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On average, how many times will I need to roll a six-sided die before I see ten ONES in total?

I'd like to be able to extend my knowledge to answer other questions in the same form, such as, "If only 15% of fish are within the legal size limit, how many fish should I need to catch before I get five keepers?

Thanks kindly

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    https://en.wikipedia.org/wiki/Negative_binomial_distribution – 2017-01-17
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    You should be able to generalize one of the good answers to cover your fishing case. You'll be correct if there are lots of fish. If there are just a few then the problem is more complicated - there might be just $15$ keepers in a pond with $100$ fish. Once you caught four you'd have a much harder time finding the last one - less than $15\%$. – 2017-01-17

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If you roll a 6-sided die 6 times, on average you'd get each number one time. 12 times (6*2) and you'd get each number 2 times. Roll it 60 times (6*10), and you'd get each number 10 times on average.

This is just intuition. Making the "on average" mathematically rigorous takes some work.

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    This is excellent intuition, but not a valid argument. End the next to last sentence with "on average", delete the last sentence, and say that the argument is just intuition and I'll upvote. – 2017-01-17
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    Not getting to delete which sentence? – 2017-01-17
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    The last one about an infinite iteration. It makes no sense. – 2017-01-17
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    Is it fine now? – 2017-01-17
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    OK if you accept my edits. – 2017-01-17
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    Thank you for making my mistake correct. – 2017-01-17
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Expected value is additive, so the expected number of times you need to roll the die to see 10 ones is 10 times the expected value of times you must roll the die to get a one. If after performing an experiment the probability of an event is $p$, the expected value of repetitions of the experiment needed for the event to occur is the inverse of the probability $1/p$. The probability of getting a one after rolling a die is $1/6$, so the expected number of times you need to roll a die to see a one is 6. Because of the additive property mentioned above, the expected number of times you need to roll the die to get 10 ones is $10 \times 6 = 60$.