**Polya urn model**: At time $0$ an urn initially contains $b$ $\tt{B}$lue balls and $r$ $\tt{R}$ed balls. At time $1$, a ball is drawn uniformly at random (removing it) from the urn, and *two* balls of the drawn color are added to the urn. This procedure is then repeated at times $2,3,4,\ldots$, each time increasing by one the total number of balls in the urn while randomly changing its composition, as follows: 


$$(x,\ y) \longrightarrow\begin{cases}
(x+1,\ y), & \text{with probability $x\over x+y$} \\[2ex]
(x,\ y+1), & \text{with probability $y\over x+y$}
\end{cases}$$


where $x$ (resp. $y$) denotes the number of blue (resp. red) balls. 


Let $T\_w$ denote the time of the first occurrence of word $w\in\{\tt{B,R}\}^\*$ in the sequence of drawn colors. (E.g., $T\_\epsilon=0$ always, and in the sequence $\tt{RBRBB}\ldots$, $T\_{\tt{B}}=T\_{\tt{RB}}=2,\ T\_{\tt{BB}}=5$, etc.)



> 
> **Question**: Is there a known *closed form* for $E(T\_w)$ in any case other than $T\_{\tt \epsilon}$, $T\_{\tt B}$, or $T\_{\tt R}$? If so, what is it, and how to derive it? 
> 
> 
> 


This was [asked on MathSE](https://math.stackexchange.com/q/2560846/16397) two weeks ago and remains unanswered. As shown there, it is straightforward (via hypergeometric series properties) that for all integers $r\ge 0$ and $b\gt 1$,
$$E(T\_{\tt{B}})\ =\ {\Large\sum}\_\limits{k=0}^\infty{(r)\_k(b)\_1\over (r+b)\_{k+1}}(k+1)\ =\ 1+{r\over b-1},\tag{1}$$
where the Pochhammer symbols are defined as follows: $$(a)\_n = \begin{cases} 1 &\text{ if }n = 0 \\
 a(a+1)(a+2)\cdots (a+n-1) &\text{ if } n > 0.
 \end{cases}$$
However, $E(T\_{\tt{BB}})$, for example, seems much less tractable: 
$$\begin{align}E(T\_{\tt{BB}})&={\Large\sum}\_\limits{j=2}^\infty{\Large\sum}\_\limits{i=j-2}^\infty{(r)\_i(b)\_j\over (r+b)\_{i+j}}\binom{i}{j-2}(i+j)\tag{2}\\[2ex]
&={\Large\sum}\_\limits{i=0}^\infty{\Large\sum}\_\limits{j=0}^i{(r)\_i(b)\_{j+2}\over (r+b)\_{i+j+2}}\binom{i}{j}(i+j+2)\tag{3}\\[2ex]
&\overset{??}{=}\ \left(1+{r\over b-1}\right)\left(2+{r\over b-2}\right)\quad(r\ge 0,\ b\ge 3).\tag{4}
\end{align}$$


Equation (4) is an unproved conjecture resulting from inspecting, for a variety of $(b,r)$ values, simulations of the urn process, as well as numerical evaluations of the double sums (2) and (3). 



Polya urn: Mean number of draws to get a specific sequence of colors?
The crucial thing you need concerns exchageability properties of the Polya urn. 


The following two procedures give the same law: 


1. Generate a sequence of $B$s and $R$s using Polya's urn as you describe, starting from $b$ blue and $r$ red balls;
2. First randomly draw $p$ from a Beta$(b,r)$ distribution. Now, given $p$, generate an i.i.d. sequence of $B$s and $R$s in which each entry is an $B$ with probability $p$ and a $R$ with probability $1-p$.


To learn more about this, search for articles about Polya's urn mentioning "exchangeability" or "de Finetti's theorem". (In the simplest case $b=r=1$, $p$ has Uniform$[0,1]$ distribution.)


Then to get the expected occurrence time of a particular word in the Polya's urn sequence, it's enough to understand the expected occurrence time in an i.i.d. sequence with probability $p$, and then average over $p$ from the Beta$(b,r)$ distribution.


For example, given $p$, the first occurrence time of $B$ is geometric with parameter $p$, and so has mean $1/p$. Then
$$
E(T\_B)=\int\_{p=0}^1 p^{-1} \frac{p^{b-1} (1-p)^{r-1}}{B(b,r)}dp
=\frac{B(b-1,r)}{B(b,r)}=\frac{b+r-1}{b-1}
$$
which agrees with your answer. 


For longer words, expected occurrence times in i.i.d. sequences can be obtained in various ways, for instance via martingales (if I remember right, there is a nice treatment of this in Williams' *Probability with Martingales*), or recursively by conditioning on early entries in the sequence. For example, given $p$, one gets that the expected occurrence time of $BB$ is $1/p + 1/p^2$. Then 
\begin{align\*}
E(T\_{BB})&=
\int\_{p=0}^1 (p^{-1}+p^{-2}) \frac{p^{b-1} (1-p)^{r-1}}{B(b,r)}dp
\\
&=\frac{B(b-1,r)}{B(b,r)}+\frac{B(b-2,r)}{B(b,r)}
\\
&=\frac{b+r-1}{b-1}+\frac{(b+r-1)(b+r-2)}{(b-1)(b-2)}
\end{align\*}
which agrees with your conjecture.


