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Conference rusure::math

Title:Mathematics at DEC
Moderator:RUSURE::EDP
Created:Mon Feb 03 1986
Last Modified:Fri Jun 06 1997
Last Successful Update:Fri Jun 06 1997
Number of topics:2083
Total number of notes:14613

1511.0. "Throwing die" by HGRD01::JOSEPHCHEUNG () Thu Oct 24 1991 20:50

    Hi,

    	There is a probability problem :

    	X and Y play as follows :

    	An unbiased cubical die, having three white faces marked
    respectively 1, 2 and 3, and three black faces with similarly marked,
    is thrown repeatedly, X scores the number of points shown on any white
    faces that appear and Y scores the number on any black faces. The game
    is won by the player whose total score first exceeds his opponent's
    score by 6 or more points. Find X's chance of winning the game after
    the first throw if the face which appeared at that throw as
    (1) a white face marked 1,
    (2) a black face marked 3.
    
    -- Joseph
T.RTitleUserPersonal
Name		DateLines
1511.1BEING::EDPAlways mount a scratch monkey.Fri Oct 25 1991 09:4363
    We can model this game with a system of equations.  Let pn be the
    probability of a player winning if they are n points ahead.  Then p6 is
    1, because the player has won.  If a player is five points ahead, there
    is a 3 in 6 chance they will roll a 1, 2, or 3 in their favor, and one
    chance each of rolling a 1, 2, or 3 against them, so p5 is 3/6 plus
    (p4+p3+p2)/6.  Similarly, we can write each of the other probabilities
    in terms of some constants plus other probabilities.  Also, let qn be
    the probability of a player winning if they are n points behind.  Then
    the equations are:
    
p6 = 1
p5 = 3/6 + (p4+p3+p2)/6
p4 = 2/6 + (p5+p3+p2+p1)/6
p3 = 1/6 + (p5+p4+p2+p1+p0)/6
p2 = (p5+p4+p3+p1+p0+q1)/6
p1 = (p4+p3+p2+p0+q1+q2)/6
p0 = (p3+p2+p1+q1+q2+q3)/6
q1 = (p2+p1+p0+q2+q3+q4)/6
q2 = (p1+p0+q1+q3+q4+q5)/6
q3 = 0/6 + (p0+q1+q2+q4+q5)/6
q4 = 0/6 + (q1+q2+q3+q5)/6
q5 = 0/6 + (q2+q3+q4)/6
q6 = 0
    
    Since p6 and q6 are constants, the remaining equations are a system of
    11 equations with 11 variables.  We could reduce this further.  For
    example, by symmetry, we expect p0 to be 1/2, and we could make quick
    substitutions for q1 through q5 by replacing them with (1-p1) through
    (1-p5), again by symmetry.  But this is the age of computers, so let's
    just stick the equations in an array and tell a computer to figure it
    out.  I multiplied all the equations by six, to make all the
    coefficients integers, and wrote them in the array below, with the
    constant terms moved to a vector on the right.

 p5 p4 p3 p2 p1 p0 q1 q2 q3 q4 q5 =
[-6  1  1  1  0  0  0  0  0  0  0] -3
[ 1 -6  1  1  1  0  0  0  0  0  0] -2
[ 1  1 -6  1  1  1  0  0  0  0  0] -1
[ 1  1  1 -6  1  1  1  0  0  0  0]  0
[ 0  1  1  1 -6  1  1  1  0  0  0]  0
[ 0  0  1  1  1 -6  1  1  1  0  0]  0
[ 0  0  0  1  1  1 -6  1  1  1  0]  0
[ 0  0  0  0  1  1  1 -6  1  1  1]  0
[ 0  0  0  0  0  1  1  1 -6  1  1]  0
[ 0  0  0  0  0  0  1  1  1 -6  1]  0
[ 0  0  0  0  0  0  0  1  1  1 -6]  0
    
    My calculator produces these numbers for the variables:
    
    	p5 = .864442127217
    	p4 = .803962460898
    	p3 = .732533889469
    	p2 = .650156412932
    	p1 = .576642335768 (X's probability of winning after gaining 1)
    	p0 = .500000000002 (which gives us an estimate of numerical error)
    	q1 = .423357664236
    	q2 = .349843587072
    	q3 = .267466110533 (X's probability of winning after losing 3)
    	q4 = .196037539105
    	q5 = .135557872785                                            
    
    
    				-- edp
1511.2BEING::EDPAlways mount a scratch monkey.Fri Oct 25 1991 15:365
    P.S., exact answers are p5 = 829/959, p4 = 771/959, p3 = 1405/1918, p2
    = 1247/1918, p1 = 79/137.
    
    
    				-- edp
1511.3exact solutionCIVAGE::LYNNLynn Yarbrough @WNP DTN 427-5663Fri Oct 25 1991 16:046
Happily, the probabilities are actually rationals and MAPLE cranks them out 
in that form, so precision is not an issue. The vector of solutions is

	829  771  1405  1247   79        58   671   513  188  130
	---, ---, ----, ----, ---, 1/2, ---, ----, ----, ---, ---
	959  959  1918  1918  137       137  1918  1918  959  959