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02 Mar 21
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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"?
02 Mar 21
@joe-shmo saidI don't fully understand the question.
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"?
What are the restrictions on his movements?
For example could he not just take 6 steps to the north?
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02 Mar 21
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@venda saidThe restrictions are 6 steps in any cardinal direction.
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?
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.
02 Mar 21
@joe-shmo saidSo after each step,he has 4 choices?
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.
I'll have to think about it!!
02 Mar 21
@joe-shmo saidFirst attempt:-
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"?
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
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02 Mar 21
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@venda saidYour 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".
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
02 Mar 21
1 edit
@joe-shmo saidOk , and I realise the probability will be a fraction and the equation will have to include the ways not to end up at "p"
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".
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02 Mar 21
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@venda saidThats right...the simplest way:
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"
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.
03 Mar 21
@athousandyoung saidIt's not as easy as that.
Total possibilities should be easy to calculate. 4^6
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!
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03 Mar 21
@joe-shmo saidI believe it's 180 +180 +20 +20=400.
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.
Still thinking about how not to return to "p"
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@venda said"It's not as easy as that"
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!
.
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.
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03 Mar 21
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@venda said400 is correct for the number of ways to get back to "P" in 6 steps!
I believe it's 180 +180 +20 +20=400.
Still thinking about how not to return to "p"
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!