If I want to know how many times I need to roll a die in oder 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?

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?