Probability you play the piano song at least 8 times before getting it correctly 4 times? Question: You are working on a difficult passage from a new piece you are learning on the piano.You wish to play it correctly 4 times before calling it a day. If you have a probability of 2/3 of playing it correctly on every attempt, and the attempts are independent (unfortunately!), what is the probability that you attempt to play it at least 8 times?
So I'm taking the compliment of in order to figure out what is the probability that you attempt to play it at least times by doing 1-prob of getting it right 4 times in under 8 attempts.
I went though this case by case. The first case is in 4 attempts I play the song correctly 4 times. 
The second case is getting it right in 5 attempts. I know automatically the fifth attempt I have to play the song correctly or I would have stopped early (would fall into the previous case) thus that leaves 4 combinations to play. 
The third case is getting it right in 6 attempts.Again the 6th attempt I must get the song correct or I would have stopped earlier which leaves 10 combinations... and so on. 
What is wrong with this approach?
 A: The approach mostly looks good, though there may be some errors in the arithmetic as you try to evaluate the various terms. So you might want to check that.
(It is often helpful to write step-by-step equations where you have a complete, correct equation at each step, rather than writing a formula at some random location on the paper, going off somewhere to get a number from it, and writing the number in some other random place on the page.
I suspect you may have the correct numbers, just written in an unexpected sequence.)
The one thing that looks like an obvious mistake is that you are adding up the cases where you play exactly four times, exactly five times, exactly six times,
exactly seven times, or exactly eight times.
The sum of those probabilities is the probability to finish in eight or fewer tries, and when subtract the sum from $1,$ you get the probability that you have to do more than eight tries, that is, nine or more.
So the first thing I would do is stop including the eighth try in your sum.
Then the sum will actually represent playing the piece correctly four times in
under eight attempts.
A: In a direct answer, I think you're on the right track.  I'll share my method, which sounds similar to yours, but it treats the "history" as an all-at-once rather than try to combine runs of 4, 5, 6, etc. separately.
The question wants to know how likely is it that you have to go at least as many as 8 tries to succeed.  In problems like this, it's useful to think of what has to happen to require an eighth try.
In your previous 7 tries, you must have succeeded precisely 3 times, no more, no less, and in no particular order.  Think about how many combinations that is, and what fraction that is to the total number of outcomes in 7 tries.
That gives you a fraction of the likelihood of getting to the 8th try, and you have a 1/2 odds of getting it right on the 8th try.  
Can you work it from there? 
