The binomial distribution fixes the number of trials and counts the successes. Turn the question around: keep trying until the first success, and count the trials. Rolling until a six, calling until someone answers, buying cereal until you find the prize.
Key idea
With independent trials, each a success with probability $p$, the trial number $X$ of the first success is geometric:
$$P(X = k) = (1 - p)^{k - 1}p, \quad k = 1, 2, 3, \ldots$$
Its mean is $E(X) = \frac{1}{p}$, and $P(X > k) = (1 - p)^k$.
Why it works
The first success on trial $k$ means exactly this sequence: $k - 1$ failures, then a success. By independence its probability is the product $(1 - p)^{k - 1}p$. "More than $k$ trials are needed" means the first $k$ trials all fail: $(1 - p)^k$. The mean $1/p$ matches intuition: if success happens one time in six, you wait six tries on average.
Worked example: First six on a given roll
Roll a fair die until a six appears. Find $P(X = 3)$.
- Two failures, then a success: $\left(\frac{5}{6}\right)^2 \cdot \frac{1}{6}$.
- $= \frac{25}{216} \approx 0.116$.
Answer: $\frac{25}{216}$
Worked example: The average wait
On average, on which roll does the first six appear?
- A six comes up on 1 roll in 6, so the wait averages 6 rolls.
- $E(X) = \dfrac{1}{p} = \dfrac{1}{1/6} = 6$.
Answer: The 6th roll.
Worked example: A long wait
What is the probability that more than 4 rolls are needed?
- More than 4 rolls are needed exactly when the first 4 rolls all miss.
- $$P(X > 4) = \left(\frac{5}{6}\right)^4 = \frac{625}{1296} \approx 0.482$$
Answer: About $0.482$.
Common mistake
Believing a success is "due" after a run of failures. The trials are independent: after 5 misses, the chance of a six on the next roll is still $\frac{1}{6}$, and the expected wait from now is still 6 more rolls. Check: does the process have a memory? Dice and coins do not.