Probability doubt

[Noel]

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 with player 3,and so on.Let X denote the number of times player 1 is winner.Find
P{X=i}, i=0,1,2,3,4.

[Amit Goyal]

Pr(X = 0) = (1/2)
Pr(X = 1) = (1/2)(1/3) = (1/6)
Pr(X = 2) = (1/2)(2/3)(1/4) = (1/12)
Pr(X = 3) = (1/2)(2/3)(3/4)(1/5) = (1/20)
Pr(X = 4) = (1/2)(2/3)(3/4)(4/5) = (1/5)

[Noel]

Thank you very much sir.could you please provide an explanation?

[Amit Goyal]

Sure. Let me give you another way of doing it which is easier to explain.
Let Y(j) be the number received by player j.
Pr(X = 0)
= Pr(Y(1) < Y(2))
= 1/2

Pr(X = 1)
= Pr(Y(3) > Y(1) > Y(2))
= 1/6

Pr(X = 2)
= Pr(Y(4) > Y(1) > max{Y(2), Y(3)})
= Pr(Y(4) > Y(1) > Y(2) > Y(3)) + Pr(Y(4) > Y(1) > Y(3) > Y(2))
= 1/24 + 1/24 = 1/12

Pr(X = 3)
= Pr(Y(5) > Y(1) > max{Y(2), Y(3), Y(4)})
= Pr(Y(5) = 5, Y(1) = 4)
= 3!/5!
= 1/20

Pr(X = 4)
= Pr(Y(1) > max{Y(2), Y(3), Y(4), Y(5)})
= Pr(Y(1) = 5)
= 4!/5!
= 1/5

[Noel]

Thank you very much sir.