Re: Probability doubt

Posted by Amit Goyal on
URL: http://discussion-forum.276.s1.nabble.com/Probability-doubt-tp7586579p7586612.html

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