What is the probability of being the last to touch an object passed around a circle of $N+1$ people? We have $n+ 1$ people numbered by $0,1,...,n$ standing in a circle.  Person $0$ has a bag of chips to start passing around.  Every time, the person $k$ who is holding the bag of chips has probability $j$ to pass to person $k+ 1$ and probability $i = 1-j$ to pass to person $n−1$ with $j >0.5$.  The game ends when all but one have held the bag, the one who hasn't held the chips gets to eat them.  
Calculate $P_k$: the probability that the person $k$ will eat the chips
The way I wanted to approach this is the following. I realize that for person $K$ to be the one, both person $k+ 1$ and person $k−1$ must have held the bag. I wanted to condition  on  whether  person $k−1$ will have touched  the bag before person $k+ 1$ or vice versa. This is my first time in probability and I am having trouble approaching the question. I know that we need to A)  compute the probability that the $k-1$ person has held the bag before the $k+1$ person. B) Consider that given $k-1$ held the bag before the $k+1$ person, Calculate is the probability that person K wins  C) Given that person $k + 1$ touches the bag before $k-1$, calculate the probability person k eventually wins. And then use those to calculate the probability of $K$ winning. But I am not sure how to calculate the individual probabilities 
 A: This is a modification of an answer to a similar question where which had $j=i=\frac12$ and all the probabilities in the answer to the original question were $P_k=\frac1n$.  Let's call movement to a position one more positive a step to the right.
So to find the probability of finishing at position $k$, we want to consider probabilities that on a biased random walk, either


*

*Event 1: Starting from $0$, the walk visits $k-1$ before visiting $-(n-k)$, then, starting from $k-1$, it visits $-(n-k)$ (i.e. a distance of $n-1$ to the left) before visiting $k$ (i.e. a distance of $1$ to the right)

*Event 2: Starting from $0$, the walk visits $-(n-k)$ before visiting $k-1$, then, starting from $-(n-k)$, it visits $k-1$ (i.e. a distance of $n-1$ to the right) before visiting $-(n-k+1)$ (i.e. a distance of $1$ to the left)
Using biased versions of the Gambler's Ruin formulae gives the sum of these probabilities as
$$P_k= \frac{1-\left(\frac{i}{j}\right)^{n-k}}{1-\left(\frac{i}{j}\right)^{n-1}}  \frac{1-\left(\frac{j}{i}\right)^{1}}{1-\left(\frac{j}{i}\right)^{n}} +\frac{1-\left(\frac{j}{i}\right)^{k-1}}{1-\left(\frac{j}{i}\right)^{n-1}} \frac{1-\left(\frac{i}{j}\right)^{1}}{1-\left(\frac{i}{j}\right)^{n}}  $$ which can be simplified to $$P_k= \dfrac{j^{k-1}i^{n-k}(j-i)}{j^{n}-i^{n}}$$
and though this might suggest $\frac00$ when $j=i=\frac12$, its limit as $j \to \frac12$ would be $\frac1n$, as in the linked answer.  
To give an example of how simple this turns out to be, suppose $j=\frac34$ then $i=\frac14$, so you are $3$ times as likely to go right as left, and if $n=4$ then 


*

*the probability the final position is $1$ is $P_1= \frac1{40}$

*the probability the final position is $2$ is $P_2= \frac3{40}$

*the probability the final position is $3$ is $P_3= \frac9{40}$

*the probability the final position is $4$ is $P_4= \frac{27}{40}$
with the probability being multiplied by $\frac j i = 3$ each time and $1+3+9+27=40$.  
