Walking a grid

Imagine a person standing at a point "P" on a plane which has four cardinal directions N,S,E,W. They take a sequence of 6 ( 1 unit ) steps in any of these directions with equal probability ( no diagonal movement ). What is the probability they end their journey back a "P"?

@joe-shmo said
Imagine a person standing at a point "P" on a plane which has four cardinal directions N,S,E,W. They take a sequence of 6 ( 1 unit ) steps in any of these directions with equal probability ( no diagonal movement ). What is the probability they end their journey back a "P"?

I don't fully understand the question.
What are the restrictions on his movements?
For example could he not just take 6 steps to the north?

@venda said
I don't fully understand the question.
What are the restrictions on his movements?
For example could he not just take 6 steps to the north?

The restrictions are 6 steps in any cardinal direction.

N,N,N,N,N,N is a perfectly legitimate sequence of steps. It doesn't get them back to "P", but its perfectly legitimate place they could end up after taking 6 steps.

@joe-shmo said
The restrictions are 6 steps in any cardinal direction.

N,N,N,N,N,N is a perfectly legitimate sequence of steps. It doesn't get them back to "P", but its perfectly legitimate place they could end up after taking 6 steps.

So after each step,he has 4 choices?
I'll have to think about it!!

@venda said
So after each step,he has 4 choices?
I'll have to think about it!!

Yeppers!

@joe-shmo said
Imagine a person standing at a point "P" on a plane which has four cardinal directions N,S,E,W. They take a sequence of 6 ( 1 unit ) steps in any of these directions with equal probability ( no diagonal movement ). What is the probability they end their journey back a "P"?

First attempt:-
6 steps.
To get back to starting point
The number of steps north must always be equal to the number of steps south.The number of steps east must always equal the number of steps west.
The number of ways to arrange nnn and sss = 6!/3! 3! =20 similarly for east and west.
The number of ways to arrange nnss ew or eewwns =6!/2! 2! =180
Therefore total ways to end up at start is 180 +20+20 =220

@venda said
First attempt:-
6 steps.
To get back to starting point
The number of steps north must always be equal to the number of steps south.The number of steps east must always equal the number of steps west.
The number of ways to arrange nnn and sss = 6!/3! 3! =20 similarly for east and west.
The number of ways to arrange nnss ew or eewwns =6!/2! 2! =180
Therefore total ways to end up at start is 180 +20+20 =220

Your on the right track. Check the number number of ways to arrange EEWWNS and NNSSEW again and don't forget I'm asking what is the probability of ending the journey at "P".

@joe-shmo said
Your on the right track. Check the number number of ways to arrange EEWWNS and NNSSEW again and don't forget I'm asking what is the probability of ending the journey at "P".

Ok , and I realise the probability will be a fraction and the equation will have to include the ways not to end up at "p"

@venda said
Ok , and I realise the probability will be a fraction and the equation will have to include the ways not to end up at "p"

Thats right...the simplest way:

P( ending at "P" ) = Ways that End at "P"/ All Ways

Like I said...you are very close...you have an addition error and figure out the ALL ways number.

@venda said
Ok , and I realise the probability will be a fraction and the equation will have to include the ways not to end up at "p"

Total possibilities should be easy to calculate. 4^6

In case you are stuck Venda. You have the number of combinations right. Sometimes these silly mistakes is hard to see:

How many ways to arrange NNSSEW?

How many ways to arrange EEWWNS?

Now...check your last summation.

@athousandyoung said
Total possibilities should be easy to calculate. 4^6

It's not as easy as that.
6!/4! = 15 combinations.
This gives you the number of possible ways for starting in each direction where you can't return after taking 6 steps within a 3*3 grid eg if you take 3 steps north there are 5 ways you can't return to "p" (n.n.n.s.e.w + 4 more).For 2 steps north it's 4 ways ,for 1 step north it's 6 ways making 15 ways for each direction of start where you can't return to "p".
I'm sure Mr. Tracks will be along soon to help out! but I'm still working on the problem!
.

@joe-shmo said
In case you are stuck Venda. You have the number of combinations right. Sometimes these silly mistakes is hard to see:

How many ways to arrange NNSSEW?

How many ways to arrange EEWWNS?

Now...check your last summation.

I believe it's 180 +180 +20 +20=400.
Still thinking about how not to return to "p"

@venda said
It's not as easy as that.
6!/4! = 15 combinations.
This gives you the number of possible ways for starting in each direction where you can't return after taking 6 steps within a 3*3 grid eg if you take 3 steps north there are 5 ways you can't return to "p" (n.n.n.s.e.w + 4 more).For 2 steps north it's 4 ways ,for 1 step north it's 6 ways making 15 ways for each directi ...[text shortened]... o "p".
I'm sure Mr. Tracks will be along soon to help out! but I'm still working on the problem!
.

"It's not as easy as that"

Actually venda, it is that easy.

You have a 6 letter string

ALL possible ending points are counted by the multiplication principle of possible directions for each step.

S1,S2,S3,S4,S5,S6

4*4*4*4*4*4 = 4^6 possible strings and hence 4^6 journey ending points.

@venda said
I believe it's 180 +180 +20 +20=400.
Still thinking about how not to return to "p"

400 is correct for the number of ways to get back to "P" in 6 steps!

ATY is correct about ALL possible paths = 4^6

thus;

P = 400/4^6 = 400/4096 = (20/64)^2 = (5/16)^2

I'll give you both partial credit!