From mathematician (and author of Basic Probability: What Every Math Student Should Know) Henk Tijms comes a choice matter of chance and chocolate:
I have 10 chocolates in a bag: Two are milk chocolate, while the other eight are dark chocolate. One at a time, I randomly pull chocolates from the bag and eat them — that is, until I pick a chocolate of the other kind. When I get to the other type of chocolate, I put it back in the bag and start drawing again with the remaining chocolates. I keep going until I have eaten all 10 chocolates.
For example, if I first pull out a dark chocolate, I will eat it. (I’ll always eat the first chocolate I pull out.) If I pull out a second dark chocolate, I will eat that as well. If the third one is milk chocolate, I will not eat it (yet), and instead place it back in the bag. Then I will start again, eating the first chocolate I pull out.
What are the chances that the last chocolate I eat is milk chocolate?
Define a function $B(d, m)$ which represents a $bag$ containing $d$ dark chocolates and $m$ milk chocolates. Applying the eating algorithm described, the function returns the probability that the last chocolate is milk chocolate.
For $B(d, m)$, $d > 0$ and $m > 0$, two cases are defined.
The probability of eating all dark chocolates then all white chocolates ($d...dm...m$) is :
$$\frac{d}{d+m} \times \frac{d-1}{d+m-1} \times \frac{d-2}{d+m-2} \times ... \times \frac{1}{m+1} = \frac{d!m!}{(d+m)!}$$
Similarly, the probability of eating all white chocolates then all dark chocolates ($m...md...d$) is :
$$\frac{m}{d+m} \times \frac{m-1}{d+m-1} \times \frac{m-2}{d+m-2} \times ... \times \frac{1}{d+1} = \frac{d!m!}{(d+m)!}$$
These probabilities are equal.
For $B(d, m)$, the probability of choosing $i$, $1 < i < d$, dark chocolates before choosing a milk chocolate is :
$$\frac{d}{d+m} \times \frac{d-1}{d+m-1} \times \frac{d-2}{d+m-2} \times ... \times \frac{d - (i-1)}{d+m-(i-1)} \times \frac{m}{d+m-i}$$
$$ = \frac{\frac{d!}{(d-i)!}}{\frac{(d+m)!}{(d+m-i)!}} \times \frac{m}{d+m-i}$$
For $B(d, m)$, the probability of choosing $j$, $1 < j < m$, milk chocolates before choosing a dark chocolate is :
$$\frac{m}{d+m} \times \frac{m-1}{d+m-1} \times \frac{m-2}{d+m-2} \times ... \times \frac{m - (j-1)}{d+m-(j-1)} \times \frac{d}{d+m-j}$$
$$ = \frac{\frac{m!}{(m-j)!}}{\frac{(d+m)!}{(d+m-j)!}} \times \frac{d}{d+m-j}$$
Therefore, $$B(d, m) = \sum_{i = 1}^{d-1}{\frac{\frac{d!}{(d-i)!}}{\frac{(d+m)!}{(d+m-i)!}} \times \frac{m}{d+m-i} \times B(d-i, m)} + \sum_{j = 1}^{m-1}{\frac{\frac{m!}{(m-j)!}}{\frac{(d+m)!}{(d+m-j)!}} \times \frac{d}{d+m-j} \times B(d, m-j)} + \frac{d!m!}{(d+m)!}$$
Notice that this answer would be exactly the same if $B(d, m)$ were defined as returning the probability that the last chocolate is dark chocolate.
Therefore, for $B(8, 2)$, the answer is : $$\frac{1}{2}$$