Puzzle time: the runt of random points
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wrote on 10 May 2021, 15:09 last edited by
Choose 10 numbers uniformly at random from the unit interval [0,1].
On average, what is the value of their minimum?
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wrote on 10 May 2021, 15:11 last edited by Horace 5 Oct 2021, 15:12
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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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I suppose the number where there's a 50/50 shot of never randomly choosing below it in 10 tries.
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wrote on 10 May 2021, 15:14 last edited by
Hint:
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Think circle, not line.
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wrote on 10 May 2021, 15:15 last edited by
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.5=(1-n)^10
=0.066967
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wrote on 10 May 2021, 15:32 last edited by
@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.
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wrote on 10 May 2021, 20:08 last edited by
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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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wrote on 10 May 2021, 23:34 last edited by jon-nyc 5 Nov 2021, 01:06
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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wrote on 11 May 2021, 00:10 last edited by
I'm sure that's right.
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wrote on 11 May 2021, 01:30 last edited by
@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.
wrote on 11 May 2021, 02:44 last edited by Horace 5 Nov 2021, 02:45@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.