How do I calculate this expected value? The problem is as follows:
Six players draw, one after another and independently, a number uniformly distributed on $[0,1]$.
A player is called a recordist if he draws a number that is larger than all the numbers drawn by players before him.
What is the expected value of the number of recordists?
Attempted solution:
I define $X_i,\ 1 \leq i \leq 6$, as the indicator of the $i$th player being a recordist.
In other words,
$$X_i = \left\{
        \begin{array}{ll}
            1 &, \text{if the $i$th player is a recordist} \\
            0 &, \text{else}
        \end{array}
    \right.
$$
Now since for all $\,i \neq j, \; \Pr(X_i \gt X_j ) = 0.5\,$, we have
$$\Pr(X_i = k) = \left\{
        \begin{array}{ll}
            \left(\frac{1}{2}\right)^{i-1} &, k=1\\
            \ \ \ \ 0 &, k=0
        \end{array}
    \right.
$$
Next, $X = \sum_{i=1}^{6} X_i$ , is the number of recordists, and so
$$
\mathbb{E} X = \mathbb{E} \left(\sum_{i=1}^{6} X_i\right) = \sum_{i=1}^{6} (\mathbb{E} X_i)
   = \sum_{i=1}^{6} \left(\frac{1}{2}\right)^{i-1}
   = \sum_{i=0}^{5} \left(\frac{1}{2}\right)^{i}
   = \frac{63}{32}
$$
This is not the correct answer and so I turn to you for help.
Thx, Gal.
 A: As in your calculation, let $X_i=1$ if Player $i$ is a recordist, and $0$ otherwise. 
Since we are dealing with a continuous distribution, the probability of a tie is $0$. Since all permutations are equally likely, the probability $i$ is a recordist is $\frac{1}{i}$. Thus the expected number is
$$1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}.$$
A: If I understand your question correctly then the answer is $\frac {1764} {720} = \frac {49} {20}$.
To each person assign a ranking.  The person who draws lowest has ranking 1, the person who draws the largest number has ranking 6.
List the rankings of a drawing in order as an array, $S$, by the order of draw; for example, one $S$ could be $\{3, 4, 1, 6, 5, 2\}$.  The number of recordists is given by the length of a set:
$$|\{i\,  \text{ s.t. } (\forall\, j < i)\, S_j < S_i\}|$$
There are $6!$ possible permutations.  With an epiphany we see that all permutations are equally likely.  There is probably some clever way to divide the $720$ permutations into cases, for example we know that all permutations starting with $6$ have only $1$ recordist, the first drawer.  You can probably recursively define the rest based on that property.
I found it easier to run the $720$ cases through a C program and got $1764$.

Calling the number of players $N$, and given that there are $N!$ permutations which forms the denominator for the probability, the numerator of this probability forms a sequence as a function of $N$ which is apparently something called "Unsigned Stirling numbers of the first kind".  Here is its entry: Online integer Encyclopedia.
They don't show any nice closed form, which means using a computer was probably the only approach for a general $N$.
