A very sleep kitten walks towards the edge of a ball pit. One step forward and they will fall in. Exactly one-third of their steps are towards the edge, while the other two-thirds lead them back to safety. What is the probability that they don't fall in?
I looked at this problem, but it didn't seem to be asking the same thing: For this probability question, should I consider him stepping back and then forward again?
I've been looking at some cases (do they step forward or back first), but am still stuck.
Edit: This problem seems to be a duplicate, so I'll post the link to the original: Random Walk of a drunk man
Any and all help is appreciated!
What you say implies that doom is $1$ step away, and safety $2$ steps away, and we want to compute the probability of reaching safety, ie she is saved !
Without going into characteristic equations, etc, can do this short problem by examining step by step
There are $4$ points where you could be, $A-B-C-D$ with $D$ signifying doom, $A$ safety and present position $C$
$P(D) = 0$ and $P(A) =1$ are the "boundary conditions", as doom must be avoided and safety ensured
From C, since we don't want doom, with one step the equation is
$c = 0.5b + 0.5*0$ [we don't want doom, and the Pr of getting to $b$ is 0.5]
From B, the equation is $b= 0.5*1 + 0.5c$ [We sure want safety, but may slide back to $c$ with $Pr = 0.5 $]
Solving the equations, we get $c = 1/3$, ie the probability of reaching safety $=1/3$
We might also intuitionally say that as there is $1$ step to doom and $2$ to safety, doom is twice as probable as safety