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Quiz Quandary III (Posted on 2014-07-15) Difficulty: 3 of 5
In a certain island, one-third of the native people are liars who always lie, one-third are knights who always tell the truth, and one-third are knaves that is, people who strictly alternate between speaking the truth and telling a lie, irrespective of order. The chances of encountering any one of the three natives on a road on the island are the same.

Four friends Art, Ben, Cal and Dan Ė who are natives of this island got the top four ranks in a certain quiz The following statements are made by each of Art, Ben, Cal and Dan.

Art
1. Exactly two of us are knights.
2. Ben got the first rank and Cal got the third rank.

Ben
1. Dan got the fourth rank.
2. Exactly one of us is a knave.

Cal
1. The absolute difference between Ben's rank and mine is 2.
2. Dan is not a liar.

Dan
1. Exactly one of us is a liar.
2. I am a knave.

Assuming no ties, determine the probability that:

(i) The absolute difference between Cal's rank and Danís rank is 1.
(ii) Exactly three of the four friends are liars.
(iii) Exactly two of the four friends are knaves.
(iv) At least one of the four friends is a knight.

*** For Artís second statement - assume the entire statement is a lie if the whole statement or any of its parts thereof is false. For example- A2 is false if in reality Ben got the first rank and Cal got the fourth rank.

No Solution Yet Submitted by K Sengupta    
Rating: 3.5000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Some Thoughts computer solution | Comment 7 of 8 |
DefDbl A-Z
Dim crlf$, types$
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()
 ChDir "C:Program Files (x86)DevStudioVBprojects looble"
 Text1.Text = ""
 crlf$ = Chr(13) + Chr(10)
 Form1.Visible = True
 DoEvents
 types$ = "0123" ' liar,second is truth,first is truth, knight
 places$ = "abcd": h$ = places$
 Do
 
   For a = 0 To 3
    a1 = a 2: a2 = a Mod 2
   For b = 0 To 3
    b1 = b 2: b2 = b Mod 2
   For c = 0 To 3
    c1 = c 2: c2 = c Mod 2
   For d = 0 To 3
    d1 = d 2: d2 = d Mod 2
    
    ktct = Abs((a = 3) + (b = 3) + (c = 3) + (d = 3))
    kvct = Abs((a = 1 Or a = 2) + (b = 1 Or b = 2) + (c = 1 Or c = 2) + (d = 1 Or d = 2))
    ctgno = 1
    For i = 1 To kvct
     ctgno = ctgno / 2
    Next
    situations = situations + ctgno
    If Abs(ktct = 2) = a1 And Abs(Mid(places$, 1, 1) = "b" And Mid(places$, 3, 1) = "c") = a2 Then
      If Abs(Mid(places$, 4, 1) = "d") = b1 Then
       If Abs(kvct = 1) = b2 Then
        br = InStr(places$, "b")
        cr = InStr(places$, "c")
        If Abs(Abs(cr - br) = 2) = c1 And Abs(d <> 0) = c2 Then
          lct = Abs((a = 0) + (b = 0) + (c = 0) + (d = 0))
          If Abs(lct = 1) = d1 And Abs(d = 1 Or d = 2) = d2 Then
            Text1.Text = Text1.Text & a & b & c & d & "   " & places$ & crlf
            dr = InStr(places$, "d")
            If Abs(cr - dr) = 1 Then pt1ct = pt1ct + ctgno
            If lct = 3 Then pt2ct = pt2ct + ctgno
            If kvct = 2 Then pt3ct = pt3ct + ctgno
            If ktct > 0 Then pt4ct = pt4ct + ctgno
            DoEvents
          End If
        End If
       End If
      End If
    End If
   Next d
   Next c
   Next b
   Next a
   permute places$
 Loop Until places$ = h$
 Text1.Text = Text1.Text & crlf & mform(pt1ct, "####0") & mform(situations, "#####0") & mform(pt1ct / situations, "#0.000000") & crlf
 Text1.Text = Text1.Text & mform(pt2ct, "####0") & mform(situations, "#####0") & mform(pt2ct / situations, "#0.000000") & crlf
 Text1.Text = Text1.Text & mform(pt3ct, "####0") & mform(situations, "#####0") & mform(pt3ct / situations, "#0.000000") & crlf
 Text1.Text = Text1.Text & mform(pt4ct, "####0") & mform(situations, "#####0") & mform(pt4ct / situations, "#0.000000") & crlf
 Text1.Text = Text1.Text & situations & crlf & "done"
End Sub

finds

ABCD   order
0000   adbc
0011   adbc
0100   adbc
0000   adcb
0011   adcb
0100   adcb
1231   bacd
3231   bacd
0000   badc
0011   badc
0100   badc
0000   bcda
0011   bcda
0100   bcda
0000   bdac
0011   bdac
0100   bdac
1020   bdca
0220   cabd
0320   cabd
0000   cadb
0011   cadb
0100   cadb
0000   cbda
0011   cbda
0100   cbda
0000   cdab
0011   cdab
0100   cdab
0000   dabc
0011   dabc
0100   dabc
0000   dacb
0011   dacb
0100   dacb
0000   dbca
0011   dbca
0100   dbca
0000   dcba
0011   dcba
0100   dcba

9.3750 1944 0.004823 6.0000 1944 0.003086 3.7500 1944 0.001929 0.8750 1944 0.000450
1944

which shows, in the first 41 lines, the possibilities. They are not all equally probable, as:

0 represents liar: truth value for two statements 00.
1 represents knave: truth value for two statements 01, in that order.
2 represents knave: truth value for two statements 10, in that order.
3 represents knightr: truth value for two statements 11.

The two knave situations count for only 1/3 of possibilities, so only half of a unit counts for a situation having one knave; only 1/4 for a situation having two knaves, etc.  These situations add up to the correct 1944, which is 3^4 * 4!, that they would be if the order of the statements didn't matter.

With these weights, the probability for part (i) is 9.375/1944, for part (ii) is 6/1944, for (iii), 3.75/1944 and for part (iv), .875/1944.  Decimal values are shown to the side above.

(Edited to show true values of the counts, rather than rounded to an integer, as the counts include the weighting consideration for knave count).

Edited on July 17, 2014, 12:49 pm
  Posted by Charlie on 2014-07-17 12:33:20

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