Puzzle time: the runt of random points
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I suppose the number where there's a 50/50 shot of never randomly choosing below it in 10 tries.
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Klaus is right.
Well, that was my answer. Official solution comes Saturday.
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One way to think about it is as follows: Pick 11 random points on a unit circumference circle (expected distance between points is 1/11). Randomly pick one as marking start/end point of unit line.
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@Horace My original intuition was similar to yours but with a different formula.
The probabilities would be additive, not multiplicative, no?
So if m is the expected minimum value, that meant that:
10 * p(x<m) = 0.5, or
p(x<m) = 0.05
which gave me 1/20.But as you pointed out that would be a median value, the mean would be higher than that.
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@Horace My original intuition was similar to yours but with a different formula.
The probabilities would be additive, not multiplicative, no?
So if m is the expected minimum value, that meant that:
10 * p(x<m) = 0.5, or
p(x<m) = 0.05
which gave me 1/20.But as you pointed out that would be a median value, the mean would be higher than that.
@jon-nyc said in Puzzle time: the runt of random points:
@Horace My original intuition was similar to yours but with a different formula.
The probabilities would be additive, not multiplicative, no?
So if m is the expected minimum value, that meant that:
10 * p(x<m) = 0.5, or
p(x<m) = 0.05
which gave me 1/20.But as you pointed out that would be a median value, the mean would be higher than that.
My formula gives the correct median. It's the chance of rolling higher than "x" 10 times in a row, so multiplicative.
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