Puzzle time: the runt of random points

[Horace]

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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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[jon-nyc]

@horace And that number would be?

[jon-nyc]

Hint:

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Think circle, not line.

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[Horace]

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.5=(1-n)^10

=0.066967

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[Horace]

@jon-nyc said in Puzzle time: the runt of random points:

Hint:

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Think circle, not line.

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Right, my formula would give the median but not the mean. Higher minimums have more room for outliers than lower minimums, which can only go to zero.

[Klaus]

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If you choose 1 number, the average of the mimimum is 0.5.
With 2 numbers, it's 0.333. With 3 it's 0.25.

With n numbers it's 1/(n+1), hence for 10 it's 0.909...
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[jon-nyc]

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]

I'm sure that's right.

[jon-nyc]

@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.

[Horace]

@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.