Random probability question - need help

Tbond

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Dec 9, 2013
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I have a probability question that hopefully someone can help answer.

If you had a machine that randomly generated a 1 or -1 at each interval, and you started at zero, what's the probability that you'd get to +14 versus -6. Is it equal probability of both?

For example, if the machine generated the following stream:

+1
-1
+1
+1
+1
-1

You'd end up at a cumulative result of 2. Given that process continuing, eventually it would hit either 14 or -6. Is the probability 50/50 that either would get hit? If the targets changed would that make any difference? For example, +10, -10.
 
Thanks for the response Romsek. I hadn't heard of Markov chains till now. Yes, the probability of hitting +1 or -1 are equal.

Is there a formula for computing the probabilities given length of sequence of N? Or do I need to do it manually by generating random samples and figuring the probabilities? If I do need to do it manually, how large a sample size would I need to have a high confidence, besides infinitity (for obvious reasons).

Thanks again!
 
I have a probability question that hopefully someone can help answer.
If you had a machine that randomly generated a 1 or -1 at each interval, and you started at zero, what's the probability that you'd get to +14 versus -6. Is it equal probability of both?
For example, if the machine generated the following stream:
+1
-1
+1
+1
+1
-1
You'd end up at a cumulative result of 2. Given that process continuing, eventually it would hit either 14 or -6. Is the probability 50/50 that either would get hit? If the targets changed would that make any difference? For example, +10, -10.
This may not help at all. If not please ignore it.
Look at this webpage. It expands \(\displaystyle {\left( {x + {x^{ - 1}}} \right)^{10}}\).
If the machine produced a string of length ten those coefficients and exponents tell us the sums and the number of ways to get each.
For example: the term \(\displaystyle 10x^8\) tells us there are 10 ways to get the sum of 8.
The term \(\displaystyle \dfrac{{120}}{{{x^4}}}\) tells us there are 120 ways to get the sum -4.
The 256 tells us there are that many ways to get a sum of zero.
Each of those are then divided be \(\displaystyle 2^{10}\) to get the probability.

So the probability of getting a sum of 8 is \(\displaystyle \dfrac{10}{2^{10}}\)

Now you can change the exponent on \(\displaystyle {\left( {x + {x^{ - 1}}} \right)^{13}}\) and click the \(\displaystyle \boxed{=}\) at the end of the input box and the you have the numbers for length 13.
 
For your original 14 versus -6 : 14 = 3/10, -6 = 7/10

General case:
a versus -b
a = 1 / (a + |b|) * |b|
b = 1 - a

Your example:
a = 1 / 20 * 6 = .3
b = 1 - .3 = .7

Thanks. This equation has proved to be the most accurate. I've tested large datasets manually and the above solution matches the results.
 
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