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Picking numbers (Posted on 2015-07-17) Difficulty: 3 of 5
Bernardo randomly picks 3 distinct numbers from the set (1; 2; 3; 4; 5; 6; 7; 8; 9) and arranges them in descending order to form a 3-digit number.
Silvia randomly picks 3 distinct numbers from the set (1; 2; 3; 4; 5; 6; 7; 8) and also arranges them in descending order to form a 3-digit number.

What is the probability that Bernardo's number is larger than Silvia's number?
Source: AMO

See The Solution Submitted by Ady TZIDON    
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Solution computer solution | Comment 3 of 7 |
Probability Bernardo's number is higher, then Silvia's number, then a tie:

0.660714285714286 .327380952380952 0.011904761904762
3108/4704   1540/4704   56/4704   
37/56   55/168   1/84    done

First row in decimal approximation; second row unreduced fraction; third row reduced fraction.

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


Private Sub Form_Load()
 Form1.Visible = True
 
 Text1.Text = ""
 crlf = Chr$(13) + Chr$(10)

 For a = 9 To 3 Step -1
 For b = a - 1 To 2 Step -1
 For c = b - 1 To 1 Step -1
   For x = 8 To 3 Step -1
   For y = x - 1 To 2 Step -1
   For z = y - 1 To 1 Step -1
      bernardo = 100 * a + 10 * b + c
      silvia = 100 * x + 10 * y + z
      If bernardo > silvia Then bct = bct + 1
      If silvia > bernardo Then sct = sct + 1
      If bernardo = silvia Then eqct = eqct + 1
      ct = ct + 1
   Next
   Next
   Next
Next
Next
Next
 Text1.Text = Text1.Text & bct / ct & Str(sct / ct) & mform(eqct / ct, " 0.000000000000000") & crlf
 Text1.Text = Text1.Text & bct & "/" & ct & "   " & sct & "/" & ct & "   " & eqct & "/" & ct & "   " & crlf
 g1 = gcd(bct, ct): g2 = gcd(sct, ct): g3 = gcd(eqct, ct)
 Text1.Text = Text1.Text & bct / g1 & "/" & ct / g1 & "   " & sct / g2 & "/" & ct / g2 & "   " & eqct / g3 & "/" & ct / g3 & "   " & " done"
 
  
End Sub

Function gcd(a, b)
  x = a: y = b
  Do
   q = Int(x / y)
   z = x - q * y
   x = y: y = z
  Loop Until z = 0
  gcd = x
End Function


  Posted by Charlie on 2015-07-17 15:11:20
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