Method of Indicators Random Variables Problem and the Hiring Problem I'm trying to do the following problem from The Probability Tutoring Book by Carol Ash.

Pick numbers at random between $0$ and $1$. The $i^{th}$ number sets a "record" if it's larger than all of it's predecessors.
For example, the sequence $.1, .04, .3, .12, .6, .5$ has $3$ record setters $(.1, .3, .6)$.
We always take the first number to be a record setter.
Find the expected number of record setters.

This problem reminded me of the hiring problem which is set up like so:
Each day a new candidate comes to interview for a job. We hire the $i^{\text{th}}$ person if that person is more qualified than everyone who came before. We interview for $n$ days. What is the expected number of people we hire?
We can proceed to do the problem by reasoning like so:
Let $X$ be the number of people we hire.
$$X = X_1 + X_2 + ... + X_n$$
where $X_i = 1$ if we hire that person and $X_i = 0$ otherwise.
Well the probability that the best person is on the $i^{th}$ day is $\frac {1}{i}$ since all of the people have the same potential to be the most qualified candidate.
Now this becomes a simple exercise in linearity of expectation :
$$E[X] = \sum_i {E[X_i]} = \sum_i {\frac{1}{i}} = O(\log(n))$$

At first I thought this reasoning would easily apply to my "record setting" problem (which according to the solutions at the end of the book it does). However, something rubs me the wrong.
Why doesn't the probability that $i^{th}$ number is the highest so far depend on what values the previous numbers are? For example,
$$P(2^{\text{nd}}\ \text{number highest}) = 1 - n_1 $$ where $n_1$ is the first number we choose and
$$P(i^{\text{th}}\ \text{number highest}) = 1 - \max(n_1, n_2, n_3, ..., n_{i-1}) $$. Suddenly my reasoning for the hiring problem, "all of the people have the same potential to be the most qualified candidate" doesn't hold up.
Can someone see where my logic went wrong? Any help would be helpful. :)
 A: 
Why doesn't the probability that ith number is the highest so far depend on what values the previous numbers are?

In the hiring problem, it does depend on the "highest so far" in the same way. Your mistake, I think, is that you forget that this highest so far is also a random variable! (People make that explicit by noting it with a capital letter.) So you marginalize over it.
Note $X_i$ the binary variable corresponding to "hire person $i$". Then you assume that the skill of person $i$ is $N_i$, a categorical variable that goes from 1 (if the $i$th person is the least skilled) to n (if the $i$th person is the best). To compute the probability to hire the second person out of $n$, you compute:
\begin{aligned}
P(X_2=1) &= \sum_i^n P(X_2=1|N_1=i)P(N_1=i) \\
         &= \frac{1}{n} (P(X_2=1|N_1=1) + \cdots + P(X_2=1|N_1=n)) \\ 
         &= \frac{1}{n} \frac{n (n-1)}{2}\frac{1}{n-1} = \frac{1}{2}
\end{aligned}
(where $P(X_2=1|N_1=i)=\frac{n-i}{n-1}$.)
I haven't use your "abstract" reasoning that says that "all of the people have the same potential to be the most qualified candidate". It is not intuitive to me, and maybe the reason you got lost is that you relied on it?
A: 
Why doesn't the probability that $i^{th}$ number is the highest so far depend on what values the previous numbers are? For example,


$$P(2^{\text{nd}} \text{ number highest}) = 1 - n_1 $$ where $n_1$ is the first number we choose ....

$n_1$ is a random variable, so that probability should be related to the expected value of the preceding scores, through the law of total expectation. $$\begin{align}\mathsf P(2^{\text{nd}}\ \text{number highest})&=\mathsf P({n_2>n_1})\\[1ex]&=\mathsf E(\mathsf P(n_2>n_1\mid n_1))\\[1ex]&=\mathsf E(1-n_1) \\[1ex]&= \tfrac 12\end{align}$$
And similarly, you can argue for...
$$\begin{align}\mathsf P(i^{\text{th}}\ \text{number highest})&=\mathsf P({n_i=\max(n_1,\ldots,n_i)})\\[1ex]&=\mathsf E(\mathsf P(n_i>\max(n_1\ldots n_{i-1})\mid n_1,\ldots,n_{i-1}))\\[1ex]&~~\vdots\\[1ex]&=\tfrac 1i\end{align}$$
