Binomial Distribution for a person walking in straight line

[CourtneyS]

Homework Statement

Can I measure the probability of a person being at a certain end location after n steps using the binomial distribution if,
probability student goes x=x+3 is 0 <= p <0.5 , x=x-1 is 0<= 0.5 p <1.

Homework Equations

x=x+3 is 0 <= p <0.5
x=x-1 is 0<= 0.5 p <1

The Attempt at a Solution

I know that I can do this for x=x+1 0<=p<0.5 and x=x-1 0.5<=p<1
And I know it doesn't work if it's position based, so for example at x=10, the probabilities changed.
But, can I do it if they move a larger distance in one direction than the other such as the example I have posted?
Thanks

 

[Ray Vickson]

Homework Statement

Can I measure the probability of a person being at a certain end location after n steps using the binomial distribution if,
probability student goes x=x+3 is 0 <= p <0.5 , x=x-1 is 0<= 0.5 p <1.

Homework Equations

x=x+3 is 0 <= p <0.5
x=x-1 is 0<= 0.5 p <1

The Attempt at a Solution

I know that I can do this for x=x+1 0<=p<0.5 and x=x-1 0.5<=p<1
And I know it doesn't work if it's position based, so for example at x=10, the probabilities changed.
But, can I do it if they move a larger distance in one direction than the other such as the example I have posted?
Thanks

I think you meant to say that the moves are $x \to x+3$ w.p. $p$ and $x \to x-1$ w.p. $q = 1-p$; here $0 < p < 1/2$. (The case $p=0$ should be excluded, because it is trivial---there is no randomness at all.)

After $n$ steps, if the student takes $k$ steps to the right and $(n-k)$ steps to the left, how far to the right has he/she moved? (Of course, a negative distance to the right is a positive distance to the left.)

I did not understand at all the rest of your post, where you talk about x = 10 and larger distances, etc.

 

[CourtneyS]

Can I say that probability for any given end point x, the probability of ending up there is :
probability = {factorial(n)/(factorial((n+(x+3))/2)*factorial((n-(x-1))/2))}*{p^(1/2(n+(x+3)) * q^(1/2(n+(x-1)))}
Where p = probability of moving x+3 and q = 1-p

 

[Ray Vickson]

Can I say that probability for any given end point x, the probability of ending up there is :
probability = {factorial(n)/(factorial((n+(x+3))/2)*factorial((n-(x-1))/2))}*{p^(1/2(n+(x+3)) * q^(1/2(n+(x-1)))}
Where p = probability of moving x+3 and q = 1-p

That is not what I get. Just solve for $k$ in terms of $x$ and substitute that into the binomial expression, but do it carefully. (Of course, I assume the starting point is $x = 0$).

BTW: it is normal in probability to use shorthand notation for binomial coefficients: instead of
\frac{a!}{b! (a-b)!}
we usually write
\binom{a}{b} \;\;\rm{or} \;\; C(a,b) \;\; \rm{or} \;\; {}_aC_b
Of course, when we actually want to compute the binomial coefficient, we fall back on the original formula in terms of factorials. However, do what makes you most comfortable; I am just offering advice that you can take or leave.