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# More random numbers

Best Answer plasmid, 22 March 2014 - 06:47 AM

If we continue to consider the question to be asking "what value for the third (or Nth) number from smallest to largest is most likely to occur" then

Spoiler for

If we instead use the definition of "expected value" as the average value after a large number of trials then
Spoiler for

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7 replies to this topic

### #1 bonanova

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Posted 16 March 2014 - 05:42 AM

Inspired by BMAD's puzzle, consider choosing four random numbers on [0,1].

If you sort them lowest to highest, what is the expected value of the third number?

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### #2 santhu221633

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Posted 17 March 2014 - 11:02 AM

.625

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### #3 bonanova

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Posted 17 March 2014 - 01:55 PM

Hi santhu, welcome to the Den. Your answer is close but not quite it.
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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
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### #4 santhu221633

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Posted 19 March 2014 - 11:35 AM

.625

Yeah i gave a crude - well approximate answer.

The answer will be  2*(x/4) + 2*(x/42) +2*(x/43)+ .....  where here x=1, so the answer is 2/3=.666

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### #5 plasmid

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Posted 20 March 2014 - 05:37 AM

I agree with your answer (if the problem is asking "what value is most likely to occur for the third number", which might be different from the expected value if it's defined as "the average outcome if you were to run a large number of trials") but that formula makes me think you took a different approach than I did.

Spoiler for how I did it
How'd you solve it to end up with that formula?

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### #6 bonanova

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Posted 20 March 2014 - 10:59 AM

Are the four expected values equally spaced?

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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell

### #7 plasmid

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Posted 22 March 2014 - 06:47 AM   Best Answer

If we continue to consider the question to be asking "what value for the third (or Nth) number from smallest to largest is most likely to occur" then

Spoiler for

If we instead use the definition of "expected value" as the average value after a large number of trials then
Spoiler for

• 0

### #8 bonanova

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Posted 24 March 2014 - 09:53 PM

It's a result that was interesting to me.

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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell

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