Your question about the chance of seeing a particular image of Toby when using the Dranzlador was a good one. Your guess of 5% was correct. I've put the latest images ("15951" and "amoeba") up, so now there is a 1 in 20 chance of seeing an image. That's exactly 5%!

Another thing that's interesting is this: How many times *on average* do you have to hit refresh before you see a particular image? If there are 20 images, it takes 20 refreshes, *on average*, to get an image. I just tried this and it took a full 83 hits to get the one you named "categoryrain" to appear but then a mere 3 more hits to get another occurrence of it. It's only *on average* that it takes 20 hits. The math behind this almost intuitive fact (of 20 refreshes) is quite interesting.

If there is a probability p that something will occur n times and a probability q that something will occur m times, then the expected value of occurences (i.e. *average*) is n*p + m*q. In our case, there are infinitely many possible occurences (1 hit, 2 hits, ..., 83 hits etc.) For example, the contribution of the 83 hit outcome to the expected value of hits is 83*(19/20)^(82)*(1/20). This is because the probability of *not* getting the image in the 1st 82 hits is (19/20)^82 and the chance you will get it on the last hit is 1/20. (The ^ symbol means "to the power of" in case you are not familiar with this notation. e.g. 2^3 = 2*2*2 = 8)

1*(1/20) + 2*(19/20)^(1)*(1/20) + 3*(19/20)^(2)*(1/20) + ...

is the expected value (i.e. *average*) number of hits to get the image of interest. This infinite series can actually be summed and shown to equal the following number:

(1/20)*(1)/(1-(19/20))^2

(The proof of this involves Calculus!)

This number simplifies to 20. You could post a comment on Math4Chip asking why this is so if you're a little shaky with fractions, powers, operator precedence etc.

Pretty amazing stuff, huh?!

- TcT