Chance of finding the Almond? A christmas tradition in Norway is to make porridge with a skinned almond hidden in it; the finder of the almond receiving a small gift, often sweets.
Today around the christmas table the question arose:
Does the order of serving impact the likelihood of receiving the almond?
Intuitively, it seems as if all participants have an equal chance of getting the almond, but is this really the case?
 A: Suppose you're serving the porridge to 10 people.  If you mix the porridge thoroughly and immediately divide it equally into 10 bowls, each bowl has a 1/10 chance of having the almond.
Now, suppose you mix the porridge thoroughly and scoop out one bowl of porridge, which is immediately eaten.  Then, you mix the porridge again and scoop out a second bowl which is immediately eaten.  If you repeat this process until all 10 bowls are served and eaten, it's natural to wonder if there is any advantage to going first, or last, or in the middle-- but in fact, it doesn't matter; each bowl still has a $\frac{1}{10}$ chance of having the almond.
Here's how to compute the odds:
Bowl 1 has a $\frac{1}{10}$ chance of having the almond.  That's because the porridge was thoroughly mixed, and you removed one tenth of the porridge for the first bowl.  (It's just the same as any bowl of porridge in scenario 1 at the top.)  This also means there is a $\frac{9}{10}$ chance Bowl 1 does not have the almond.
What are the odds for Bowl 2?  If Bowl 1 does not have the almond (a $\frac{9}{10}$ chance), then the almond is in the remaining porridge.  We are removing one ninth of the remaining porridge for Bowl 2, so there will be a $\frac{1}{9}$ chance the almond goes from the remaining porridge to Bowl 2.  We multiply to combine these two probabilities, giving $\frac{9}{10} \cdot \frac{1}{9} = \frac{1}{10}$.
Bowl 3 can only have the almond if it is not in Bowl 1 (a $\frac{9}{10}$ chance) and if it is not scooped from the remaining porridge into Bowl 2 (a $\frac{8}{9}$ chance) and if it is scooped from the remaining porridge into Bowl 3 (a $\frac{1}{8}$ chance, because there are 8 servings remaining at that point).  We multiply to combine these three probabilities, giving $\frac{9}{10} \cdot \frac{8}{9} \cdot \frac{1}{8} = \frac{1}{10}$.
This pattern continues.  For example, for Bowl 7, the probability of getting the almond is $\frac{9}{10} \cdot \frac{8}{9} \cdot \frac{7}{8} \cdot \frac{6}{7} \cdot \frac{5}{6} \cdot \frac{4}{5} \cdot \frac{1}{4} = \frac{1}{10}$.  (This is just combining the odds that the almond is not scooped out into the first 6 bowls, but is scooped out into Bowl 7, when only 4 portions remain.)
The chances are the same, no matter how you dish out the porridge or when it is eaten (assuming the porridge is well-mixed, and that everyone gets the same amount). The only difference is that if you watch everyone else eat their porridge one-by-one, you are spreading out the probability of losing into several stages, but your overall odds of winning remain $\frac{1}{10}$.
