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#1
10-16-2005, 12:16 PM
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Random number between 1 and 100
If I take batches of 9 random numbers between 1 and 100, what will the average of the highest number in each batch be? I'd love to see some calculations so I can do it for 8 etc, it would help me to stop asking silly questions.
Regards Mack 
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#2
10-16-2005, 12:50 PM
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Re: Random number between 1 and 100
Are the 9 numbers unique?
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#3
10-16-2005, 12:53 PM
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Re: Random number between 1 and 100
[ QUOTE ]
If I take batches of 9 random numbers between 1 and 100, what will the average of the highest number in each batch be? I'd love to see some calculations so I can do it for 8 etc, it would help me to stop asking silly questions. [/ QUOTE ] Let X100 be 1 if some number of the 9 is 100, 0 otherwise. Let X99 be 1 if some number of the 9 is 99+, 0 otherwise. Let X98 be 1 if some number of the 9 is 98+, 0 otherwise. ... The maximum of the 9 is X1+X2+X3+...+X100. By the linearity of expected value (regardless of dependence), E(X1+X2+X3+...+X100)=E(X1)+E(X2)+E(X3)+...+E(X100) . E(Xn) = Probability at least one number is in the range n to 100 = 1- Probability all 9 numbers are from 1 to n-1 = 1-((n-1)/100)^9. So, the expected value of the maximum is 1+(1-1/100^9)+(1-(2/100)^9)+...+(1-(99/100)^9) ~ 90.4925 For batches of 8 numbers, the average maximum is about 89.3822. For batches of 100 numbers, the average maximum is about 99.4279 ~ 100-(1/e + 1/e^2 + 1/e^3 + ...) = 100-1/(e-1). For batches of n*100 numbers, the average maximum is about 100-1/(e^n-1).
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#4
10-16-2005, 12:58 PM
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Re: Random number between 1 and 100
No the ball gets replaced when it is picked each time, so there is a possibility of the same ball being picked again.
Mack
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#5
10-16-2005, 01:07 PM
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Re: Random number between 1 and 100
My guess was an unspecified number slightly higher than 89, just thank god you don't work in a tall building I built. [img]/images/graemlins/grin.gif[/img]
Thank You Mack
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#6
10-16-2005, 02:09 PM
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Re: Random number between 1 and 100
If you are willing to take the median instead of the average (mean), the problem is much easier. The median is exp[ln(0.5)/9]*100 = 92.59. Substitute 8 for 9 and you get the answer for batches of 8. In this case the mean is lower than the median because it's possible to get a batch with a much lower maximum, but you can't get one higher than 100.
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Linear Mode
