You are Here: Home >< Maths

# Partitioning and expectation watch

1. An email is sent on the network in which the recipients (0,1,2,3,4,5} are in communication.
1 can send to 4 and 2
2 to 1,3,5
3 to 0,2,5
4 to 1, 5
5 to 0,2,4
0 to 3 and 5
If a message is sent to 2,3,4,5 it is forwarded randomly to a neighbour (even if this means a repeat). 0 and 1 never forward messages.
Let ek be the expected number of time that a message starting at k is passed on.
Find e4.

So:

E(X) = E(X|A)P(A)

I think I need to partition this but I'm unsure on the partition.

Let X be the number of times a message is sent on
E(X) = E(X|1st move is to 0)P(1st move is to ) + E(X|1st move is to 1)P(1st move is to 1) + E(X|1st move is to 2)P(1st move is to 2) + E(X|1st move is to 3)P(1st move is to 3) + E(X|1st move is to 4)P(1st move is to 4) + E(X|1st move is to 5)P(1st move is to 5)

I'm not sure how I'd work these out using a general k to start at.

If I assume I start at 4, as I'm trying to find e4 then I get
e4 = 0 + 1x(1/2) + 0 + 0 + e5 (1/2)
2e4 = 1 + e5

e5 = 1x(1/2) + 0 + e2 (1/3) + 0 + e4 (1/3)

I don't really think I'm going about this the right way, I would have thought I need to find a formula for starting at a general k but I don't know how.
2. Are 5 and 0 considered neighbours? This will make a difference to the answer and isn't clear to me from the wording of the question.

My first impression is that simply working out e_4 is going to be easier than trying to find e_k.

For e_4 we have e.g.
P(0 times forwarded)=1/2 (i.e. 4 is to forward to the neighbour 3, but 4 cannot send to 3)

However if 4 is to forward to neighbour 5 (probability 1/2) I need to know if 0 and 5 are considered neighbours or not to continue.
3. (Original post by Kate2010)
I don't really think I'm going about this the right way, I would have thought I need to find a formula for starting at a general k but I don't know how.
What you're doing looks fine. You can't really expect to find a formula for 'general' k (since each state is somewhat unique in terms of which neighbours it has), but if you carry on, you'll find you get 4 linear equations relating . All you have to do is solve them, which will be a bit tedious, but not too bad I think.
4. (Original post by nota bene)
Are 5 and 0 considered neighbours? This will make a difference to the answer and isn't clear to me from the wording of the question.
Sorry nota bene, I don't think I explained it very well. My question gives a diagram and I tried to explain it in words but the diagram goes a bit like this:

1----2---3
/ --- / -- /
4---5---0

where neighbours are considered to be those connected by an edge.
(Ignore the dashes in the middle line, I couldn't get it to space out otherwise).

(Original post by DFranklin)
What you're doing looks fine. You can't really expect to find a formula for 'general' k (since each state is somewhat unique in terms of which neighbours it has), but if you carry on, you'll find you get 4 linear equations relating . All you have to do is solve them, which will be a bit tedious, but not too bad I think.

Setting up a system of simultaneous equations I got e4 = 2/3 so I think I have either set it up wrong or made a slip somewhere with the calculations.

These were my equations:

e0 = 0
e1 = 0
e2 = 1 + (1/3)(e1+e3+35)
e3 = 1 +(1/2)(e0+e2)
e4 = 1 + (1/2)(e1+e5)
e5= 1 + (1/3)(e0+e2+e4)
5. I certainly don't get e4 = 2/3, so I suspect you've made a calculation error. (As a matter of course you should substitute your values for e1,e2,e3,e4 into your equations to check).
6. (Original post by DFranklin)
I certainly don't get e4 = 2/3, so I suspect you've made a calculation error. (As a matter of course you should substitute your values for e1,e2,e3,e4 into your equations to check).
Ok I've tried it again and subbed my values in to check and I now get:
e0 = e1 = 0, e2 = e5 = 8/3, e3 = e4 = 7/3
Is it ok for it not to be a whole number?
Or have I set up the equations incorrectly?
7. Yes it's fine for it not to be a whole number (and as far as I remember, those are the numbers I got).

### Related university courses

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: January 27, 2010
The home of Results and Clearing

### 2,956

people online now

### 1,567,000

students helped last year
Today on TSR

### Took GCSEs this summer?

Fill in our short survey for Amazon vouchers!

### University open days

1. Bournemouth University
Wed, 22 Aug '18
2. University of Buckingham
Thu, 23 Aug '18
3. University of Glasgow
Tue, 28 Aug '18
Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams