% Exercise with Markov chains. Determine probability
% of winning a simplified game of baseball with different
% batting orders.
%---------------------------------------------------------
%states: runs, outs, onbase
% 1 0,0,0
% 2 0,0,1
% 3 0,1,0
% 4 0,1,1
% 5 0,2
% 6 1,0,0
% 7 1,0,1
% 8 1,1,0
% 9 1,1,1
% 10 1,2
% 11 2
%---------------------------------------------------------
% Transition matrix for A:
% A makes an out with probability .6, a base hit with
% probability .3, and a homerun with probability .1.
PA=zeros(11,11);
PA(6,1) = .1; PA(2,1) = .3; PA(3,1) = .6;
PA(11,2) = .1; PA(7,2) = .3; PA(4,2) = .6;
PA(8,3) = .1; PA(4,3) = .3; PA(5,3) = .6;
PA(11,4) = .1; PA(9,4) = .3; PA(5,4) = .6;
PA(5,5) = 1;
PA(11,6) = .1; PA(7,6) = .3; PA(8,6) = .6;
PA(11,7) = .4; PA(9,7) = .6;
PA(11,8) = .1; PA(9,8) = .3; PA(10,8) = .6;
PA(11,9) = .4; PA(10,9) = .6;
PA(10,10) = 1;
PA(11,11) = 1;
% Transition matrix for B:
% B makes an out with probability .7, a base hit with
% probability .1, and a homerun with probability .2.
PB=zeros(11,11);
PB(6,1) = .2; PB(2,1) = .1; PB(3,1) = .7;
PB(11,2) = .2; PB(7,2) = .1; PB(4,2) = .7;
PB(8,3) = .2; PB(4,3) = .1; PB(5,3) = .7;
PB(11,4) = .2; PB(9,4) = .1; PB(5,4) = .7;
PB(5,5) = 1;
PB(11,6) = .2; PB(7,6) = .1; PB(8,6) = .7;
PB(11,7) = .3; PB(9,7) = .7;
PB(11,8) = .2; PB(9,8) = .1; PB(10,8) = .7;
PB(11,9) = .3; PB(10,9) = .7;
PB(10,10) = 1;
PB(11,11) = 1;
% Initial state vector: no runs, no outs, no one on base.
x0 = zeros(11,1); x0(1) = 1;
% By analysis, each batter hits at most twice before the
% game is over.
BABAx0 = PB*PA*PB*PA*x0,
ABABx0 = PA*PB*PA*PB*x0,
fprintf('Probability of winning if A hits first = %f\n',BABAx0(11))
fprintf('Probability of winning if B hits first = %f\n',ABABx0(11))