Prove that the probability of drawing a white ball at the $i_{th}$ step is $\dfrac {w}{w+b}$ In an urn, there are $w$ white and $b$ black balls. Balls are drawn from the urn without replacement. Prove that the probability of drawing an urn at the $i_{th}$ step is $\dfrac {w}{w+b}$

Attempt:
I am trying to obtain a formal proof to this statement. Although, Andre Nicolas has given a justification Here by arguing that all possibilities are equally likely, still, I was wondering if a formal proof is possible. I attempted it in this way:

Let $X_i$ be a random variable that takes the value $1$ when a white ball is drawn at the $i_{th}$ step and the value $0$ when a black ball is drawn.
Now, $P(X_i=1) = \sum_{x_k \in \{0,1\}} P(X_1=x_1,\cdots,X_{i-1}=x_{i-1},X_i=1)$
The above statement is certainly true for $i=1$.
Let us assume that the statement is true for $i-1$, i.e.
$P(X_{i-1}=1)=\dfrac{w}{w+b}=\sum_{x_k \in \{0,1\}} P(X_1=x_1,\cdots,X_{i-2}=x_{i-2},X_{i-1}=1)$
Then, $P(X_i=1) =  P(X_1=x_1,\cdots,X_{i-1}=1,X_i=1) + P(X_1=x_1,\cdots,X_{i-1}=0,X_i=1)=$
$ \dfrac {w}{b+w} \cdot  \dfrac {w-1}{b+w-1} + [ 1 -\dfrac {w}{b+w} ] \cdot  \dfrac {w}{b+w-1} =\dfrac {w}{b+w}$

Is the proof correct? Is a proof other than induction also possible? Thanks a lot for reading through.

 A: It is often tricky to say when a proof is wrong when the result is right...  :) 
 However, personally I would not accept this proof.  The induction assumption is:
$$P(X_{i-1} = 1) = {w \over w+b} = \color{red}{\sum P(X_1 = x_1, \dots, X_{i-2} = x_{i-2}, X_{i-1} = 1)}$$
However your induction step uses:
$$P(X_i=1) =  \color{blue}{\sum P(X_1=x_1,\cdots,X_{i-1}=1,}\color{green}{X_i=1)} + \sum P(X_1=x_1,\cdots,X_{i-1}=0,X_i=1)=\dots$$
The blue-green thing is not the same as the red thing, even though the blue part of the blue-green thing is "the same as" the red thing.  So you cannot just use the induction assumption.
I am not saying your equations are wrong - indeed every equal sign you used is correct, in the sense that they do link things which are equal.  I am just saying you are not using induction correctly.
Indeed, if I may guess your intention, towards the end you are equating
$$\color{blue}{\sum P(X_1=x_1,\cdots,X_{i-1}=1,}\color{green}{X_i=1)} = \color{blue}{w \over w+b} \cdot \color{green}{w-1 \over b+w-1}$$
where the green part is meant to be understood as $\color{green}{P(X_i = 1 \mid X_{i-1} = 1)}$.  But it is hard to justify this using induction or any other logic you've explicitly mentioned.
