Work out how big the Nitrogen Dice Jackpot has to be for play to be profitable

[actmyname]

No, because each number in the streak costs another unit.

This is where I disagree. If you develop the distribution based on two outcomes: either hitting the streak of n lucky numbers or not, then you can follow a negative binomial distribution where the 'success' is not hitting the jackpot and the 'failure' is reaching the streak. It doesn't matter how many numbers you hit if they don't reach the consecutive n-in-a-row. That's why the cost is the same and the probability is based accordingly, as the complement of pⁿ.

For the 6 10 sided dice example: If you roll 2 of the specific numbers in a row and lose, you lose 2000 satoshis. If your roll 5 of the specific numbers in a row and lose, you lose 5000 satoshis. The failed attempts have different costs, and the various probabilities for hitting each one must be factored in to determine the average cost per attempt.

I don't disagree with this but to calculate E(X) you can simplify it with basing f(x) on "successfully entered n-in-a-row streak" or "did not enter n-in-a-row streak" and calculating X₁f(x₁) + X₂f(x₂)

Does anybody else want to weigh in on this? (I'm not too great at statistics so there's possibly some miscalculations in there)