Distribution of the number of times a player wins a game with random numbers Five distinct numbers are randomly distributed to players numbered
$1$ through $5$. Whenever two players compare their numbers, the one
with higher one is declared the winner. Initially, players $1$ and $2$
compare their numbers; the winner then compares with player $3$,
and so on. Let $X$ denote the number of times player $1$ is winner. Find $$P(X = i), \quad \text{ for }\ i = 0, 1, 2, 3, 4$$
 A: In the following I denote the numbers that are distributed with $1,2,3,4$ and $5$ since the only thing that matters is their order and not their exact values. In total, there are $5!=120$ ways for these numbers to be distributed among the $5$ Players, each of these ways being equally probable. So, $$P(X=i)=\frac{\text{ways in which Player 1 wins exactly $i$ games}}{120}$$ for $i=0,1,2,3,4$. 


*

*$i=4$. This is the most straightforward case, since Player $1$ can win all $4$ comparisons if and only if he has the highest number and the other numbers are distributed randomly among the others, i.e. $$P(X=4)=\frac{1\cdot4!}{120}=\frac{24}{120}$$ 

*$i=3$. For Player $1$ to win exactly $3$ comparisons (and hence lose the $4$-th and last comparison) he must have the second highest number, and the Player with the highest number must be Player $5$. This gives only $1$ choice for the numbers of Players $1$ and $5$ and the other $3$ numbers may be distributed randomly among the $3$ remaining Players, hence $$P(X=3)=\frac{1\cdot3!\cdot1}{120}=\frac{6}{120}$$ 

*$i=2$. I will spare me this one, because it is the most cumbersome. I will skip to $i=1$ and $i=0$ and use the fact that all probabilities must add up to $1$ and return to find this by complementarity.

*$i=1$. For Player $1$ to win exactly $1$ comparison (and hence lose the second comparison) he must have a higher number than Player $2$ and a lower number than Player $3$. This can happen in following ways (notation (no of Player $1$/ no of Player $2$ / no of Player $3$): $(2/1/3-4 \text{ or } 5)=3$ ways, $(3/1 \text{ or }2/ 4 \text{ or }5)=4$ ways, $(4/ 1,2 \text{ or }3/ 5)= 3$ ways. For any of these ways we must multiply with $2!$ for the ways to fill the last two places. So, in total, this gives 
$$P(X=1)=\frac{10\cdot2!}{120}=\frac{20}{120}$$ 

*$i=0$. Player $1$ will win exactly $0$ comparisons if he has a lower number than Player $2$. So, the possible ways for this to happen are: $(1/ \text{ any })= 4$ ways, $(2/3,4 \text{ or } 5)=3$ ways, $(3/ 4 \text{ or }5)=2$ ways and $(4/ 5)=1$ way. For each of these ways we must multiply with $3!$ for the ways to distribute the rest of the numbers to Players $3$ to $5$. So, this gives $$P(X=0)=\frac{10\cdot3!}{120}=\frac{60}{120}$$
So, by complementarity $$P(X=2)=1-P(X\neq 2)=\frac{120-24-6-20-60}{120}=\frac{10}{120}$$

