A one dimensional random walk

[amal_]

Hello! I am having a problem with knowing the number of ways a boy can walk to be at a certain position in the following exercise ,I would very much appreciate some help.
"A boy stands on a street corner and tosses a fair die .If 1 or 2 occurs, he walks one block west ;otherwise he walks one block east. At his new position he repeats the procedure. What is the probability that after having tossed the coin 6 times he is
(a) at his starting point;
(b)2 blocks from his starting point;
(c)4 blocks from his starting point."

Now i realize that the probability of the boy walking west is 2/6=1/3 and of him walking east is 4/6=2/3.
Being at his starting point means that at the end he would have walked 3 blocks west and 3 blocks east meaning that the probability is (1/3)^3*(2/3)^3 ,but there are plenty of ways he could have walked these 6 blocks ,I am having trouble knowing the number of ways he could have walked these blocks .
For the second question -being 2 blocks from the starting point -means that he could either be 2 blocks west or two blocks east which have different probabilities in this case .
Overall I am wondering if there is an easier approach to solve these kind of problems or maybe a generic way to find out how to calculate these probabilities let's say if he tossed the coin n times.

 

[Dr.Peterson]

Being at his starting point means that at the end he would have walked 3 blocks west and 3 blocks east meaning that the probability is (1/3)^3*(2/3)^3 ,but there are plenty of ways he could have walked these 6 blocks ,I am having trouble knowing the number of ways he could have walked these blocks .

You're saying that he might have walked any permutation of WWWEEE, right? Have you learned how to count those?

For the second question -being 2 blocks from the starting point -means that he could either be 2 blocks west or two blocks east which have different probabilities in this case .

So, you can separately calculate two cases.

Overall I am wondering if there is an easier approach to solve these kind of problems or maybe a generic way to find out how to calculate these probabilities let's say if he tossed the coin n times.

You may be on your way to such a general method!

 

[pka]

Hello! I am having a problem with knowing the number of ways a boy can walk to be at a certain position in the following exercise ,I would very much appreciate some help.
"A boy stands on a street corner and tosses a fair die .If 1 or 2 occurs, he walks one block west ;otherwise he walks one block east. At his new position he repeats the procedure. What is the probability that after having tossed the coin 6 times he is
(a) at his starting point;
(b)2 blocks from his starting point;
(c)4 blocks from his starting point."

You need to review what you posted. For one, you start by tossing a die the end by flipping a coin. I am reasonably sure that you mean die in both places. If the walk consists of [imath]WWWEEE[/imath], three blocks west followed by three blocks east the the lad is back at the starting point.
Any rearrangement of that string say [imath]EWEEWW[/imath] brings him back to the starting point. There are [imath]\dfrac{6!}{(3!)^2}=20[/imath]
ways to rearrange that walk. You are also correct that each of those walks occurs with the probability [imath]\left(\dfrac{2}{6}\right)^3\left(\dfrac{4}{6}\right)^3[/imath]
Beyond that, I am not sure as to what your questions mean.

 

[amal_]

You need to review what you posted. For one, you start by tossing a die the end by flipping a coin. I am reasonably sure that you mean die in both places. If the walk consists of [imath]WWWEEE[/imath], three blocks west followed by three blocks east the the lad is back at the starting point.
Any rearrangement of that string say [imath]EWEEWW[/imath] brings him back to the starting point. There are [imath]\dfrac{6!}{(3!)^2}=20[/imath]
ways to rearrange that walk. You are also correct that each of those walks occurs with the probability [imath]\left(\dfrac{2}{6}\right)^3\left(\dfrac{4}{6}\right)^3[/imath]
Beyond that, I am not sure as to what your questions mean.

I am sorry for the confusion, I did mean die throughout the exercise.
Would you please mind explaining how did you find that there are 20 ways to rearrange the string?

 

[amal_]

You're saying that he might have walked any permutation of WWWEEE, right? Have you learned how to count those?

Yes, that's what i meant. I did learn how to to calculate the permutation.
I think we can rearrange WWWEEE in 6! way, since we are rearranging 6 elements from 6 without restrictions, would that be correct?

 

[pka]

Yes, that's what i meant. I did learn how to to calculate the permutation.
I think we can rearrange WWWEEE in 6! way, since we are rearranging 6 elements from 6 without restrictions, would that be correct?

No it does not work that way. That requires permutations with symbols repeated. Consider the word [imath]TENNESSEE[/imath]
That word has nine letters but several repeat or are not distinct. With subscripts [imath]TE_1N_1N_2E_2S_1S_2E_3E_4[/imath] all nine letters are distinct. So it can be arrange in [imath]9![/imath] ways. In each one of those arrangements the E's appear in [imath]4![/imath] ways.
Thus [imath]TENNESSEE[/imath] can be arranged in [imath]\dfrac{9!}{(4!)(2!)^2}[/imath] ways.
The name [imath]MISSISSIPPI[/imath] can be arranged in [imath]\dfrac{11!}{(4!)^2(2!)}[/imath] ways
The walk [imath]WWWEEE[/imath] can be arranged in [imath]\dfrac{6!}{(3!)^2}[/imath] ways.

 

[Dr.Peterson]

Yes, that's what i meant. I did learn how to to calculate the permutation.
I think we can rearrange WWWEEE in 6! way, since we are rearranging 6 elements from 6 without restrictions, would that be correct?

That's how many ways you can arrange ABCDEF.

One way to see the correct method in this case is that there are 6 locations in the string, and you can choose any 3 to be W, with the rest E. So there are [imath]{6\choose 3}=\frac{6!}{3!3!} = 20[/imath] arrangements.

 

[sky2rain]

That's how many ways you can arrange ABCDEF.

One way to see the correct method in this case is that there are 6 locations in the string, and you can choose any 3 to be W, with the rest E. So there are [imath]{6\choose 3}=\frac{6!}{3!3!} = 20[/imath] arrangements.

And this is saying, no matter which direction he goes, the farthest direction he may travel is 3 blocks and for the other 3 moves he has to travel the opposite direction. And for the opposite direction he can travel only at most 3 blocks for it would exhaust the possibility of getting back to the starting point. So neither direction he moves cannot be more than 3 or less than 3 (since if he moves less than 3 out of 6 moves in any direction the opposite will be more than 3).

And the nature in doing that is no matter at what point you decide to travel in a direction you deem to be +1, the opposite will be -1, and the sum of equal moves of one direction and opposite direction is always 0, no matter how you decide to arrange your walk pattern. So for all 20 ways of your moving pattern with equal number of opposite directions, you always end up arriving the starting point, and (1/3)^3*(2/3)^3*20, about 22% possibility.

 

[sky2rain]

Here is a more intuitive approach to this problem: Basically this is the same as if he flip a coin according to die outcome. Safe to conclude that there exists a permutation with “repetition” and not “without replacement”. The entire sample space for flipping coin 6 times has 2^6 events - a permutation with repetition - and if the guy base his decision on a fair coin flip the probability for each outcome is in plain sight 1/(2^6). However it’s not so you will analyze each situation and multiply the probability 6 times for each outcome.



Now we analyze the RV of the guy arriving the start point after 6 moves. Basing on common sense, we conclude that the farthest he could travel in one direction is 3-blocks for him to reach the starting point and we call it the first case, and this could have only 1 outcome which is he must travel back 3-blocks in opposite direction.



Then analyze what happens if he walks one direction two blocks. He can’t keep walking in the same direction further otherwise it overlaps with the above case, so he’s one block away and with 3 moves left. 2^3 for both directions have 16 outcomes which is probably still cumbersome to analyze , and we further split it to two subcases: he walks the opposite direction (2 blocks away) or keeps walking in the same direction (back to start) and there is only 2 moves left. For 1st sub-case which puts him 2 blocks away. it’s obvious that he has only one move which is to travel 2 blocks again opposite to arrive at the start point and for 2nd sub-case he can travel either direction then opposite to get back. So for both subcases there are total of 3 outcomes.



At last case we analyze if he travels a direction only one block. He can’t travel the same direction further since it overlaps with the 2nd case. So he must travel back to the starting point and be with 4 moves left. And no matter which direction he travels it is identical to with what happens with the 2 subcases in the 2nd case. So there are total 6 outcomes for him for he may travel both directions on the fourth move.



1+3+6=10, and for both directions there are 20 outcomes