We randomly draw balls one by one without replacement from an urn with $w$ white balls and $b$ black balls until the urn is emptied. We call the event of a consecutive appearance of two balls of distinct colors a color switch. What is the expected number of color switches in this process?
We can model this as a random walk on a lattice. I do not think the probability of a color switch at a pair of given consecutive draws starting at the $i$'th draw is independent of $i$. Am I right?
On
Inspired by Graham Kemp's answer, I come up with a direct answer.
Consider the set of sequences where the $i$'th draw initiates a color switch (2 possibilities). Cutting out $i$'th and $i+1$'st draws, with two possible subsequences WB and BW, from each of these sequences generates a unique sequence of draws from $b-1$ black balls and $w-1$ white balls. Conversely inserting WB and BW between the $i-1$'th and $i$'th draws of each sequence of draws from $b-1$ black balls and $w-1$ white balls generates two distinct sequences of draws from $b$ black balls and $w$ white balls. The $(b,w)$ sequences thus generated are all distinct. The $(b-1,w-1)$ sequences are all equally probable. Thus the total number of $(b,w)$ sequences with a color switch initiating at draw $i$ is $2 {b+w-2\choose b-1}$. The total number of $(b,w)$ sequences is $b+w\choose b$. The probability is then $2\frac b{b+w}\frac w{b+w-1}$.
On
Here is a more direct solution than my first one. The probability of getting a black ball on any draw is $\frac b{b+w}$ because by labeling the balls there is an equal probability of a specifically labeled ball as any other emerging out of any given draw. This can also be seen by using the characteristic function of a labeled ball falling on the $i$'th draw and the linearity of the expectation function. By the same token, given the first of two consecutive draws being a black ball, the immediate second draw is equivalent to have a white ball out of a specific draw for $b-1$ black balls and $w$ white balls ensemble. That conditional probability is then $\frac w{b+w-1}$. There are $2$ sequences to realize the color switch, namely WB and BW. So the probability of a color switch occurring starting on any arbitrary draw is $2\frac b{b+w}\frac w{b+w-1}$.
On
There are $b+w-1$ consecutive pairs.
Number the draws chronologically and for $i=1,\dots,b+w-1$ let $X_i$ take value $1$ if pair $(i,i+1)$ produces a switch. Let $X_i$ take value $0$ otherwise.
Then $X:=\sum_{i=1}^{b+w-1}X_i$ denotes the total number of switches and the $X_i$ have equal distribution.
With linearity of expectations and symmetry we find:$$\mathbb EX=(b+w-1)P(X_1=1)=(b+w-1)\frac{2wb}{(b+w)(b+w-1)}=\frac{2wb}{b+w}$$
Actually this is not more than a work-out of the answer of Graham.
addendum
The balls are placed on the spots $1,2,\dots, b+w$.
For $i\in\{1,2,\dots,b+w-1\}$ let $W_i$ denote the event that a white ball will cover spot $i$ and let $B_i$ denote the event that a black ball will cover spot $i$.
Then: $$P(W_i\cap B_{i+1})=P(W_i)P(B_{i+1}\mid W_i)=\frac{w}{b+w}\frac{b}{b+w-1}$$
To understand this let it land that all original balls have equal probability to become the ball that covers spot $i$, so that for each of them this probability is $\frac1{b+w}$.
Then - since $w$ of these balls are white - the probability that a white ball will do that is $\frac{w}{b+w}$, which is expressed in $P(W_i)=\frac{w}{b+w}$.
Under the condition that this indeed happens there are $b+w-1$ equiprobable candidates left for covering spot $i+1$ and $b$ of them are black so that $P(B_{i+1}\mid W_i)=\frac{b}{b+w-1}$
Similarly we find that: $$P(B_i\cap W_{i+1})=\frac{b}{b+w}\frac{w}{b+w-1}$$
Then: $$P(X_i=1)=P(W_i\cap B_{i+1})+P(B_i\cap W_{i+1})=\frac{2wb}{(b+w)(b+w-1)}$$
This for every $i$.
No, it is identical for all $i$ in $1$ to $b+w-1$.
The probability for a colour switch after draw $i$ (call this event $S_i=1$), is the probability that draw $i$ is one colour and draw $i+1$ is the other. $$\forall i{\in}\{1,..,b{+}w{-}1\}~~, ~~\mathsf P(S_i{=}1)=\dfrac{\ldots}{\ldots~\ldots}$$
That will help you find the expectation.
For the variance, note that events $S_i{=}1, S_j{=}1$ are not going to be independent.$$\mathsf P(S_i{=}1,S_j{=}1)=\begin{cases}\mathsf P(S_i{=}1) &:& i=j, i{\in}\{1,..,b{+}w{-}1\},j{\in}\{1,..,b{+}w{-}1\}\\\ldots &:& \lvert i-j\rvert=1, i{\in}\{1,..,b{+}w{-}1\},j{\in}\{1,..,b{+}w{-}1\}\\\ldots &:&\lvert i-j\rvert>1, i{\in}\{1,..,b{+}w{-}1\},j{\in}\{1,..,b{+}w{-}1\}\\0&:& i{\notin}\{1,..,b{+}w{-}1\} \lor j{\notin}\{1,..,b{+}w{-}1\} \end{cases}$$