## FW. Walk Around the Clock

A quick puzzle from one of our listeners, which has a surprising and interesting solution:

Standing at 4:00, repeatedly flip an unfair coin, moving clockwise if it comes up heads, and counterclockwise if it comes up tails.

What is the probability that you visit every hour 1-11, before you reach 12:00?

In our next post we’ll ask a seemingly unrelated question that turns out to help quite a bit…

## 5 Comments »

1. ### Steven H. Noble said,

July 2, 2009 at 8:49 pm

I’m guessing if I’m trying to find the null space of an 11×11 matrix to solve this problem I’m missing out on a more straight forward solution?

2. ### strauss said,

July 6, 2009 at 4:24 pm

I don’t quite see how that would be helpful– what’s your idea? I can see one way to use an 11×11 matrix, but there’s a simpler approach (which, as one might guess, uses the next post!)

3. ### Shawn said,

April 10, 2012 at 12:39 am

The probability has gotta be pi/4.

4. ### strauss said,

April 18, 2012 at 9:53 am

Gosh it’s been a long time since I’ve thought about this puzzle; but I’m pretty certain that the answer isn’t pi/4! A quick experimental run gives success about 9% of the time, with a fair coin.

(Note too the problem asks for the general solution, too, with a possibly weighted coin)

Let me try to recall how the heck this problem works. The vague idea, which the next post is a clue, is to express the likelihood of success from a given position in terms of the success from neighboring positions– it turns out it’s a special weighted average. But missing all the details… (There is some cleverness needed to set it up right in the first place)

5. ### Shawn said,

June 5, 2012 at 4:12 pm

Well, it seems that if the probability of the coin coming up heads is 0 or 1, the only movement will be clockwise or counterclockwise respectively, so it will always be 4 3 2 1 12 or 4 5 6 7 8 9 10 11 12. Whatever function models this probability should yield an output of 0 if the input is 0 or 1.

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