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
computer aided solution
