Question

Five distinct numbers are randomly distributed to players numbered$1$through $5.$ Whenever two players compare their numbers, the one with the higher one is declared the winner. Initially, players$1$and $2$ compare their numbers; the winner then compares her number with that of player $3,$ and so on. Let $X$ denote the number of times player $1$ is a winner. Find$PX=i,i=0,1,2,3,4.$

Short answer

$P(X=0)=\frac{1}{2}$

$P(X=1)=\frac{1}{6}$

$P(X=3)=\frac{1}{20}$

$P(x=2)=\frac{1}{12}$

$P(x=4)=\frac{1}{5}$

Step-by-step solution

Step 1:Given information

Five distinct numbers are randomly distributed to players numbered $1$through $5$. Whenever two players compare their numbers, the one with the higher one is declared the winner. Initially, players $1$and $2$compare their numbers; the winner then compares her number with that of player $3$, and so on. Let $X$denote the number of times player$1$ is a winner.

Step 2Explanation

Observe that the first and the second player have similar probability to obtain any number. Utilizing the principle of that symmetry, we keep that

$P(X=0)=\frac{1}{2}$

Event $X=1$means that the first player has got a greater number than the second player, but not the third player. So, select any three numbers out of five of them and say that the minimal number out of these three proceeds to the second player, the mean number to the first one, and the biggest to the third one. Permute the remaining two numbers on the remaining two people. Hence

$P(X=1)=\frac{\left(\begin{array}{c}5\\ 3\end{array}\right)·2!}{5!}=\frac{1}{6}$

Step 3:Explanation

Event $X=2$means that the first player has got greater number than the second and the third player, but not than the fourth player. So, choose any four numbers out of five of them and say that the minimal number and the next minimal out of these four go to the second and the third player (and permute them), third number to the first one and the largest to the fourth player. Give remaining number to the last person. Hence

$P(X=2)=\frac{\left(\begin{array}{l}5\\ 4\end{array}\right)·2!}{5!}=\frac{1}{12}$

Step 4Explanation

Event $X=3$means that the first player has got greater number than the second, the third, and the forth player, but not than the fifth player. So, permute these five numbers as follows: give the highest to the last person, the second highest to the first, and permute remaining numbers on the remaining people. Hence

$P(X=3)=\frac{3!}{5!}=\frac{1}{20}$

Event $X=4$means basically that the first player has won all the battles, i.e., he has got the greatest number.

Hence

$P(X=4)=\frac{4!}{5!}=\frac{1}{5}$

Final answer

$P(X=0)=\frac{1}{2}$

$P(X=1)=\frac{1}{6}$

$P(X=3)=\frac{1}{20}$

$P(X=2)=\frac{1}{12}$

$P(X=4)=\frac{1}{5}$