Question

In a knockout tournament ${{2}^{n}}$ equally skilled players; ${{S}_{1}},{{S}_{2}}........{{S}_{n}}$ are participating. In each round players are divided in pairs at random and the winner from each pair moves in the next round. If ${{S}_{2}}$ reaches the semi-finals then find the probability that ${{S}_{1}}$ wins the tournament.

Answer 1

Hint: Focus on the point that ${{S}_{2}}$ has reached the semi-finals and we know that only 4 out of the ${{2}^{n}}$ players can reach the semi-finals and one out of this four wins the tournament. So, the probability of ${{S}_{2}}$ winning is $\dfrac{1}{4}$ . Now the left out chances, i.e., $\dfrac{3}{4}$ is the probability that someone other than ${{S}_{2}}$ wins the tournament and rest all are equally skilled players, so their probability of winning would be same.

Complete step-by-step answer:
Probability can be mathematically defined as $=\dfrac{\text{number of favourable outcomes}}{\text{total number of outcomes}}$ .
Now, let’s move to the solution to the above question.
As it is given that ${{S}_{2}}$ has reached the semi-finals and we know that only 4 out of the ${{2}^{n}}$ players can reach the semi-finals and one out of this four wins the tournament. So, the probability of ${{S}_{2}}$ winning is $\dfrac{1}{4}$ .
Now the left out chances, i.e., $\dfrac{3}{4}$ is the probability that someone other than ${{S}_{2}}$ wins the tournament and rest all are equally skilled players, so their probability of winning would be same. Also, we know that there were a total of ${{2}^{n}}$ players, so if we separate ${{S}_{2}}$ , we can say that the probability of winning of 1 out of the ${{2}^{n}}-1$ players is $\dfrac{3}{4}$ . Now the probability that the one winning out of this ${{2}^{n}}-1$ players is ${{S}_{1}}$ is equal to $\dfrac{3}{4}$ multiplied by the probability of ${{S}_{1}}$ winning among this ${{2}^{n}}-1$, i.e., $\dfrac{3}{4}\times \dfrac{1}{\left( {{2}^{n}}-1 \right)}$ , as there are ${{2}^{n}}-1$ players and all have equal chances to win.
Therefore, the answer to the above question is $\dfrac{3}{4}\times \dfrac{1}{\left( {{2}^{n}}-1 \right)}$ .

Note:The key to the above question is the interpretation of the fact that ${{S}_{2}}$ reaching the semi-finals means that the probability of ${{S}_{2}}$ winning is $\dfrac{3}{4}$ . Also, for verifying your answer you can try taking some small values of n like 2, 3 and manually form the tournament fixtures and check with the arrived result by putting the value of n.