# Why do we sometimes subtract from the exponent in a probability formula?

I am unsure of why we sometimes subtract from the exponent in a probability formula.

For example, for this question:

A six-sided die is tossed. If it turns up a 1 or 2 the player wins. If it turns up 5 or 6 the player loses. If it turns up 3 or 4 the die is repeatedly tossed until either the same number as the first toss turns up (in which the player wins) or a 5 or 6 turns up in which case the game is lost.

Let $$N$$ represent the number of tosses until the game stops. Find a formula for $$P(N = n)$$.

The answer for $$n = 1$$ is $$2/3$$, but the answer for n greater or equal $$2$$ is $$\frac{1}{6}\left(\frac{1}{2}\right)^{n-2}$$

I'm guessing the numbers $$1/6$$ and $$1/2$$ are the probabilities for success and failure? I'm also confused about why you need to subtract two from $$n$$?

Suppose $$n > 1$$. How is it possible that $$N = n$$? We must get a $$3$$ or $$4$$ on the first toss (probability $$1/3$$), then the next $$n-2$$ tosses can't be $$5, 6$$, or the first number that was rolled (probability $$(1/2)^{n-2}$$), then the final toss must be either $$5, 6$$, or the same as the first toss (probability $$1/2$$). The probability of this sequence of events occurring is $$(1/3) \cdot (1/2)^{n-2} \cdot (1/2)$$.