Probability of Meeting Again in a Random Walk

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SUMMARY

The discussion centers on calculating the probability that two individuals, starting at the origin and taking random steps along the x-axis, will meet again after N steps. The probability for one individual is defined by the formula Wn(n1) = N!/[(n1!)(N-n1)!]*p^n1*q^(N-n1), where p and q are both 1/2. To find the combined probability for two individuals, participants suggest considering their relative positions and potentially using integrals to account for all possible meeting scenarios over N steps.

PREREQUISITES
  • Understanding of random walks in probability theory
  • Familiarity with combinatorial mathematics and factorial notation
  • Knowledge of basic probability concepts, including independent events
  • Ability to work with integrals in the context of probability distributions
NEXT STEPS
  • Explore the concept of relative position in random walks
  • Study combinatorial probability and its applications in random processes
  • Learn about the use of integrals in calculating probabilities for continuous distributions
  • Investigate the implications of symmetric random walks on meeting probabilities
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Mathematicians, statisticians, and students studying probability theory, particularly those interested in random walks and combinatorial analysis.

mathlete
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The question:

"Two men start out together at the origin, each having a 1/2 chance of making a step to the left or right along the x-axis. Find the probability that they meet again after N steps."

It then says it may help to consider their relative position but I don't see how that would help.

The probability for one person is Wn(n1) = N!/[(n1!)(N-n1)!]*p^n1*q^(N-n1)

where N is total steps, n1 is steps to the right, p = q = 1/2 (for this problem). I just don't know how to combine/adjust it for two people.

Also, I'm not sure, but would this involve an integral from 0 to N steps at some point (to cover all cases)?
 
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i commented on the other post... hopfully it can help you
 

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