How many random numbers must be drawn on average to make the sequence fall for the first time? 
Consider a game in which random numbers are constantly selected with equal probability in the interval [0, 1] until the first time the number selected is smaller than the previous one.
Then, how many random numbers need to be drawn on average for this to happen?

I saw the answer elsewhere, but it was explained in a way that I had trouble understanding, so have to re-ask here, if anyone could explain in a more understandable way.
BTW, this is the thought process I had when handling this question, not sure at which step the error was introduced:


*

*Probability of needing to draw 2 times: $\frac{1}{2}$

*Probability of needing to draw 3 times: On the basis that the first two is increasing (probability is $\frac{1}{2}$), the probability that the third time decreasing is again $\frac{1}{2}$, so the final probability of needing to draw 3 times is $(\frac{1}{2})^2$

*...

*Probability of needing to draw n+1 times: $(\frac{1}{2})^n$

*Therefore the average number of times needed to draw = $\sum_{n=1}^\infty \frac{n+1}{2^n}=3$

 A: What you state in your second bullet is not okay.
Under the condition $X_1<X_2$ the probability on $X_2<X_3$ is not $\frac12$ but is less than $\frac12$.
You seem to confuse it with the unconditional probability of $X_2<X_3$.
Hint for solution:
If $N$ denotes the number of selections needed then:$$N>n\iff X_1<X_2<\cdots<X_n$$so that:$$P(N>n)=\frac1{n!}$$
A: Let's reframe the question. We have a sequence of random variables $\left(X_n\right)_{n=1}^\infty$, each of which is drawn i.i.d. from $[0, 1]$ with a flat probability distribution. Let $N$ be the least positive integer such that $X_N < X_{N-1}$, what is $\mathbb{E}(N)$?
We can see that if $N > 3$, then $X_3 > X_2$ and $X_2 > X_1$, so $X_3 > X_1$.
In fact, for any positive integers $m$ and $n$ such that $N > m > n$, we have $X_m > X_n$.
Thus, $$\begin{aligned}\mathbb{P}(N \ge 3) &= \mathbb{P}(X_3 > X_2 \land X_2 > X_1) \\&\neq \mathbb{P}(X_3 > X_2)\mathbb{P}(X_2 > X_1)\end{aligned}$$
Because, by knowing that $X_2 > X_1$, it becomes less likely that $X_3 > X_2$. The events are not independent, $\mathbb{P}(X_3 > X_2 \mid X_2 > X_1) < 0.5$.
This is intuitive if we take it ad absurdum. If I tell you that a random number $X$ between 0 and 1 was larger than 99 out of 99  other i.i.d. random numbers we drew from the same distribution, it's intuitive that $X$ is likely very close to 1, and so it's quite unlikely another random number between 0 and 1 will be larger again.
