in reply to Re: Generating random 6 digit numbers
in thread Generating random 6 digit numbers

Whatever you do, don't generate a random number and then try to verify whether you've used it before. That's a recipe for disaster. It will seem ok at first, but once your halfway through your range, you are going to (statistically) have to generate 450,000 random numbers before you find one that you haven't used before! (I just KNOW a mathematician is gonna correct that last bit:)

If you have an urn with 50 blue marbles (representing already seen), and 50 red marbles (not yet seen), do you really think that the average number of draws (with replacement) that you need to make in order to choose a red marble is anywhere near 50? (remember, at this point the probability of drawing a red marble on a given draw is still a relatively high 0.5)
:-)

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Re: Re: Re: Generating random 6 digit numbers
by BrowserUk (Patriarch) on Aug 25, 2002 at 19:07 UTC

    I knew it:) I just KNEW!

    Ok. Worst case scenario, you pick 51 balls before getting a red one. Best case, 1. so that makes the average number 26?

    In the original example, ~225,000 before getting an unused one?

    Update:

    Bah! Having submited this, I suddenly remember drawing dozens and dozens of bell-shaped graphs. I've a sneaky suspicion they've something to do with this!

    I'm gonna stick to nice comfortable metrics like "lots" in future:^p

    What's this about a "crooked mitre"? I'm good at woodwork!
[reply]
      Actually, worst case scenario would be not drawing a red marble before the heat death of the universe, but that would be extraodinarily improbable. Actually, with 50 blue and 50 red marbles, the average number of draws to get a red marble is just 2. In fact, the probability of 10 or more draws without finding a red marble is less than 0.001. When the ratio of blue to red is 90:10, the average number of draws to pull a red is only 10 (and the probability of more than 50 draws without a red is around .005).
[reply]

        Now I remember why I always liked that sage old adage about "damned statistics":)

        What's this about a "crooked mitre"? I'm good at woodwork!
[reply]
      You nailed the reason why the average was wrong by speaking of the bell-shaped graphs, its more of a weighted average. The chance of not getting the ball you want after 1 pull is 1/2. After 2 pulls it is (50/100)*(50/99), the chance of you missing it the first time times the chance of you missing it the second time (99 because one ball is now out of the running). The chance of you pulling 49 balls out and still not getting the one you want is (50/100)*(50/99)*(50/98)*...*(50/51) which is about 2.9*10^(-9). It *could* happen, it just likely will not.

      On a side note, this is my first node that's not a question! I'm so happy I could cry! Would that people would ask more math based questions so I could actually answer them ^_^. ~Adam
[reply]

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