JGrando_Assign10
John Grando
February 17, 2018
Grando 10 Homework
Smith is in jail and has 1 dollar; he can get out on bail if he has 8 dollars. A guard agrees to make a series of bets with him. If Smith bets A dollars, he wins A dollars with probability .4 and loses A dollars with probability .6. Find the probability that he wins 8 dollars before losing all of his money if
- he bets 1 dollar each time (timid strategy).
Answer:
Smith’s betting chances can be represented in a transition matrix:
(tm_a <- matrix(c(1, 0, 0, 0, 0, 0, 0, 0, 0, 0.6, 0, 0.4, 0, 0, 0, 0, 0, 0, 0, 0.6,
0, 0.4, 0, 0, 0, 0, 0, 0, 0, 0.6, 0, 0.4, 0, 0, 0, 0, 0, 0, 0, 0.6, 0, 0.4, 0,
0, 0, 0, 0, 0, 0, 0.6, 0, 0.4, 0, 0, 0, 0, 0, 0, 0, 0.6, 0, 0.4, 0, 0, 0, 0,
0, 0, 0, 0.6, 0, 0.4, 0, 0, 0, 0, 0, 0, 0, 0, 1), ncol = 9, byrow = TRUE))## [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9]
## [1,] 1.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0
## [2,] 0.6 0.0 0.4 0.0 0.0 0.0 0.0 0.0 0.0
## [3,] 0.0 0.6 0.0 0.4 0.0 0.0 0.0 0.0 0.0
## [4,] 0.0 0.0 0.6 0.0 0.4 0.0 0.0 0.0 0.0
## [5,] 0.0 0.0 0.0 0.6 0.0 0.4 0.0 0.0 0.0
## [6,] 0.0 0.0 0.0 0.0 0.6 0.0 0.4 0.0 0.0
## [7,] 0.0 0.0 0.0 0.0 0.0 0.6 0.0 0.4 0.0
## [8,] 0.0 0.0 0.0 0.0 0.0 0.0 0.6 0.0 0.4
## [9,] 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 1.0From this week’s reading, we found that the probability for each initial state can be found by computing B = NR…
So, R is:
(rm_a <- matrix(c(0.6, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0.4), nrow = 7, byrow = FALSE))## [,1] [,2]
## [1,] 0.6 0.0
## [2,] 0.0 0.0
## [3,] 0.0 0.0
## [4,] 0.0 0.0
## [5,] 0.0 0.0
## [6,] 0.0 0.0
## [7,] 0.0 0.4and \(N = {\left( I - Q \right)}^{-1}\):
qm_a <- matrix(c(0, 0.4, 0, 0, 0, 0, 0, 0.6, 0, 0.4, 0, 0, 0, 0, 0, 0.6, 0, 0.4,
0, 0, 0, 0, 0, 0.6, 0, 0.4, 0, 0, 0, 0, 0, 0.6, 0, 0.4, 0, 0, 0, 0, 0, 0.6, 0,
0.4, 0, 0, 0, 0, 0, 0.6, 0), ncol = 7, byrow = TRUE)
(nm_a <- solve(diag(7) - qm_a))## [,1] [,2] [,3] [,4] [,5] [,6] [,7]
## [1,] 1.6328311 1.054718 0.6693101 0.4123711 0.2410785 0.1268834 0.05075337
## [2,] 1.5820777 2.636796 1.6732752 1.0309278 0.6026963 0.3172086 0.12688343
## [3,] 1.5059477 2.509913 3.1792228 1.9587629 1.1451229 0.6026963 0.24107851
## [4,] 1.3917526 2.319588 2.9381443 3.3505155 1.9587629 1.0309278 0.41237113
## [5,] 1.2204600 2.034100 2.5765266 2.9381443 3.1792228 1.6732752 0.66931007
## [6,] 0.9635210 1.605868 2.0340999 2.3195876 2.5099128 2.6367962 1.05471848
## [7,] 0.5781126 0.963521 1.2204600 1.3917526 1.5059477 1.5820777 1.63283109So, NR, or B, is:
(bm_a <- round(nm_a %*% rm_a, 3))## [,1] [,2]
## [1,] 0.980 0.020
## [2,] 0.949 0.051
## [3,] 0.904 0.096
## [4,] 0.835 0.165
## [5,] 0.732 0.268
## [6,] 0.578 0.422
## [7,] 0.347 0.653So, starting with 1 dollar in this scenario, there is a 2.0% chance of winning 8 dollars before losing all his money.
- he bets, each time, as much as possible but not more than necessary to bring his fortune up to 8 dollars (bold strategy).
Answer:
Again, we can solve this with a transition matrix; however, it will look slightly different this time.
Transition matrix:
(tm_b <- matrix(c(1, 0, 0, 0, 0, 0, 0, 0, 0, 0.6, 0, 0.4, 0, 0, 0, 0, 0, 0, 0.6,
0, 0, 0, 0.4, 0, 0, 0, 0, 0.6, 0, 0, 0, 0, 0, 0.4, 0, 0, 0.6, 0, 0, 0, 0, 0,
0, 0, 0.4, 0, 0, 0.6, 0, 0, 0, 0, 0, 0.4, 0, 0, 0, 0, 0.6, 0, 0, 0, 0.4, 0, 0,
0, 0, 0, 0, 0.6, 0, 0.4, 0, 0, 0, 0, 0, 0, 0, 0, 1), ncol = 9, byrow = TRUE))## [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9]
## [1,] 1.0 0 0.0 0 0.0 0 0.0 0 0.0
## [2,] 0.6 0 0.4 0 0.0 0 0.0 0 0.0
## [3,] 0.6 0 0.0 0 0.4 0 0.0 0 0.0
## [4,] 0.6 0 0.0 0 0.0 0 0.4 0 0.0
## [5,] 0.6 0 0.0 0 0.0 0 0.0 0 0.4
## [6,] 0.0 0 0.6 0 0.0 0 0.0 0 0.4
## [7,] 0.0 0 0.0 0 0.6 0 0.0 0 0.4
## [8,] 0.0 0 0.0 0 0.0 0 0.6 0 0.4
## [9,] 0.0 0 0.0 0 0.0 0 0.0 0 1.0R matrix:
(rm_b <- matrix(c(0.6, 0.6, 0.6, 0.6, 0, 0, 0, 0, 0, 0, 0.4, 0.4, 0.4, 0.4), nrow = 7,
byrow = FALSE))## [,1] [,2]
## [1,] 0.6 0.0
## [2,] 0.6 0.0
## [3,] 0.6 0.0
## [4,] 0.6 0.4
## [5,] 0.0 0.4
## [6,] 0.0 0.4
## [7,] 0.0 0.4N Matrix:
qm_b <- matrix(c(0, 0.4, 0, 0, 0, 0, 0, 0, 0, 0, 0.4, 0, 0, 0, 0, 0, 0, 0, 0, 0.4,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0.6, 0, 0, 0, 0, 0, 0, 0, 0, 0.6, 0, 0, 0, 0, 0, 0,
0, 0, 0.6, 0), ncol = 7, byrow = TRUE)
(nm_b <- solve(diag(7) - qm_b))## [,1] [,2] [,3] [,4] [,5] [,6] [,7]
## [1,] 1 0.4 0 0.16 0 0.0 0
## [2,] 0 1.0 0 0.40 0 0.0 0
## [3,] 0 0.0 1 0.24 0 0.4 0
## [4,] 0 0.0 0 1.00 0 0.0 0
## [5,] 0 0.6 0 0.24 1 0.0 0
## [6,] 0 0.0 0 0.60 0 1.0 0
## [7,] 0 0.0 0 0.36 0 0.6 1B Matrix:
(bm_b <- round(nm_b %*% rm_b, 3))## [,1] [,2]
## [1,] 0.936 0.064
## [2,] 0.840 0.160
## [3,] 0.744 0.256
## [4,] 0.600 0.400
## [5,] 0.504 0.496
## [6,] 0.360 0.640
## [7,] 0.216 0.784So, starting with 1 dollar in this scenario, there is a 6.4% chance of winning 8 dollars before losing all his money.
- Which strategy gives Smith the better chance of getting out of jail?
Answer:
The bolder strategy (b) gives the better chance of getting out of jail for te starting state of $1. Also, below is a graphical representation of the probability of getting out of jail for each state.
library(ggplot2)
ggplot(data = rbind(data = data.frame(type_val = "Conservative", start_val = seq(1,
7), val = bm_a[, 2]), data.frame(type_val = "Bold", start_val = seq(1, 7), val = bm_b[,
2])), aes(x = start_val, y = val, color = type_val)) + geom_line() + scale_x_continuous(breaks = c(1,
2, 3, 4, 5, 6, 7)) + scale_y_continuous(breaks = c(seq(0.1, 0.8, by = 0.1))) +
labs(x = "Starting Amount", y = "Proability of Getting Out of Jail", color = "Strategy") +
ggtitle("Jail Question Plot") + theme(plot.title = element_text(hjust = 0.5)) +
theme(plot.background = element_rect(fill = "lightgrey"), panel.background = element_rect(fill = "white"),
panel.grid.major.x = element_line(color = "lightgrey"), panel.grid.major.y = element_line(color = "lightgrey")) +
theme(axis.text = element_text(size = 12, color = "grey55"), axis.title = element_text(size = 14,
color = "grey55"), title = element_text(size = 14, color = "grey55"))
Here we see that the bold strategy give a better probability of getting out of jail regardless of the starting amount.