The book asks the question:
A bag contains $a$ white and $b$ black balls. Balls
are chosen from the bag according to the following
method:
A ball is chosen at random and is discarded.
A second ball is then chosen. If its color is
different from that of the preceding ball, it is
replaced in the bag and the process is repeated
from the beginning. If its color is the same, it is
discarded and we start from step 2.In other words, balls are sampled and discarded until a change in
color occurs, at which point the last ball is returned to the urn and
the process starts anew. Let $P_{a,b}$ denote the probability that the last
ball in the bag is white. Prove that $P_{a,b}=0.5$Hint: Use induction on $k=a+b$.
I tried doing this and got :
Denote the probability that in step 2 the 2 balls were in the same color by $p$, then with probability $p$ we are returning to step 1 with $k-2$ balls and with probability $1-p$ we are returning to step 1 with $k-1$ . From the induction hypothises we get that:
$P_{a,b}=0.5p + 0.5(1-p)=0.5$.
Did I do this right ?
Best Answer
No. Hint: Try using any other number instead of $0.5$; the proof will "work" just the same.
[Edit:]
Since the problem is incompletely stated and there was some confusion in the comments due to your apparent misreading of step $2$, here's a complete solution.
Note that step $2$ doesn't return to step $1$ in both cases, as you seem to have assumed; it gets repeated until a different colour is drawn.
We need to add the assumption that $a$ and $b$ are positive; if one of them is zero the probability clearly cannot be $1/2$.
The base case is $k=2$, i.e. $a=b=1$. In this case the probability is $1/2$ by symmetry.
Now assume that the claim holds for all $k'\lt k$. (This is called complete induction or strong induction, as opposed to the form of induction where the claim is only assumed for one value of $k$.) Three different things can happen by the time we return to step $1$. We can exhaust the white balls, we can exhaust the black balls, or we can get a colour change and return the last ball. In the last case, the probability is $1/2$ by the induction hypothesis. Thus, the overall probability is $1/2$ if the two other cases are equally likely. The probability of exhausting the white balls is $1$ in $\binom{a+b}a$, and the probability of exhausting the black balls is $1$ in $\binom{a+b}b$; since these two binomial coefficients are equal, the probabilities of exhausting either colour are indeed the same.