On Tuesday, me & Rosie got Hugh's sheet for the week. Question 3 was the question about the mean number of transmissions required to sent a frame, given a failure rate of p.

We worked this out, but you said we needed to show our working out, and
that it was a lucky guess. I looked at the answers afterwards and
thought that it was *too* lucky a guess. I thought I'd e-mail you the
working out to prove that we weren't cheating or being lucky (and
because I'm surprised we came up with the simplest solution on the first
go).

First I figured out that no frame is guaranteed to get there, by writing down the probability of each frame of getting there given a p of 50%:

- Frame 1: 50%
- Frame 2: 75%
- Frame 3: 87.5%

This of course goes on forever halving (with any probability >0 it never gets there). Rosie pointed out that the question said 'mean number' of frames, so I realised this meant that you only need to send 2 frames, and the 'mean' (average) probability is that 1 will get there (2 * 50% = 100% rather than 75%). Rosie then did the maths and the equation.

- If p is the probability of failure, then success (the opposite) is 1-p (this is a common probability formula)
- If you need to send 2 frames on a 50% (or 0.5) probability (as above) to get 1 frame there successfully, then the equation is 1/(1-p) - this is the number of frames needed to send to get a mean guarantee of delivery of 1 frame

It's fairly simple. I think it's interesting that we bypassed 2 pages of working out (and the original equation) to get to the same answer (I didn't even understand the original equation, never mind the working out).

*We got the tutor (a PhD student) to come over and see if it was correct. "This is impossible!" he cried, "This takes 5 pages of calculation!"*