Three friends - let\'s call them X, Y, and Z ... Can some please show me, step b
ID: 3271275 • Letter: T
Question
Three friends - let's call them X, Y, and Z ...
Can some please show me, step by step, how to do (a) and (b)?
Three friends _let's call them X, Y, and Z - like to play pool (pocket billiards). There are some pool games that involve three players, but these people like to play 9-ball, which is a game between two players with the property that a tie is not possible (there's always a winner and a loser in any given round). Since all three of these friends can't play 9-ball at the same time, they use a simple rule to decide who plays in the next round: loser sits down. For example, suppose that, in round 1, X and Y play; then if X wins, Y sits down and the next game is between X and Z. Question: in the long run, which two players square off against each other most often? Least often? So far what I've described is completely realistic, but now we need to make a (strong simplifying assumption. In practice people get tired and/or discouraged, so the probability that (say) X beats Y in any single round is probably not constant in time, but let's pretend that it is, to get a kind of baseline analysis: let 0Explanation / Answer
a) time homogenious markov chain means when markov operator doesnot change accross the transition ...here we are assuming that X beats Y at any stage is fixed that is P_xy and the other probability also as it is from differnt transition stage so it is a time homogenious markov chain
b) here 3 state space (XY) (XZ) (YZ)
if we look at the (1,1) th element of the transition matrix that is transition from (XY) to (XY) is 0 because when X and Y are playing that means some one lose and the other one win the game cant end with a draw so it is 0
(1,2 ) is (XY) to (XZ) that y loses and the probability is P_XY
(1,3) is 1-Pxy ( row sum is 1)
(2,1) is (XZ) to (XY) that z loses so prob is Pxz and the si,ilar arguments for other elements ...
so thats why our transition matrix is look like above
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