A Puzzle

[StevenG]

For general n, we can use the same reasoning to get 1/(n+1) for the first and last intervals but we need to think again to get 1/(n+1) for all the intervals.

I bashed out the general case many years age, working out the probability functions. It's not difficult, just tedious (lots of factorials and summations), but the result is true.

[kenberg]

Explanation of why the expected length of the intervals is 1/(n+1).

 

 

We have independent uniformly distributed A1, ... , An rvs on [0,1 and we let me let X1, ... Xn be the same set of numbers listed in increasing order X1<X2< ...<Xn. We claim that the expected values of the various lengths, (X1-0), of (x2-X1), ... of (1-Xn) , are all 1/(n+1).

 

Suppose that A1, ..., An , and Y are all independent and uniformly distributed on [[0,1]. As we are about to draw these numbers, we can agree that the number Y is as likely to be the smallest as it is to be the second smallest or the third smallest etc. So, before we know any of the values, we can say P(X3<Y<X4) is 1/(n+1). Let's use 3 to represent any number, since bbo is not really set up for complex notation. We will just show that the expected value of x4-X3 is 1/(n+1).

 

So I am supposing that we we agree that before any numbers are drawn, P(X3<Y<X4)=1/(n+1).

 

Now we compute this another way: We first draw X1...Xn. Knowing these values, but not yet having drawn Y, then

P(X3<Y<X4 conditioned on the values of X1...Xn) = X4-X3.

We can now calculate P(X3<Y<X4) for all choices of X1...Xn as follows: If we take the expected value of P(X3<Y<X4 conditioned on the values of X1...Xn) for all choices of X1...Xn, we get P(X3<Y<X4), which we know to be 1/ (n+1). So the expected value of the left side of the equation

P(X3<Y<X4 conditioned on the values of X1...Xn) = X4-X3

we get 1/(n+1). Thus, E(X4=X3) is the same number, 1/(n+1).

 

 

So, for each interval, the expected value of the length is 1/(n+1).

[kenberg]

I bashed out the general case many years age, working out the probability functions. It's not difficult, just tedious (lots of factorials and summations), but the result is true.

 

Thanks If you go direct, you get a lot of binomial coefficients. But going as I did gives the result without these nasty factorials. The cost is that it uses some conditional expectations in a manner that might not be immediately certain, although it seems right.

[gwnn]

Was that a reply to me? I accepted a long time ago that each of the four intervals have an expected value of 15 minutes, in fact I said that I consider it to be self-evident.

 

My point is not that any given interval doesn't have an expected value 15 minutes, my point is that the interval you get to has an expected value larger than 15 minutes. It is most likely that you get to the interval that just happened to had been the longest interval. It is least likely that you get to the interval that was chosen to be the shortest interval. Do you dispute this? If not, did you take this into account?

Edited May 19, 2014 by gwnn

[kenberg]

Was that a reply to me? I accepted a long time ago that each of the four intervals have an expected value of 15 minutes, in fact I said that I consider it to be self-evident.

 

My point is not that any given interval doesn't have an expected value 15 minutes, my point is that the interval you get to has an expected value larger than 15 minutes. It is most likely that you get to the interval that just happened to had been the longest interval. It is least likely that you get to the interval that was chosen to be the shortest interval. Do you dispute this? If not, did you take this into account?

 

 

I concede. With one train per hour the expected waiting time appears to be 7/12, not 1/2. So right, I am wrong. (for n=1 I tried a couple of ways and got 7/12 both times but I am now wary...).

 

I thought the approach I was taking would cope ok with the varying lengths of the intervals that I arrived at, but apparently not. I was aware that we get to longer intervals more frequently, I thought that way I was doing it dodged that,. Not so.

 

So I need to rethink a little. It looked good but indeed it was not.

[kenberg]

Where are we? Is there an agreed upon answer yet? I did not follow some of the previous arguments, and I agree that what I said was wrong. Surely if there is an answer for 3 there is an answer for n. I apologize for being wrong but worse (or maybe better) i am short of time to think it through.

 

For that matter, is it even agreed that the a reasonable interpretation is that we are averaging for a guy arriving at some random time during one hour?

 

I am sorry, and embarrassed, for jumping the gun here. It seemed nice and easy.

[gwnn]

I thought we all agreed that the waiting time is 15 minutes until you said it was 3/16-th of an hour (11 minutes and 15 seconds). All you need to see this is that you are just interested in the length of the first interval, and it doesn't matter whether you arrive at the top of the hour or somewhere in between, as long as these three trains and the next three will arrive at the same minutes (i.e. the first train and the fourth will be exactly an hour apart, etc). At least I don't see why it would matter. So the waiting time is 1/(n+1) (if we don't know when the last train left, i.e. lamford's second problem).

[gwnn]

Out of curiosity I went back to my old friend Excel and ran a few tests. The expected values of the intervals are (if we sort them before averaging):

1st: 0.0624+/-0.00016

2st: 0.1461+/-0.00023

3rd: 0.2709+/-0.00013

4th: 0.5206+/-0.00023

 

I hope I didn't mess up the error bars, but anyway, I'm surprised that the expected value of the longest interval is more than 30 minutes long. I'm kind of reminded of the inequality thread here (distributing stuff randomly accounts for more inequality than we would normally expect - though obviously I'm not saying that stuff are distributed randomly in real life!).

[kenberg]

I thought we all agreed that the waiting time is 15 minutes until you said it was 3/16-th of an hour (11 minutes and 15 seconds). All you need to see this is that you are just interested in the length of the first interval, and it doesn't matter whether you arrive at the top of the hour or somewhere in between, as long as these three trains and the next three will arrive at the same minutes (i.e. the first train and the fourth will be exactly an hour apart, etc). At least I don't see why it would matter. So the waiting time is 1/(n+1) (if we don't know when the last train left, i.e. lamford's second problem).

 

Ah!!

 

"as long as these three trains and the next three will arrive at the same minutes

 

This is exactly what I was not assuming.

 

I was assuming the trains between 2 and 3 arrived at random times A1, A2, A3 and the trains arriving between 3 and 4 arrive at random times B1, B2, B3 with all six arrival times independent.

 

I agree that there is no problem if B1 is one hour later than A1. I thought that the point was that the arrival times between 3 and 4 had nothing to do with the arrival times between 2 and 3. it gets easier if there are, effectively, three trains going in a circle at a constant speed of one cycle an hour.

 

I guess I have not understood this problem from the beginning. Whatever was originally intended, I like my version.