Bye (for now?)

Got too much work on at the moment, so am having a break from RHP for at least the next few months.

Since this is the only forum I have really posted in I thought I'd say bye and leave you with a couple of (hard) puzzles!!

1) A knight starts at the corner of a chessboard, and at each time t = 1, 2, 3, 4, ... it moves once, with all possible moves equally likely. Let T be the first time the knight returns to its starting corner. What is the expected value of T?

2) A monkey is typing on a typewriter, with each letter A, B, C, ... , Z equally likely. How long before you expect to see the sequence ABRACADABRA?

Enjoy!

An ugly incomplete solution for question 1

I can solve it with a system of 64 equations in 64 variables, in some mathematical software. Basically we need to write a number in each square of the chessboard, such that at a1 we have 0, and at each other square the number is 1 more than the average of the numbers which are at a knight's-move distance. Then the solution is the number at b3 (or c2) plus 1. (I think)

I managed to solve problem 2 in the binary case - the monkey types 0 or 1, and we wait for a specific string of 0's and 1's. But I didn't solve yet the original problem. Maybe you should try the binary case first.

Originally posted by SPMars
Got too much work on at the moment, so am having a break from RHP for at least the next few months.

Since this is the only forum I have really posted in I thought I'd say bye and leave you with a couple of (hard) puzzles!!

1) A knight starts at the corner of a chessboard, and at each time t = 1, 2, 3, 4, ... it moves once, with all possible moves equall ...[text shortened]... ... , Z equally likely. How long before you expect to see the sequence ABRACADABRA?

Enjoy!

Originally posted by David113
I managed to solve problem 2 in the binary case - the monkey types 0 or 1, and we wait for a specific string of 0's and 1's. But I didn't solve yet the original problem. Maybe you should try the binary case first.

No. The waiting time depends on which string of 0's and 1's you are waiting for.

Example:

If you wait for 01 or 10 then the expected waiting time is 4; but if you wait for 00 or 11 then the expected waiting time is 6.

Originally posted by SPMars
Got too much work on at the moment, so am having a break from RHP for at least the next few months.

Since this is the only forum I have really posted in I thought I'd say bye and leave you with a couple of (hard) puzzles!!

1) A knight starts at the corner of a chessboard, and at each time t = 1, 2, 3, 4, ... it moves once, with all possible moves equall ...[text shortened]... ... , Z equally likely. How long before you expect to see the sequence ABRACADABRA?

Enjoy!

Originally posted by David113
No. The waiting time depends on which string of 0's and 1's you are waiting for.

Example:

If you wait for 01 or 10 then the expected waiting time is 4; but if you wait for 00 or 11 then the expected waiting time is 6.

See proof here:

http://www.qbyte.org/puzzles/p082s.html