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

[gumpzilla]

This is limited to distributions where the only parameter is x, where x is the probability that they are drawing from numbers greater than 50?

It seems unlikely to me that you're going to do much better, as even the conditional probabilities seem like they're going to have to take into account the piecewise nature of your distributions. Using your example, 30% of the time beating player 1 is going to be completely uncorrelated with whether or not I can beat player 2. The calculation, at some level, is going to have to reflect that.

[CallMeIshmael]

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This is limited to distributions where the only parameter is x, where x is the probability that they are drawing from numbers greater than 50?

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right

[ QUOTE ]
It seems unlikely to me that you're going to do much better, as even the conditional probabilities seem like they're going to have to take into account the piecewise nature of your distributions. Using your example, 30% of the time beating player 1 is going to be completely uncorrelated with whether or not I can beat player 2. The calculation, at some level, is going to have to reflect that.

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Yeah.

I spent like two days and I couldnt come up with anything better than the algorithm in the OP. I was hoping some of hte more math savy out there might know some kind of trick, but I fear you are right.