The drunk bird, chance of never coming home

Question

There is an old saying that a drunk man will stumble his way home eventually, but a drunk bird may be lost forever. Model the bird as a walker on an infinite 3D lattice, moving one unit along one of the six axis directions each step, each with probability $1/6$ , forever.

Roughly what is the probability the bird never returns to its starting point?

Accepted Answer

The 3D symmetric walk is transient, with eventual-return probability equal to Polya's constant, about 0.3405. The chance of never returning is just the complement: P(	ext{never returns}) = 1 - 0.3405 \approx 0.6595, about 66%. That is the mathematical version of the saying: the man on a 2D street grid is recurrent (returns with certainty), while the bird in 3D space is transient and is more likely than not to be lost for good. Takeaway. Because the 3D walk is transient, escape is the likely outcome, roughly a 0.66 chance of never coming home.

The drunk bird, chance of never coming home

Your answer

Discussion

Learn the concepts

Related questions