Question:
Consider the game of tennis when deuce is reached. If a player wins the next point, he has advantage. On the following point, he either wins the game or the game returns to deuce. Assume that for any point, player A has probability .6 of winning the point and player B has probability .4 of winning the point.
Answer:
The transition matrix for this problem is:
(tm <- matrix(c(1, 0, 0, 0, 0,
0, 1, 0, 0, 0,
0.6, 0, 0, 0.4, 0,
0, 0, 0.6, 0, 0.4,
0, 0.4, 0, 0.6, 0
), ncol = 5, byrow = TRUE))
## [,1] [,2] [,3] [,4] [,5]
## [1,] 1.0 0.0 0.0 0.0 0.0
## [2,] 0.0 1.0 0.0 0.0 0.0
## [3,] 0.6 0.0 0.0 0.4 0.0
## [4,] 0.0 0.0 0.6 0.0 0.4
## [5,] 0.0 0.4 0.0 0.6 0.0
Question:
Answer:
From this week’s reading, we found that the probability for each initial state can be found by computing B = NR…
So, R is:
(rm <- matrix(c(0.6, 0, 0,
0, 0, 0.4
), nrow = 3, byrow = FALSE))
## [,1] [,2]
## [1,] 0.6 0.0
## [2,] 0.0 0.0
## [3,] 0.0 0.4
and \(N = {\left( I - Q \right)}^{-1}\):
qm <- matrix(c(0, 0.4, 0,
0.6, 0, 0.4,
0, 0.6, 0
), ncol = 3, byrow = TRUE)
(nm <- round(solve(diag(3) - qm), 4))
## [,1] [,2] [,3]
## [1,] 1.4615 0.7692 0.3077
## [2,] 1.1538 1.9231 0.7692
## [3,] 0.6923 1.1538 1.4615
So, NR, or B, is:
(bm <- round(nm %*% rm, 3))
## [,1] [,2]
## [1,] 0.877 0.123
## [2,] 0.692 0.308
## [3,] 0.415 0.585
and t is:
(tm <- round(nm %*% matrix(rep(1,3), ncol = 1), 4))
## [,1]
## [1,] 2.5384
## [2,] 3.8461
## [3,] 3.3076
Question:
Answer:
The expected duration at deuce, which is state 4, is 3.8461 games and the probability that B will win is 30.8%.