This came up in a conversation earlier as a number of us are participating in a World Cup sweepstakes.
In order to draw lots for the World Cup round of 16 – we created two buckets – each containing the 8 teams on either side of the competition bracket.
Because there are 8 people involved, each person received 2 teams – 1 team from each bucket.
Now assuming that each team is as likely to reach the final as any other – and the two teams being on either side of the competition – this seems to be two independent probabilities:
So we worked out the probability of any given person making it to the World Cup Final being:
$$P(A~\text{or}~B)=P(A\cup B)=P(A)+P(B)-P(A\cap B)$$
$$P(A\cup B) = \frac{1}{8} + \frac{1}{8} – \left(\frac{1}{8} \times \frac{1}{8}\right)$$
$$P(A\cup B) = 0.25 – 0.015625$$
$$P(A\cup B) = 0.234375$$
After this was calculated, we wanted to understand by how much we changed the probabilities of any given person reaching the final with either of their two team by creating the two buckets instead of just drawing two teams from a bucket of 16.
Looks at this, because there's a chance of the two teams knocking each other out before the final – this seems to be dependent probabilities – but we're kind of unsure how we should approach this.
I guess we would be looking first at the chances that each team is on either side of the competition – and then from there, the chances of the teams meeting before the final.
Does anyone know the best method to calculate this?
Best Answer
Your first calculation of $\frac18+\frac18-\frac18 \times \frac18$ for two buckets looks correct. An alternative approach to get the same answer would be to look at the complement of the probability neither reaches the final $$1-\left(\frac78\right)\left(\frac78\right)=0.234375$$
For a single bucket you use the same sort of idea $$1-\left(\frac{14}{16}\right)\left(\frac{13}{15}\right)\approx 0.2416667$$ or if you prefer your original approach: $\frac18+\frac18-\frac{2 \times 1}{16 \times 15}$ or $\frac1{8 \choose 1}+\frac1{8 \choose 1}-\frac1{16 \choose 2}$