# Solving a Problem

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!"