Code sent to your email.

He heads for the security elevators and goes down [removed by administrator]... When he returns he has the answers he needs.

So that is the reason for his superhuman capabilities... As if I did not know.

]]>Was that supposed to be funny? Eric is much less talkative than you are and rarely even explains where those magical answers come from. He heads for the security elevators and goes down [removed by administrator]... When he returns he has the answers he needs. DL, knows nothing.

Please email me the code and I will bring over pizza.

]]>Please refer to the site as B and the CAS as M.

Why don't you start a thread for us beginners on how to use it?

That is what the computer math thread is for?! Also as you have been told, I am not an expert. Why don't you ask the Colonel or DL?

Was M involved a lot?

Of course, it is unthinkable to leave him out.

]]>That is a nice generalization of the problem from [site removed]. Was Math involved a lot? Why don't you start a thread for us beginners on how to use it?

]]>Also, you can use the formula which will be easier than the series.

where lower = 50 and higher = 99.

]]>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:

Are you sure about the last part? Because my calculation (based on your calculations, actually!!!) results to

bobbym wrote: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.

Well, it is quite a bit

So what is this big number in the denominator of the second fraction? Where does it come from?

]]>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.

Well, it is quite a bit

]]>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 think you will see how involved it is. 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.

]]>I am unable to submit any working until the expiration time of the contest problem. That will be in 1 day and 19 hours from now. Since I submitted my solution he has been watching this forum, I guess to dq me.

I can say that it is done using "spot the pattern" and experimental mathematics. It yielded a small, tight formula for any number of heads.

]]>