Math Is Fun Forum

Well, I looked at the two parts separately, because they are virtually independent. For one, there are 8 possibilities for one and 16 for the other. Then I just combine them.

Hi;

Hi anna_gg;

Do not worry, it gets worse.

A shortcut consists of playing spot the pattern and going on from there proving it later by induction or whatever.

From some small problems of say 3 Hits and 4 Misses, I could guess at this formula:

This was followed by a large amount of empirical testing so that at least there is a chance it is correct.

Now for the last question. Since we have a formula we can just sum it:

Hi;

The contest is a continuing weekly for total points so it will never be over until I have the most points or die trying.

They want me to submit solutions but it is much easier for me to do the problems than to explain what I did. I will  try to explain my thoughts as I was doing it.

We call a hit H and a miss M.

The probabilty of  H hits and M misses is always the same no matter how they are ordered. That is to say {H, M, H, T, H} has the same probability of occuring as {M,H, H, H, M}. No matter what arrangement of H,H,H,M,M,M,H,M... that produces 50 H's and 50 M's. This is because the numerators and denominators of the fractions produced are the same just in different order.

We only need to know how many arrangements are there.

The first two do not need to be figured. They are always {H,M} or {M,H}. So we are really only looking for 49 H's and 49 M's.

If this is not extremely confusing we can move on to the second question.