Hint: for 1, let the direction be a coin flip: heads you go clockwise, tails counter. Where you wind up only depends on the number of heads and tails. What numbers of heads will get you back?
For 2, the important observation is that if he goes to zero, he comes back to 1 and you are where ...
counterclockwise. It is important to notice that the end state only depends on how many heads you get, not the order of heads and tails. I am suggesting you need to figure out what number of heads will result in returning to A after 10 flips.
For two flips, you are back at A if one is heads and one is tails. For three flips, you can't be back at A at all. For four flips, what do you need to be back at A? What happens new for six?
No, for six 3+3 is one way. Six heads is another, as is six tails. So for six flips, the chance you are back at A is the chance of 0 plus the chance of 3 plus the chance of 6.
No, C(6,3) is not a probability. It is one factor in the probability. Yes, if you have eight flips and start with four heads, you could get four tails to get back to A. You could also get two heads. Your original question had a fixed number of tosses. Arriving back at A after specifically 10 tosses (you might have come back before) requires one of three possible numbers of heads in the 10. Just as I said for six tosses, you need to add the probabilities of each of those numbers.
That is right if you divide by $2^6$ or $2^{10}$. No, 3+7 is a partition of $10$ into two parts, but it doesn't get you back to A. Please draw a hexagon, label a corner A, get a counter, and move it around to see how this works. You clearly have not understood it.
@RossMillikan Ok! Thank you for your help. For the mosquito problem, I had 4 as well, but it was incorrect. What method did you use to get 4?
@RossMillikan "A mosquito is walking at random on the nonnegative number line. She starts at 1. When she is at 0, she always takes a step 1 unit to the right, but, from any positive position on the line, she randomly moves left or right 1 unit with equal probability.
What is the expected number of times the mosquito will visit 0 before the first time she visits 4?"
@RossMillikan I had a similar problem in my textbook. It was almost exactly the same, except that the mosquito started at 0 rather than 1, and the question asked: "What is the expected number of times the mosquito will visit 0 before the first time she visits FIVE?"
The answer there was 4 for that problem @RossMillikan
I didn't get 4. I suggested defining a function $V(n)$ which is the expected number of times you visit zero before you get to 4 starting at $n$ . Clearly $V(4)=0$ You can write a set of equations for $V(0),V(1),V(2),V(3)$ because ifyou areat 2, you have half chance to go to 1 ad half chance to go to 3. Is this the approach used for your books problem? If so, follow along.
@RossMillikan I believe I would have the system of equations: $a_0 = a_1, a_1 = 1/2 (a_0+1)+1/2(a_2), a_1 = 1/2a_1 + 1/2a_3, a_3 = 1/2 a_2 + 1/4 a_3$
@RossMillikan in my book, it is almost exactly the same except for in the system of equation, the alo have $a_{4} = \frac{1}{2}*a_{3}$, but they substituted it into the equation $a_{3} = (\frac{1}{2}*a_{2})+(\frac{1}{4}*a_{3})$, to get $a_{3} = (\frac{1}{2}*a_{2})+(\frac{1}{4}*a_{3})$ anyway
@RossMillikan Would the function be this? $a_{n} = (\frac{1}{2}*a_{n-1})+(\frac{1}{2}*a_{n+1})$
@RossMillikan you still there?
@RossMillikan For the mosquito problem...is it TWO?
@RossMillikan This is what I know. a_0 = a_1 = 4. a _2 = 3. a _ 3 = 2. a _4 = 1.
@RossMillikan a_n denotes the expected number of visits to 0, given that the mosquito is at position n.
and i have the system of equations: a_0 = a_1+1 ; a_1 = (a_0 + a_2) / 2 ; a_2 = (a_3+a_1) / 2 ; a_3 = a_2/ 2 ? and solve for a_1? @RossMillikan @RossMillikan
@RossMillikan why a_1, instead of a_0? because the mosquito starts at point 1?
@RossMillikan However, we would have to have a given a_0, a_1, a_2, a_3 to start with. otherwise, if i do substitution, there will always be two variables
That is right, we want a_1 because that is where it starts. You are right we need to solve for all the a's to get the answer. You have four equations in four unknowns.
Your firs equation is a-)=1+a_1, so if a_1 > 0, a_0 > 1. In fact, it should be easy to see that a_0 is the greatest of them all, because it is farthest from a_4.
But the question asks"What is the expected number of times the mosquito will visit 0 before the first time she visits 4"?
@RossMillikan
@RossMillikan we want to find the expected number of times for 0, not 4
in a_0, the mosquito is already at 0, isn't it? so wuldn't it just be one count? @RossMillikan
@RossMillikan If you are busy or occupied currently, please let me know
@RossMillikan Oh, I just reread the problem. If the mosquito is at 0, it HAS to go to 1. It can go back to 0 from 1, or continue on {2, 3, 4..}. Is it ... 2 then?
no, nevermind
i think i'm confusing myself. some clarification would be appreciated. @RossMillikan @RossMillikan @RossMillikan
If it were already at zero, that is one count, but it could come back for another.
Your first equation says that if you are at zero, the expected total number of times to visit zero is one (this time) plus the expected number of times if you start at 1. This is because leaving zero you will be at 1 and you still have that expected number to come.
Just use substitution down from the top. The first equation lets you eliminate a_0, then when you put it into the second you can eliminate a_1, and so on.
That lets you substitute the right side for a_0 in the second. You have eliminated a_0 from the second. Now you have three equations in three unknowns, which is progress. Then you can solve the second for a_1= something and plug that into the third.
Discussion between Math is Life and R…
Discussion between Math is Life and R…