Question

a man is standing at point A . he moves one unit in every step he takes . he can move either in horizontal or vertical direction,even backwards or forwards,left or right.. whats is the probability that he is 1 unit away from A after 9 steps ?

Answer 1

wow so many possibilities...there are 4^9 different routes he can take
question then is to find how many of those routes end up on one of the 4 spaces 1 unit away from A

Answer 2

yeah!! this ques is amazing!!

Answer 3

in this question the order in which the moves are made do not matter
..... is it ok to say there are \(\Large 4^9/9!\) routes he can take?

Answer 4

hmm no, 9! is bigger than 4^9 so that is impossible
yes the order matters when talking about the different routes he can take

Answer 5

the number of spots he can end up do not depend on order but im not sure what that number is

Answer 6

ok i wrote a program and simulated this a million times and the probability he ends up 1 unit from A is about 24.2%

Answer 7

seems kinda high...im gonna double check my work

Answer 8

ohh..well i dont have the ans to this ques..my sir had asked this in the class..this was some 4-5 weeks ago..then we all forgot about it :P
i remembered it yesterday and tried all night but couldnt build up any method..
so,,i dont know the ans..0.242 might be correct..

Answer 9

I think i can figure out the number of possible routes that would bring you back to A after 8 steps

Answer 10

no it checks out...so answer is around 25%, may be some error due to random number generator

Answer 11

@PaxPolaris , that sounds great

Answer 12

then what ....?

Answer 13

weird, the same simulation says that there is probability of 0 he ends up back at A in 9 steps

Answer 14

true

Answer 15

not possible with 9

I said 8

Answer 16

oh i see now because its odd you never end up at starting place
thats why probability of being close to A is so high then

Answer 17

24.2% is right on the money...

the exact answer is:
\[\huge{3969 \over 4^7}\]

http://www.wolframalpha.com/input/?i=4%28%28%289choose5%29%2B9%288choose4%29%29%2B4%28%289choose4%29*5%2B%289choose3%29%286choose2%29%29%2B%289choose3%29%286choose2%29%284choose2%29%29%2F4^9

Answer 18

didnt get you sir,,how do you arrive at this expression?

Answer 19

i am very impressed, i have no idea how you came up with that
haha thats why i studied computational math...i'll run the simulations while you work on the theories

Answer 20

sir i really wanna know how you came up with the soln,,please..

Answer 21

While the order in which you take the steps may matter as far as probability is concerned...
It does not matter to where you end up..

If order doesn't matter - you can think that he takes 8 steps that cancel each other out and bring him back to the origin ‘A’, and a Final step that takes him to the endpoint.

Let's call this Final deciding step: X.

In the other 8 steps - there have to be as many up(u) as down(d), and as many Left(L) as right(R).

So the total number of u & L is 4 ...(excluding X which can be anything)

Answer 22

So, I set out to calculate the possibilities when we have 4,3,2,1, or 0 u.

When we have 4u and no L: uuuu .... (and 4d)
& X is u or d .... {the possible ways to order the 9 steps are:\[\large 2\times {9\choose5}\]
+ plus if X is L or F:\[\large 2\times {9\choose1}{8\choose4}\]

And then muliply rhe whole thing, since the number of ways is the same if we have LLLL

uuuu / LLLL : \[\Large 2\times \left[ 2\cdot {9\choose5} + 2\cdot9{8\choose4} \right]\]

Answer 23

*** that should say:
And then multiply the whole thing by 2, since the number of ways is the same if we have LLLL

Answer 24

am getting to it..i should give it an independent try again..thank you so much sir ^_^

Answer 25

then do the same for:
uuuL / uLLL \[\Large 2\times \left[ 2\cdot{9\choose4}{5\choose1}{4\choose1}+2\cdot{9\choose3}{6\choose2}{4\choose1} \right]\]

and for:
uuLL\[\Large4\cdot{9\choose3}{6\choose2}{4\choose2} \]

finally add the three up, and divide by 4^9

Answer 26

hmmn,,to be frank i didnt read your latest post...as i said,,i'd like to try on my own first,,i'll refer to it when i fail ( which i surely will).. hmmn..
and again,,a big thank you ^_^