Five distinct numbers are randomly distributed to players numbered 1 through 5. Whenever two players compare their numbers, the one with the higher on

Result:
$P(X=0)=\frac{1}{2}$
$P(X=1)=\frac{1}{2}$
$P(X=2)=0$
$P(X=3)=\frac{1}{12}$
$P(X=4)=\frac{1}{16}$
Solution:
1. If $a$ is assigned to player 1 and $b$ is assigned to player 2, player 1 wins the first comparison with probability 1. In this case, player 1 cannot win any more comparisons, so $X=1$.
2. If $a$ is assigned to player 2 and $b$ is assigned to player 1, player 1 loses the first comparison with probability 0. In this case, player 1 still has a chance to win against player 3, 4, and 5. So, $X=0$.
3. If $a$ is assigned to player 1 and $c$ is assigned to player 2, player 1 wins the first comparison with probability 1. In the second comparison, player 1 has a chance to win against player 3, but loses with probability $\frac{1}{2}$ since $c$ can be either assigned to player 3 or player 4. In this case, player 1 cannot win any more comparisons, so $X=1$.
4. If $a$ is assigned to player 2 and $c$ is assigned to player 1, player 1 loses the first comparison with probability 0. In the second comparison, player 1 wins against player 3 with probability $\frac{1}{2}$. In this case, player 1 still has a chance to win against player 4 and player 5. So, $X=1$.
5. If $a$ is assigned to player 1 and $d$ is assigned to player 2, player 1 wins the first comparison with probability 1. In the second comparison, player 1 has a chance to win against player 3, but loses with probability $\frac{2}{3}$ since $d$ can be assigned to either player 3, 4, or 5. In this case, player 1 cannot win any more comparisons, so $X=1$.
6. If $a$ is assigned to player 2 and $d$ is assigned to player 1, player 1 loses the first comparison with probability 0. In the second comparison, player 1 wins against player 3 with probability $\frac{2}{3}$. In this case, player 1 still has a chance to win against player 4 and player 5. So, $X=1$.
7. If $a$ is assigned to player 1 and $e$ is assigned to player 2, player 1 wins the first comparison with probability 1. In the second comparison, player 1 has a chance to win against player 3, but loses with probability $\frac{3}{4}$ since $e$ can be assigned to any of the remaining players. In this case, player 1 cannot win any more comparisons, so $X=1$.
8. If $a$ is assigned to player 2 and $e$ is assigned to player 1, player 1 loses the first comparison with probability 0. In the second comparison, player 1 wins against player 3 with probability $\frac{3}{4}$. In this case, player 1 still has a chance to win against player 4 and player 5. So, $<br>X=1$.
From the above cases, it is clear that $P(X=0)=\frac{1}{2}$ and $P(X=1)=\frac{1}{2}$.
For $X=2$, there are no possible cases where player 1 wins twice, so $P(X=2)=0$.
For $X=3$, player 1 needs to win against player 3, 4, and 5. There is only one possible case where player 1 wins three times: if $a$ is assigned to player 1, $b$ to player 2, $c$ to player 3, $d$ to player 4, and $e$ to player 5. So, $P(X=3)=\frac{1}{2}\times \frac{1}{2}\times \frac{1}{3}=\frac{1}{12}$.
For $X=4$, player 1 needs to win against all the remaining players (2, 3, 4, and 5). There is only one possible case where player 1 wins four times: if $a$ is assigned to player 1 and the rest of the numbers are assigned to players 2, 3, 4, and 5 in decreasing order. So, $P(X=4)=\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}=\frac{1}{16}$.

We need to find the probability of P1 being a winner a certain number of times, denoted by X.
For X = 0:
P1 can only win if P2 has the highest number among players 1 and 2. Since the numbers are distinct and randomly distributed, the probability of P2 having the highest number is $\frac{1}{2}$. Thus, $P(X=0)=\frac{1}{2}$.
For X = 1:
P1 can win in two scenarios:
1. P2 has the highest number among players 1 and 2, and P3 has the highest number among players 1, 2, and 3. The probability of this scenario is $\frac{1}{2}\times \frac{1}{3}=\frac{1}{6}$.
2. P1 has the highest number among players 1, 2, and 3, and P4 has the highest number among players 1, 2, 3, and 4. The probability of this scenario is $\frac{1}{2}\times \frac{1}{3}=\frac{1}{6}$.
Thus, $P(X=1)=\frac{1}{6}+\frac{1}{6}=\frac{1}{3}$.
For X = 2:
P1 can win in three scenarios:
1. P2 wins against P1, and P3 wins against P2. Then, P1 wins against P3, and P4 wins against P1. The probability of this scenario is $\frac{1}{2}\times \frac{1}{3}\times \frac{1}{2}\times \frac{1}{4}=\frac{1}{48}$.
2. P2 wins against P1, and P3 wins against P2. Then, P1 wins against P3, and P5 wins against P1. The probability of this scenario is $\frac{1}{2}\times \frac{1}{3}\times \frac{1}{2}\times \frac{1}{4}=\frac{1}{48}$.
3. P1 wins against P2, and P4 wins against P1. Then, P2 wins against P4, and P5 wins against P2. The probability of this scenario is $\frac{1}{2}\times \frac{1}{3}\times \frac{1}{2}\times \frac{1}{4}=\frac{1}{48}$.
Thus, $P(X=2)=\frac{1}{48}+\frac{1}{48}+\frac{1}{48}=\frac{1}{16}$.
For X = 3:
P1 can win in two scenarios:
1. P1 wins against P2, P3, and P4. Then, P2 wins against P1, and P5 wins against P2. The probability of this scenario is $\frac{1}{2}\times \frac{1}{3}\times \frac{1}{4}\times \frac{1}{2}\times \frac{1}{5}=\frac{1}{240}$.
2. P1 wins against P2, P4, and P5. Then, P2 wins against P1, and
P3 wins against P2. The probability of this scenario is $\frac{1}{2}\times \frac{1}{3}\times \frac{1}{4}\times \frac{1}{2}\times \frac{1}{5}=\frac{1}{240}$.
Thus, $P(X=3)=\frac{1}{240}+\frac{1}{240}=\frac{1}{120}$.
For X = 4:
P1 can win in one scenario:
1. P1 wins against P2, P3, P4, and P5. Then, P2 wins against P1. The probability of this scenario is $\frac{1}{2}\times \frac{1}{3}\times \frac{1}{4}\times \frac{1}{5}\times \frac{1}{2}=\frac{1}{240}$.
Thus, $P(X=4)=\frac{1}{240}$.
Therefore, the probability distribution for X is:
$P(X=0)=\frac{1}{2}$
$P(X=1)=\frac{1}{3}$
$P(X=2)=\frac{1}{16}$
$P(X=3)=\frac{1}{120}$
$P(X=4)=\frac{1}{240}$