network has a certain degree of latency between nodes, and as a result miners do not know immediately when another miners finds a block, but rather have to wait something like 1-10 seconds before learning about a new block. Now, at first glance,this might not seem like a big deal because everyone has this issue. But the reality is that network latency results in bigger miners winning a slightly higher share of blocks than their hashrate would suggest, and smaller miners get less. This is due to the fact that a miner effectively gets a head start on the next block after mining a block, and bigger miners therefore get more of these headstart a, gaining a slight advantage. The smaller one's hashrate is, the more exacerbated this effect becomes. I don't know of a foolproof way of representing this effect, though. A good starting point might be to say that a small miners is only 90-95% efficient, but such a factor would have to be empirically tuned to be as accurate as possible.

bigreddmachine
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This is also a contest of who can guess first/throw a dice of the correct value within a given time based on the previous block (right now is 2 minutes) and you have to start guessing again in the next block. Thus, if your hash rate as a solo miner or we can say guess per second/throwing a dice per second can only do a guess/throw of 24,000 in 2 minutes (@200 h/s) and there is a possible of billion guesses you'll have a minuscule of probability of guessing/throwing the dice of the required value.

However, this does not mean that you will never guess the correct value if you are very lucky. But right now the Network H/s just went to the roof at 80MH/s!

Gundamlancer
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