A while ago I got this question on my exam, anyone got an idea how to solve this?
Smarties are a chocolate candy that come in k different colors. Suppose that we do not know k.
a. We draw three smarties from a box and observe two different colors. What is the most likely value for k0, given the smarties we have drawn.
Answer: 2
b. We draw a fourth smartie and find a third color. What is now the most likely value for k0, given the smarties we have drawn.
Answer: 5
For part (a), the probability of drawing two different colours in a selection of $3$ is $$\binom{3}{1} k(k-1)(\frac{1}{k})^3 = \frac{3k(k-1)}{k^3}= \frac{3}{k}-\frac{3}{k^2}$$ The maximum value of this probability occurs when $\frac{d}{dk} \bigg(\frac{3}{k}-\frac{3}{k^2}\bigg)=0 \Rightarrow k=2$ and hence the most likely value of $k$ is $2$.
For part (b), the probability of finding a third different colour after drawing $3$ in part (a) is given by $$(\frac{3}{k}-\frac{3}{k^2})(\frac{k-2}{k})=\frac{3}{k}-\frac{9}{k^2}+\frac{6}{k^3}$$ The maximum value of this probability occurs when $k=5$ and hence the most likely value of $k$ is $5$.