TITLE: At time n, randomly choose a natural number ≤n. How long is it until a single number is chosen three times?
QUESTION [10 upvotes]: To clarify, the number ≤n is chosen uniformly at random at each step, and n chooses from the natural numbers beginning with 1.
I wish to determine the expected value of $n$ at which a natural number is chosen three times (for the first time). (I would also ideally like to know how to calculate E(a number being chosen $y$ times))
I calculated $\Pr(\text{hitting a number thrice at } n=x)$ for some low values, but it rapidly becomes a lot to do by hand.
\begin{align*}
\Pr(n=1) &= \Pr(n=2) = 0 \\
\Pr(n=3) &= 1/6 \\
\Pr(n=4) &= 1/6 \\
\Pr(n=5) &= 19/120
\end{align*}
The inspiration for this question refers to the card game Hearthstone and a particular card interaction. See
http://hearthstone.gamepedia.com/Grim_Patron
http://hearthstone.gamepedia.com/Bouncing_Blade
The title of this question and the phrasing used throughout is not how I conceived the question, but a reformulation that mjqxxxx posted.

REPLY [11 votes]: It turns out the numerical findings about $\mathbb{E}[N_3]$ by David E is not a coincidence, it is exact!
$$\mathbb{E}[N_3] = \frac{1}{1-\sin(1)}$$
Let $X_1, X_2, \ldots $ be a sequence of random variables. For each $n$,
we will assume $X_n$ take value from the set $\langle n \rangle \stackrel{def}{=}\{ 1, 2, \ldots, n \}$ with uniform probability. 
After the $n^{th}$ iteration, consider following random variables:

$Y_{m} = \# \{ i \in \langle n \rangle : X_i = m \}$ be the number of times the number $m$ is chosen.
$Z_{m} = \# \{ i \in \langle n \rangle : Y_i = m \}$ be the number of numbers which has been chosen $m$ times.

For those configuration where none of the number has been chosen more than twice,
$Z_{k}$ are not independent from each other. In fact, if we know $Z_0 = p$, we will have
$$Z_1 = n - 2p,\quad Z_2 = p\quad\text{ and }\quad Z_k = 0, \forall k > 2$$
For such a configuration, it is clear at the $(n+1)^{th}$ iteration, the probability that

$Z_0$ remains unchanged is $\frac{p+1}{n+1}$.
$Z_0$ increases by $1$  is $\frac{n-2p}{n+1}$.
some number get picked $3$ times is $\frac{p}{n+1}$.

If we define

$f_{n,p} = \mathbb{P}\left[ Z_0 = p, Z_1 = 2n-p, Z_2 = p, Z_k > 0 \land \forall k > 2 \right]$
$f_n(z)  = \sum_{p=0}^n f_{n,p} z^p$,
$F(z,t)  = 1 + \sum_{n=1}^\infty f_n(z) t^n$

We find $f_{n,p}$ satisfy following recurrence relation:
$$
f_{n+1,p} 
= \frac{p+1}{n+1}f_{n,p} + \frac{n-2(p-1)}{n+1}\begin{cases} f_{n,p-1}, & p > 0\\0, &p = 0\end{cases}\\
\implies 
(n+1) f_{n+1}(z) = 
\left\{ \left(1 + z\frac{\partial}{\partial z}\right) 
+ z\left( n - 2z\frac{\partial}{\partial z}\right) \right\} f_n(z)
$$
Multiply each term by $t^n$ and sum and notice $f_1(z) = 1$, we find:
$$\begin{align}
& \frac{\partial}{\partial t} ( F(t,z) - 1 - t )
= \left\{ 
\left(1 + z\frac{\partial}{\partial z}\right)
+ 
z\left( t\frac{\partial}{\partial t} - 2z\frac{\partial}{\partial z}\right)
\right\} ( F(z,t) - 1)\\
\iff &
\left\{ (1 - zt)\frac{\partial}{\partial t} 
- (1 + z(1-2z)\frac{\partial}{\partial t}\right\} F(z,t) = 0\\
\iff &
\left\{ (1 - zt)\frac{\partial}{\partial t} 
- z(1-2z)\frac{\partial}{\partial t}\right\} \left( \frac{z}{1-2z} F(z,t) \right) = 0
\end{align}
$$
Using method of characteristics, one can show that the general solution of last PDE has the form
$$F(z,t) = \frac{1-2z}{z} \varphi\left(\frac{1-\sqrt{1-2z}}{1+\sqrt{1-2z}}e^{t\sqrt{1-2z}}\right)$$
where $\varphi(\cdots)$ is an arbitrary function. We can determine $\varphi(\cdots)$ by setting to $t = 0$, we have
$$\frac{1-2z}{z} \varphi\left(\frac{1-\sqrt{1-2z}}{1+\sqrt{1-2z}}\right) = F(z,0) = 1$$
After a little bit of algebra, we get
$$F(z,t) = \frac{4u^2 e^{tu}}{((1+u) - (1-u) e^{tu})^2}
\quad\text{ where }\quad
u = \sqrt{1-2z}
$$
Notice for each $n$, $f_n(1) = \sum_{p=0}^n f_{n,p}$ is the probability that none of the numbers has been chosen more than twice. This means
$f_{n-1}(1) - f_n(1)$ is the probability that some "new" number get picked the third times. So the number we want is
$$\mathbb{E}[N_3] = (1-f_1(1)) + \sum_{n=2}^\infty n (f_{n-1}(1) - f_{n}(1))
= 1 + \sum_{n=1}^\infty f_n(1) = F(1,1)$$
Substitute this back to our expression of $F(z,t)$, we get
$$\begin{align}
\mathbb{E}[N_3] 
& = \frac{4i^2 e^i}{((1+i) - (1-i) e^i)^2} = \frac{1}{1-\sin(1)}\\
& \approx 6.307993516443740027513521739824160128971342...
\end{align}
$$
There are other interesting statistics one can gather from $F(z,t)$. For example,
the generating function for the "survival" probability $f_n(1)$ is
$$P_{survival}(t) \stackrel{def}{=} 1 + \sum_{n=0}f_n(1)t^n = F(1,t) = \frac{1}{1-\sin(t)}$$
and that for the probability for "death at $3^{rd}$ strike" at step $n$ is
$$\begin{align}
P_{3^{rd} strike}(t) & \stackrel{def}{=} (1-f_1(1)) t + \sum_{n=2}^\infty (f_{n-1}(1) - f_{n}(1)) t^n\\
& = (t - 1) P_{survival}(t) + 1 = \frac{t-1}{1-\sin(t)} + 1
\end{align}$$
If one throw the last expression to a CAS and ask it to compute the Taylor expansion, one get
$$\begin{align}
P_{3^{rd} strike}(t) = 
& \frac{1}{6}t^3 + \frac{1}{6}t^4 + \frac{19}{120}t^5 + \frac{47}{360}t^6
  +\frac{173}{1680}t^7 + \frac{131}{1680} t^8 +\frac{20903}{362880}t^9\\
& +\frac{75709}{1814400}t^{10} + \frac{1188947}{39916800}t^{11}
  +\frac{2516231}{119750400}t^{12} + \cdots
\end{align}$$
The coefficient for $t^k$ matches the numbers $Pr[N_3 = k]$ listed in another answer.
Update 1
Let's become Don Quixote and challenge the harder problem of computing $\mathbb{E}[N_y]$ for $y > 3$.
Similar to $y = 3$, one can setup a PDE for the generating functions.
However, if one only want $\mathbb{E}[N_y]$, there is no need to solve the PDE completely. One can use the method of characteristics and express $\mathbb{E}[N_y]$
in terms of solutions of some ODE.
The derivation is a mess. I'll spare you all boring details and here is the recipe:

The case for $y = 4$ is straight forward, one can show that
$$\mathbb{E}[N_4] = \frac{\rho^3 + 3\rho + 2}{6}
\;\;\text{ where }\;\;
\rho \;\;\text{ satisfies }\;\;
1 = \int_1^\rho \frac{6ds}{s^3 + 3s + 2}
$$
For general $y$, setup a system of ODE in $y+1$ varibles $\psi, t, z_1, z_2, \ldots, z_{y-1}$:
$$
\frac{d\psi}{d\tau} = z_1 \psi,\quad
\frac{dt}{d\tau} = t z_1-1\quad
\quad\text{ and }\quad
\frac{dz_k}{d\tau} = 
\begin{cases}
-z_k(z_k - z_{k+1}), & k < y-1\\
-z_k^2, & k = y-1
\end{cases}
$$
If one start integrate this system of ODE using initial values
$$( \psi, t, z_1, \ldots, z_{y-1} ) = (1, 1, \ldots, 1 )\quad\text{ at }\quad \tau = 0$$
to the point $\rho$ where $t(\rho) = 0$, then $\psi(\rho)$ will be equal to the
number $\mathbb{E}[N_y]$ we seek.

Following are some numerical results. The label $\verb/R1/$ and $\verb/R2/$ indicate which recipe has been used to compute the result. The number behind 
the label $\verb/R2/$ is the maximum time-step used in numerically integration.
$$\begin{array}{c:l:l}
y & \mathbb{E}[N_y] & \verb/methodology/\\
\hline
2 & 6.30799351644374 & \verb/exact/ = \frac{1}{1-\sin 1}\\
2 & 6.3079930650     & \verb/R2 /(10^{-4})\\
2 & 6.3079935164     & \verb/R2 /(10^{-6})\\
\hline
3 & 13.77250982352477 & \verb/R1/ \\
3 & 13.7725078084     & \verb/R2 /(10^{-4})\\
3 & 13.7725098234     & \verb/R2 /(10^{-6})\\
\hline
4 & 29.1475420469     & \verb/R2 /(10^{-4})\\
4 & 29.1475467696     & \verb/R2 /(10^{-6})\\
\hline
5 & 60.5714357748     & \verb/R2 /(10^{-4})\\
5 & 60.5714459868     & \verb/R2 /(10^{-6})\\
\hline
6 & 124.4243032167    & \verb/R2 /(10^{-4})\\
6 & 124.4243208364    & \verb/R2 /(10^{-6})
\end{array}
$$
As one can see, $\mathbb{E}[N_y]$ seems to approximately double for each increment of $y$.
Update 2
It turns out part of the complicated ODE in Update 1 can be solved explicitly.
For any given $y \ge 3$ and $k = 1,2,\ldots,y-1$, we have
$$z_{k}(s) = \frac{e_{y-k-1}(s)}{e_{y-k}(s)}
\quad\text{ where }\quad
e_m(x) = \sum_{k=0}^{m} \frac{x^k}{k!}$$
The new simplified recipe is
$$\mathbb{E}[N_y] = e_{y-1}(\rho)
\quad\text{ where }\quad 
\rho \quad\text{ is root of the equation }\quad
1 = \int_0^\rho \frac{ds}{e_{y-1}(s)}
$$
Following is some numerical results computed using this new recipe.
All numbers has been truncated to 6-decimals places for easy comparison. 
As one can see from the $3^{rd}$ column, $2^y$ is a reasonable decent 
$1^{st}$ order approximation for $\mathbb{E}[N_y]$ as $y$ grows.
$$\begin{array}{r:r:l}
y & \mathbb{E}[N_y]  & \mathbb{E}[N_y]/2^y\\
\hline
3 &     6.307993 & 0.788499\\
4 &    13.772509 & 0.860781\\
5 &    29.147546 & 0.910860\\
6 &    60.571445 & 0.946428\\
7 &   124.424320 & 0.972065\\
8 &   253.615739 & 0.990686\\
9 &   514.170899 & 1.004240\\
10&  1038.407593 & 1.014069\\
11&  2091.272044 & 1.021128\\
12&  4202.932580 & 1.026106\\
13&  8433.748402 & 1.029510\\
14& 16903.678242 & 1.031718\\
15& 33849.909944 & 1.033017
\end{array}
$$