Little math question

[CallMeIshmael]

OK, imagine you are playing a game against one person

You are both going to be dealt a number. Your number is is going to be in the range [50,100], and will be any real number, from an identically distributed continous variable.

The opponent will be dealt a number. There is a 30% chance that his number will be taken from [0,50] and a 70% chance from [50, 100]. (again, assume random continous, with a constantly increasing integral over the two ranges).

What are the chances that my number is going to be higher?

This seems easy, 30% + (1/2)70% = 65%.


Now, suppose I am playing against two players. The first has the same distribution as my opponent above, and the second is 60% chance to be taken from [50,100] and 40% from [0,50].

What are the chances that I am dealt a higher number than both?

One way to do it, is: (0.3)*(0.4) + 1/2(0.3)*(0.6) + 1/2(0.7)*(0.4) + 1/3(0.7)*(0.6)

(the first term represents the times when both of their numbers are in the [0,50] range, the second and third are when exactly one opponent has a chance to beat you, and the fourth occurs when both are in the [50,100] range)

But, as you add more players (say, third, fourth and fifth), the above method a breaking it down into discreet categories becomes rather long.

Im wondering if there is a better way to do this.


Specifically, we should be able to break this down into a formula such as:


P(highest number) = prob(beating p1)*prob(beating p2|beaten p1)*prob(beating p3|beaten p1,p2)...


Can anyone come up with a formula for prob(beating pN|beaten p1..N-1)??

thanks