DATA605-A10

Because the timid approach would have me make an 8x8 matrix…Instead We’ll use this formula found in the textbook to hack our specific probability.

M = 8
z = 1
q = .6
p = .4

timid = ((q/p)^z-1) / ((q/p)^M-1)
timid

## [1] 0.02030135

We shall use a transition matrix for this question.

\[ P= \left(\begin{array}{ccccc} & M_1 & M_2 & M_4 & M_0 & M_8\\ z_1& 0 & .4 & 0 & .6 & 0\\ z_2& 0 & 0 & .4 & .6 & 0\\ z_4& 0 & 0 & 0 & .6 & .4\\ z_0& 0 & 0 & 0 & 1 & 0\\ z_8& 0 & 0 & 0 & 0 & 1\\ \end{array}\right) \]

\[ solved = \left(\begin{array}{cc} &M_0&M_8\\ z_1&.936 & .064\\ z_2&.84 & .16 \\ z_4&.6 & .4 \end{array}\right) \]

bold = .064
timid < bold

## [1] TRUE

R has decreed bold to be greater than timid!
In all seriousness, to interpret our findings…
Starting at 1 dollar for each scenario; the bold strategy has a 6% success rate, where as the timid strategy has a 2% success rate. It would be wiser for our prisoner to bet boldly.