Beamy Heteros

I flip a fair, two-sided coin ten times. For each head I flip, I throw a red rock into a pond. For every tail I flip, I do the same to a white rock. Several hundred years from now, a fisherman is fishing in said pond, and begins catching rocks. Naturally, whenever he catches a rock, he throws it directly back in the pond. Furthermore, he catches ten consecutive red rocks in ten tries. Supposing that he moves around the pond in a random way, and throws his rocks back randomly, what is he probability that all ten rocks in the pond are red?

Also, why is this puzzle called 'Beamy Heteros'?

Originally posted by royalchicken
I flip a fair, two-sided coin ten times. For each head I flip, I throw a red rock into a pond. For every tail I flip, I do the same to a white rock. Several hundred years from now, a fisherman is fishing in said pond, and begins catching rocks. Naturally, whenever he catches a rock, he throws it directly back in the pond. Furthermore, he catches te ...[text shortened]... that all ten rocks in the pond are red?

Also, why is this puzzle called 'Beamy Heteros'?

Originally posted by royalchicken
I flip a fair, two-sided coin ten times. For each head I flip, I throw a red rock into a pond. For every tail I flip, I do the same to a white rock. Several hundred years from now, a fisherman is fishing in said pond, and begins catching rocks. Naturally, whenever he catches a rock, he throws it directly back in the pond. Furthermore, he catches te ...[text shortened]... that all ten rocks in the pond are red?

Also, why is this puzzle called 'Beamy Heteros'?

Fjord is correct! Good work, but why 'Beamy Heteros'?

Originally posted by royalchicken
Fjord is correct! Good work, but why 'Beamy Heteros'?

Originally posted by fjord
No clue, so a wild guess:
The fisherman caught only homos (equals); the heteros were left at peace. Wouldn't you beam?

Originally posted by fjord
7%

Let A(n) be the event 'exactly n red ones are caught'. Let B(n) be the event 'there are exactly n red ones in the ponds'. We are looking for P(B(10)|A(10)), ie the probability that ten rocks are red given that ten red rocks are drawn. So, by something called Bayes' theorem,

P(A(10)|B(10)) = P(B(10)|A(10))*P(A(10))/P(B(10))

which simplifies things, because the probability that we get 10 red ones given that there are ten in the pond (the bit on the left of the equation) is obviously just 1. So now we've got:

1 = P(B(10)|A(10))*P(A(10))/P(B(10))

Furthermore, since the probability that we throw a red rock in on any given trial is 1/2 (flip of a coin), P(B(10)) = 2^-10 = 1/1024. Thus:

1/1024 = P(B(10)|A(10))*P(A(10))

Now, P(A(10)), the probability that ten red ones are picked, will be equal to the average probability that this happens, specifically:

P(A(10)) = SUM(n=1 to 10) P(B(n))P(A(10)B(n))

= (1024^-1)*SUM(n=1 to 10) (10 choose n) (n/10)^10

If we substitute this in the other equation, the 1/1024s cancel, and by evaluating that sum we find that P(B(10)|A(10)) = 1/14.26 or about 7%.

The answer to the title is in this post.

Originally posted by royalchicken
[So, by something called Bayes' theorem,
The answer to the title is in this post.[/b]

Good work 😀. This thread can at last be laid to rest.

Originally posted by royalchicken
Good work 😀. This thread can at last be laid to rest.

Well, a search of your posts in the past few days does turn up 'Chincky Oracle", which is almost an anagram of my name.

Originally posted by royalchicken
Well, a search of your posts in the past few days does turn up 'Chincky Oracle", which is almost an anagram of my name.

Which is okay though, because your first spelling anagrams to scores of interesting things.