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 Random^Random (Posted on 2014-10-29)
Let A and B each be random real numbers chosen from the uniform interval (0,1).

Call Z the tenths place digit of AB.

Find the probability distribution of Z.

 See The Solution Submitted by Jer Rating: 5.0000 (1 votes)

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 re: towards an analytic solution - numeric integration | Comment 3 of 9 |
(In reply to towards an analytic solution by Charlie)

Actually that would be 1 -  Integral{0 to 1} ln(.9)/ln(A) dA.

It comes out to 0.2870992757... (this is for digit 9) per the .9 in the formula.

via:

DefDbl A-Z
Dim crlf\$
Function mform\$(x, t\$)
a\$ = Format\$(x, t\$)
If Len(a\$) < Len(t\$) Then a\$ = Space\$(Len(t\$) - Len(a\$)) & a\$
mform\$ = a\$
End Function

ChDir "C:\Program Files (x86)\DevStudio\VB\projects\flooble"
Text1.Text = ""
crlf\$ = Chr(13) + Chr(10)
Form1.Visible = True
DoEvents

lp9 = Log(0.9)
n = 100000
For x = 1 / (2 * n) To 1 - 1 / (2 * n) Step 1 / n
tot = tot + Exp(lp9 / x)
Next
DoEvents

Text1.Text = Text1.Text & crlf
Text1.Text = Text1.Text & crlf
n = 100000
tot = 0
For x = 1 / (2 * n) To 1 - 1 / (2 * n) Step 1 / n
tot = tot + Exp(lp9 / x)
Next
Text1.Text = Text1.Text & x & " " & 1 - tot / n

Text1.Text = Text1.Text & crlf
Text1.Text = Text1.Text & crlf
n = 10000000
tot = 0
For x = 1 / (2 * n) To 1 - 1 / (2 * n) Step 1 / n
tot = tot + Exp(lp9 / x)
Next
Text1.Text = Text1.Text & x & " " & 1 - tot / n

Text1.Text = Text1.Text & crlf & "done"
End Sub

showing

0.28709927571142

0.287099275716361

as approximations using increments of 1/100,000 and 1/10,000,000 respectively.

 Posted by Charlie on 2014-10-29 15:32:44

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