I've been using the formula
$$p(x,N)=\frac{N!}{(\frac{N+x}{2})!(\frac{N-x}{2})!} p^{1/2(N+x)} q^{1/2(N-x)}$$
to determine the probability for a dog who walks in a straight line and can either move $x_{n+1}=x_n+1$ (to right) or x_{n+1}=x_n-1 ( to left) for different values of $p, N,$ and $x$.
$p$ is probability of success (in this case probability of moving to right),
$N$ is total number of steps, and
$x$ is final position after $N$ steps.
How could I change this so that instead of steps to the right being $x_{n+1}=x_n+1$, steps to right is $x_{n+1}=x_n+4$, but steps to left is $x_{n+1}=x_n-1$?
For example, say the probability of $x_{n+1}=x_n+4$ is $p=0.3$, therefore probability of going $x_{n+1}=x_n-1$ is $q=0.7$. How could I change this formula for that case?? (I am assuming I can still use binomial distribution for steps that aren't of equal length but if this isn't the case someone let me know)
Thank you
yes you can still use the binomial distribution! My advise to you is to start with the binomial distribution in its most general form:
$$ p(M,N) = \frac{N!}{M!(N-M)!} p^M q^{(N-M)} $$
where $p$ is the probability of success, $q$ the probability of failure, $N$ the total number of steps and $M$ the number of successes.
Now you can make specific choices. For example that success corresponds to $x = 4$ and failure to $x = -1$. Then the total length traveled equals $4M + (N-M)-1$. You are free to adjust your probabilities $p$ and $q$. Ultimately you can rewrite your equation in terms of $x$ instead of $M$.