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31 or more (Posted on 2017-10-21) Difficulty: 4 of 5
There several versions of "1 to 31" game, involving one die and two players (say you and your opponent), both gifted mathematicians.
My choice is the version defining as a final goal being the first to get a score
of 31 or more.

Here we go:
First: You throw the die and note the outcome.
Then: Your opponent makes a quarter turn* and add this outcome to the earlier sum.
Then: You make a quarter turn and add... and so on...
Until: Whoever gets first the score of 31, or more WINS...

Obviously - there are no draws.
*The quarter turn gets you one of the 4 neighboring faces - never the opposite one!

Design the optimal strategy, - for each of the players and evaluate their chances of victory, provided each makes the best possible moves and that the first number
is a perfectly random digit between 1 and 6.

No Solution Yet Submitted by Ady TZIDON    
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Solution computer aided solution | Comment 1 of 4
The possible choices given specific up pips:

top     choices
1       2,3,4,5
2       1,3,4,6
3       1,2,5,6
4       1,2,5,6
5       1,3,4,6
6       2,3,4,5

Due to the symmetry, there are only three sets of four choices.

When presented with 25, there's 2/3 probability of your winning on the roll. When presented with 26 through 30, your roll is sure to be a winner.

When presented with 24, the best you can hope for is to be able to use a 1; that has 2/3 probability and enables you to reduce your opponent's chances to 2/3, rather than 1. That makes your probability of a win (2/3)*(1/3) = 2/9.

Rather than calculate by hand, it makes sense to use a computer:

The following table was produced by computer program and shows two lines for each total coming into your play.

The first line shows the total so far and your conditional best numeric plays depending on the roll of the die: the first if you roll a 1 or a 6, the second if you roll a 2 or a 5 and the third if you roll a 3 or a 4.

The optimal strategy at any point is to use the line beginning with the number as it exists going into your turn. There will be two different numbers to the right (one repeated). You must play one of them (one or both will always be possible). If both are possible, use the one that has the higher probability listed below it.

Under the given optimal plays is shown the probability you'll win if you play the optimal number shown. As the probability is 1/3 that you'll have each of the given options (actually 1/3 for one option, 2/3 for the other, due to the duplication of one of the options) and therefore probabilities, the last number is the overall probability you'll win given that the total was the number shown at the left just before you roll.


30        5        6        6
   1.000000 1.000000 1.000000   1.000000000

29        5        6        6
   1.000000 1.000000 1.000000   1.000000000

28        5        6        6
   1.000000 1.000000 1.000000   1.000000000

27        5        6        6
   1.000000 1.000000 1.000000   1.000000000

26        5        6        6
   1.000000 1.000000 1.000000   1.000000000

25        5        6        6
   0.000000 1.000000 1.000000   0.666666667

24        5        1        1
   0.000000 0.333333 0.333333   0.222222222

23        2        1        1
   0.333333 0.777778 0.777778   0.629629630

22        2        1        2
   0.777778 0.370370 0.777778   0.641975309

21        3        3        2
   0.777778 0.777778 0.370370   0.641975309

20        4        4        2
   0.777778 0.777778 0.358025   0.637860082

19        5        4        5
   0.777778 0.370370 0.777778   0.641975309

18        5        6        6
   0.370370 0.777778 0.777778   0.641975309

17        3        6        6
   0.362140 0.370370 0.370370   0.367626886

16        4        1        1
   0.362140 0.632373 0.632373   0.542295382

15        2        1        2
   0.632373 0.457705 0.632373   0.574150282

14        3        3        2
   0.632373 0.632373 0.457705   0.574150282

13        4        4        2
   0.632373 0.632373 0.425850   0.563531982

12        5        4        5
   0.632373 0.457705 0.632373   0.574150282

11        5        6        6
   0.457705 0.632373 0.632373   0.574150282

10        3        6        6
   0.436468 0.457705 0.457705   0.450625751

 9        4        1        1
   0.436468 0.549374 0.549374   0.511738838

 8        2        1        2
   0.549374 0.488261 0.549374   0.529003220

 7        3        3        2
   0.549374 0.549374 0.488261   0.529003220

 6        4        4        2
   0.549374 0.549374 0.470997   0.523248426

 5        5        4        5
   0.549374 0.488261 0.549374   0.529003220

 4        5        6        6
   0.488261 0.549374 0.549374   0.529003220

 3        3        6        6
   0.476752 0.488261 0.488261   0.484424632

 2        4        1        1
   0.476752 0.515575 0.515575   0.502634103

 1        2        1        2
   0.515575 0.497366 0.515575   0.509505544
   
You'll see the game favors the first person to take a "quarter turn" except when the initial roll to start the game was 3. Based on the equal likelihood of each of the initial values from 1 through 6, the first quarter-turn player's probability of winning when playing optimally is 0.512969858, the average of the values when encountering 1 through 6 on the initial throw.
 
DefDbl A-Z
Dim crlf$, pWin(30, 3), stratChoose(30, 3)


Private Sub Form_Load()
 Form1.Visible = True
 Text1.Text = ""
 crlf = Chr(13) + Chr(10)
 
 For nIn = 30 To 1 Step -1
   For roll = 1 To 3
     maxWin = 0
     For choice = 1 To 6
       If choice <> roll And choice <> 7 - roll Then
         newTot = nIn + choice
         If newTot >= 31 Then
           probWin = 1
         Else
           probWin = 1 - pWin(newTot, 0)
         End If
         If probWin >= maxWin Then
           stratChoose(nIn, roll) = choice: pWin(nIn, roll) = probWin
           maxWin = probWin
         End If
       End If
     Next
     pWin(nIn, 0) = (pWin(nIn, 1) + pWin(nIn, 2) + pWin(nIn, 3)) / 3
   Next
 Next nIn
 
 For nIn = 30 To 1 Step -1
   Text1.Text = Text1.Text & mform(nIn, "#0")
   For i = 1 To 3
     Text1.Text = Text1.Text & mform(stratChoose(nIn, i), " #######0")
   Next
   Text1.Text = Text1.Text & crlf
   Text1.Text = Text1.Text & "  "
   For i = 1 To 3
     Text1.Text = Text1.Text & mform(pWin(nIn, i), " 0.000000")
   Next
   Text1.Text = Text1.Text & mform(pWin(nIn, 0), "   0.000000000") & crlf
   Text1.Text = Text1.Text & crlf
 Next
 
 Text1.Text = Text1.Text & crlf & " done"
  
End Sub

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

Edited on October 21, 2017, 5:03 pm
  Posted by Charlie on 2017-10-21 16:59:55

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