Compute the expectation of X

When a certain basketball player takes his first shot in a game with probability 1/3. If he misses his first 2 shots, his third shot will go with probability 1/5. If he misses his first 3 shots, his fourth shot will go with prob. 1/6. If he misses his first 4 shots, the coach will remove him from the game. Assume that the player keeps shooting until he succeeds or he is removed from the game. Let X denote the number of shots he misses until his first success or until he is removed from the game.

Compute $\mathbb{E}[X]$

If I separate this into indicator functions then the expectation is the sum of the probabilities of each event happening. This would mean thst if I find the pmf and then sum the probabilities, I would get the expectation. My concern is that this would mean the expectation is $1$. Is that a reasonable expectation? The pmf I found is as follows:

$$\mathbb{P}(X=1)=\mathbb{P}(\text{misses 1st and makes second})=\frac{2}{3}*\frac{1}{4}$$ $$\mathbb{P}(X=2)=\mathbb{P}(\text{misses 1st, 2nd and makes third})=\frac{2}{3}*\frac{3}{4}*\frac{1}{5}$$ $$\mathbb{P}(X=3)=\mathbb{P}(\text{misses 1st,2nd,3rd and makes fourth})=\frac{2}{3}*\frac{3}{4}*\frac{4}{5}*\frac{1}{6}$$ $$\mathbb{P}(X=4)=\mathbb{P}(\text{misses 1st,2nd,3rd,4th)}=\frac{2}{3}*\frac{3}{4}*\frac{4}{5}*\frac{5}{6}$$

All of this sums to $1$ as a pmf should.

Edit: I forgot that there is also the probability that he misses 0 shots. So I should add $\mathbb{P}(X=0)=\frac{1}{3}$

Now it should add up to 1.