Sampling random numbers from a distribution

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SUMMARY

The probability that one random number sampled from a given probability distribution is greater than another, specifically P(x1 > x2), is always 0.5 under the assumption that the probability of both numbers being equal, P(x1 = x2), is negligible. This conclusion arises from the symmetry of the sampling process, where there is no inherent bias in the order of selection. Therefore, regardless of the specific distribution, the relationship holds true as long as P(x1 = x2) = 0.

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TarskiMonster
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Let's say we have a given probability distribution. We sample two random numbers from this distribution, say x1 and x2. What is the probability that x1 > x2? Is it always 0.5? Does it even depend on the distribution? Sorry if it appears trivial. I just can't seem to wrap my mind around this.


Thanks!
 
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Well, you can make the easy argument from symmetry. Since there is nothing special about picking x1 first and x2 second, P(x1>x2) = P(x1<x2). If we assume P(x1=x2) is negligible, clearly P(x1>x2) = P(x1<x2) = 0.5.
 
It will always be 0.5 as long as

<br /> \Pr(X_1 = X_2) = 0<br />
 

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