Can the bird return home?(about probability)

[far far away]

Here's a question makes me confused.
A bird leaves its home to find some food. It won't return if it doesn't find any food. Suppose the probability it returns in the first day is 1/2 and returns in the second day is 1/4 and returns in the third day is 1/8 ... returns in the nth day is (1/2)^n.
The total probability it will return is 1/2+1/4+1/8+1/16+...(1/2)^n+...=1. That means the bird absolutely will be back someday in the future. But if the bird doesn't come back in the first day then the probability it would come back would be 1/2 and that means it has the chance to not come back. Moreover, if the bird doesn't come home in the first few days then the chance it come back would become really small and that means it nearly won't come back. But how could the bird will be back and it won't be back at the same time

I was studying one dimensional random walk. I think the question is kind of similar. Thanks

 

[BvU]

Hello fafa,

There is s slight difference between the biological case and the mathematical abstraction. But you can't point out a situation where the bird is back and not back at the same time...

In the mathematical abstraction the probability of the bird dying hasn't been taken into account. For every n you can calculate the non-zero probability the bird hasn't returned yet. Even for 100 years ...
And if it doesn't find food, it's more likely to die as well...

 

[jbriggs444]

But if the bird doesn't come back in the first day then the probability it would come back would be 1/2 and that means it has the chance to not come back

If we do not know whether the bird returns the first day then its probability of returning on the second day is 1/4 -- that's a 0.5 probability that it fails to find food on the first day multiplied by a 0.5 probability that it then succeeds on the second day. If we do know that the bird fails to return on the first day then its probability of returning on the second day is 1/2 -- that's just the 0.5 probability that it finds food on the second day.

The happenstance that the bird fails to return on the first day increases the probability that it will return on the second day (because it then has a higher probability of even bothering to look for food on the second day).

This is an exercise in conditional probability. The probability of A knowing B, denoted by p(A|B) is given by the probability that A and B both occur divided by the probability that B occurs. That is $p(A|B) = \frac{p(A \& B)}{p(B)}$.

 

[far far away]

If we do not know whether the bird returns the first day then its probability of returning on the second day is 1/4 -- that's a 0.5 probability that it fails to find food on the first day multiplied by a 0.5 probability that it then succeeds on the second day. If we do know that the bird fails to return on the first day then its probability of returning on the second day is 1/2 -- that's just the 0.5 probability that it finds food on the second day.

The happenstance that the bird fails to return on the first day increases the probability that it will return on the second day (because it then has a higher probability of even bothering to look for food on the second day).

This is an exercise in conditional probability. The probability of A knowing B, denoted by p(A|B) is given by the probability that A and B both occur divided by the probability that B occurs. That is $p(A|B) = \frac{p(A \& B)}{p(B)}$.

I got it ! Thanks !