Probability in random selection - Exposing corruption in a Sweepstake We are running a sweepstake for the football world cup and the first three people to choose "drew" the past three winners while in a closed room.
Just for fun, I'd like to show the probability of this happening by chance?
As an extension, what is the probability that each of these 3 got the country that they got.
 A: You can't reasonably treat an even specified post hoc as if it were an event specified before the observation. 
That is, the usual probability computation assumes the event was specified before the draw ... yet I imagine such a specific question ("what's the chance these three people will draw the previous three winners?") wasn't even in your mind at that point. 
If the data that generates the hypothesis is also used to evaluate its surprise value (calculated in the usual way), the results are useless. 
It's like tipping a large bucket-full of (distinguishable) dice on the floor and saying "Oh, look, die #0000 got a '1', and die #0001 got a '6', and die #0002 got a '2', and ... , and die #9999 got a '5' ... wow, what are the chances of that?". The correct probability is actually 1, but by treating the event as if it had been specified before the observation, you'll compute it as $(1/6)^{10000}$.
And so an utterly unextraordinary outcome (the dice all landed on some set of faces), is treated as astoundingly unlikely, because we make the mistake of not recognizing our hypothesis is based on the data.
A: You have an urn with $32$ white balls in which are written the names of the teams participating in the 2014 World Cup. Suppose that you painted with red ink the balls corresponding to the last $3$ champions: Spain, Italy, and Brazil. If each ball has the same probability of being selected, the probability of drawing $3$ balls without replacement and getting $k$ red ones (champions) is hypergeometric. The probabilities for $k=0,1,2,3$ are
dhyper(0:3, 3, 29, 3)
[1] 0.7366935484 0.2455645161 0.0175403226 0.0002016129

Suppose that you want to consider just two possibilities: under model $M_1$, the draw is fair as described above, that is 
$$
  \mathrm{Pr}(K=k\mid M_1)= \frac{{3\choose k}{32-3\choose3-k}}{{32\choose 3}} \;I_{\{0,1,2,3\}}(k) \, .
$$
Under model $M_2$, the selection is "fixed", that is
$$
  \mathrm{Pr}(K=k\mid M_2) = I_{\{3\}}(k) \,.
$$
If you assign prior probabilites $\mathrm{Pr}(M_1)$ and $\mathrm{Pr}(M_2)$ to the models -- and you don't have to necessarily make $\mathrm{Pr}(M_1)=\mathrm{Pr}(M_2)=1/2$ -- then, after observing $K=3$, Bayes's Theorem quantifies your a posteriori belief in model $M_1$ relatively to model $M_2$:
$$
  \mathrm{Pr}(M_1\mid K=3)=\frac{\mathrm{Pr}(K=3\mid M_1) \mathrm{Pr}(M_1)}{\mathrm{Pr}(K=3\mid M_1) \mathrm{Pr}(M_1) + \mathrm{Pr}(K=3\mid M_2) \mathrm{Pr}(M_2)} \, .
$$
When you make $\mathrm{Pr}(M_2)$ small you're giving the benefit of doubt to your three friends, and vice-versa. For example, if a priori you don't believe very much that your friends are cheaters, and you assign $\mathrm{Pr}(M_1)=99\%$ and $\mathrm{Pr}(M_2)=1\%$, after observing $K=3$ you will believe that $\mathrm{Pr}(M_1\mid K=3) \approx 1.95\%$, which is pretty small.
