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

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!
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Re: Re: Re: Re: Generating random 6 digit numbers
by Anonymous Monk on Aug 25, 2002 at 19:39 UTC
    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).
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      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!
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        Blame it on the example. I feel certain you would have gotten it had the question begun: If you have an urn with one red ball and one blue ball...

        :-)

        -sauoq
"My two cents aren't worth a dime.";
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Re: Re: Re: Re: Generating random 6 digit numbers
by Adam Kensai (Sexton) on Aug 26, 2002 at 12:46 UTC
    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
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