c4countdown

Paul Howe wrote:(you have to know what an expectation is)

Gavin Chipper wrote:

Paul Howe wrote:(you have to know what an expectation is)

Or you could just tell us what it means. (The expected time for the ant to get from S to E is just the average time it would take if you ran it loads of times.)

As for the puzzle itself, I think you can only manage it in an odd number of goes, so 3, 5, 7 etc.

The probability of achieving it in 3 goes would be 1 * 2/3 * 1/3 because anything can happen the first time, and then it requires moving in either of 2 of the 3 possible ways and then just one specific way. So that would be 2/9. If we have failed after three goes we will automatically be two moves away from our goal. And the same goes for if we have failed after 5, 7, 9 etc.

When we are two away, the probability of making it in two moves would be 2/3 * 1/3, so 2/9 again.

So I would say that since it can be done in 3, 5, 7, 9, 11 moves etc. and at each point there is a 2 in 9 (1 in 4.5) chance of succeeding, the average would be the 4.5th one along, so 10. My answer is 10 minutes.

Paul Howe wrote:Oddly enough I had a completely different solution in mind and didn't think of doing it like that, but it all looks right to me, well done

Dinos Sfyris wrote:Can the ant double back on itself?

Adam Dexter wrote:What would happen if he kept toddling around one face, i.e. infinite time?

Gavin Chipper wrote:

Adam Dexter wrote:What would happen if he kept toddling around one face, i.e. infinite time?

He'd die of old age.

Adam Dexter wrote:What would happen if he kept toddling around one face, i.e. infinite time?

Charlie Reams wrote:

Adam Dexter wrote:What would happen if he kept toddling around one face, i.e. infinite time?

The probability of that is infinitesimally small.

Howard Somerset wrote: P (B to E) = 2/3
P (B to A) = 1/3

Dinos Sfyris wrote:

Howard Somerset wrote: P (B to E) = 2/3
P (B to A) = 1/3

I think you have these the wrong way round. Otherwise good effort.

Howard Somerset wrote:I'd not seen a question like this before, so having opened this thread for the first time yesterday evening and read just the first post, I took it to bed with me, and came up with an answer before finally dropping off to sleep. And having now looked at Gavin's post, which gives the same solution by a much neater method, I think I could've had 30 minutes longer sleep if I'd done it that way.

Anyway, my method:

We've got four essentially different vertices. S, E, A (adjacent to S) and B (adjacent to E).

P (S to A) = 1
P (A to B) = 2/3
P (A to S) = 1/3
P (B to E) = 2/3
P (B to A) = 1/3
As Dinos has pointed out, last two lines should read
P (B to A) = 2/3
P (B to E) = 1/3

That much is the same as Gavin's.

Now-
Expected time S to A: Obviously 1 min.

Expected time A to B:
Prob of doing it in 1 is 2/3
Failing this, we get back to A in 2, so prob of doing it in 3 is 1/3 x 2/3
Failing this, we get back to A in a further 2, so prob of doing it in 5 is (1/3)^2 x 2/3
etc.

Expected time A to B is therefore
1 x 2/3 + 3 x (1/3) x 2/3 + 5 x (1/3)^2 x 2/3 + 7 x (1/3)^3 x 2/3 + ...
A bit of simple algebra turns this into an infinite geometric sequence, giving an expected time of 2 mins.

Expected time B to E:
Prob of doing it in 1 is 1/3
Failing this, we take 1 min to get back to A, and then have already calculated that the expected time to get back from A to B is a further 2 mins, so we're ready to go again after 3 mins. So prob of doing it in 4 is 2/3 x 1/3 [There is clearly a flaw in the wording of my argument at this point, because it's not possible to do it in 4 mins, but we're working on expectations, and an expectation is not always a value that can actually occur, so I felt happy to proceed.]
Similarly, prob of doing it in 7 is (2/3)^2 x 1/3
etc,

Expected time B to E is therefore
1 x 1/3 + 4 x (2/3) x 1/3 + 7 x (2/3)^2 x 1/3 + 10 x (2/3)^3 x 1/3 + ...
which sums to 7.

Expected time S to E is therefore 1 + 2 + 7 = 10 mins.

Is that the same method as yours Paul, or do we now have three methods?

Kai Laddiman wrote:
[ Quote ]

[ Edit ]



(Kai's Cryptic Smilies)

David Williams wrote:I have to say, Howard, that although your answer is correct I can't quite persuade myself that adding the expected times together is valid. Probably it is, but I don't think it's self-evident.