jordan

Nicely donce VM. Assuming all your calculations are correct, that made a lot of sense to me. I'm glad you took the time to explain all of your steps or else that would have all been over my head.

MrL, Do you have any examples of what he gets wrong? As far a I knew he is pretty well respected so I would like to know a little reasoning behind you claim.

Seems like you need some unit of time in there since velocity=m/sec.

I am having a little trouble following the variables in your equations. It would be very helpful if you could lable them for me.

Why are there only 40? If the equation is really long, don't worry about finding and posting it. I was just currious whether it was an equation or some other method. I'm sure I'll run across it some day, though.

wolfon, would you mind explaining about your equation in which you calculated the lowest number (post 29)?

I would agree that the open universe theory prevails. YT, the only problem I see with the spring analogy is that the forces you are envisioning are different than the ones the universe would experience. If its mass is too great for the speed at which matter is moving, it will colapse. I don't know what would prevent it from having the exact same mass and speed the next time, creating an infinite loop. Mabey you can let me know what you were thinking of.

I believe this is called the closed universe theory. It proposes that the mass of the universe is such that it will eventualy colapse in on it self, effectivly causing a "Big Bang" and creating a new universe. The converse is the open universe theory in which matter continues to disperse and cool.

Emmon, you are assuming that the order matters. If it did, each ball (now changed to 42 of them) would have the odds you said, but since order doesn't matter, the odds are less. The way I see it now is this, The machine can pick any 3 numbers it wants. To find the number of possibilities, you use the equation: n! ------ r!((n-r)!) were n=number of balls (42) and r=number drawn(3) The answer I get is 11480. As I said earlier, joining two sets of three on a ticket of six yeilds 20 sets of three. So deviding the number of sets that can be drawn (11480) by the number each ticket holds (20) means that you would need 574 tickets to guarentee a win. wolfson, I didn't understand your last two post. If you could explain them to me (where you got the equations and what the variables are) that would be helpful as I am determined to get an answer now.

Can you give me any more. wolfson, you said there is an equation?

This is a very good question. From an evolution standpoint, a being evolves to the betterment of itself. Your question asks how two can evolve for the betterment of both of them. I don't have any ideas on this one.

New idea.... Deviding by 2 wouldn't be enough. By placing three sets together, you have made more than two different sets. I came up with 20. A ticket with numbers 1-2-3-4-5-6 would not only have 1-2-3 and 4-5-6 covered like I origionaly thought. It would cover 1-2-4 1-2-5 1-2-6 etc. I got 20 sets of three on each ticket, although I'm not sure if there is overlapping or not. If each ticket does have 20 sets of three, you 163 looks a lot more reasonable.

Your last post was on the right track, Pinch. What you need is to find how many possiblilities there are for picking three numbers from 42. You found 14 for picking two numbers from six (I think you missed 5-6, making the total 15 tickets). This would take a lot longer for three numbers from 42, however. (Mabey one of your previous posts covered an equation for this, I don't have a lot of time to check now) When you do that, it seems like you could devide your total by 2 as a ticket is 6 numbers, or 2 sets of 3 numbers. By doing this you have covered every possibility the machine can draw. You certainly won't make a profit though.

There seems to be enough information now. It's just too late at night for me to be thinking about this stuff. I know there is some formula for finding the number of possibilities for choosing a line of 6 numbers from 49 numbers. If someone could post relativly soon so I could get an answer that would be appreciated.

So 6 numbers are randomly drawn from the numbers 1 through 49. Does the order in which they are drawn matter? Are the numbers replaced after being drawn? (I would asume not) I only ask because my initial guess is that there are 1.7x10^9 possibilities for lines, meaning you will have to make more than 168 lines to ensure a winning ticket. For the first number drawn, you have one number and the pool has 49, giving you a 1 in 49 chance of getting the number. On the second number you have a 1 in 48 chance as the pool decreases by one. I am pretty sure your odds are multiplied in a situation like this. So the result would be: 49x48x47x46x45x44 which equals about 1x10^10. Your odds are now 6 in 1x10^10 or 1.7x10^9 This is just an idea though. It makes sense to me right now so I thought I would give my input.

I am not sure what you mean by "lines of numbers", perticularly in your last sentence.

You said -2 is prime. This is because its factors are 1 and -2 (itself). We can't include -1 and 2, just like we don't say the factors of 5 are -5, -1, 1, and 5. Therefore it is prime. Does that sound right? Also, does anyone know how they test a number such as blike's 2^20996011 - 1 .

blike, do you know how they test numbers like that. dave, that is what I thought.

Why do lists of primes always begin with 2?

Can negatives be prime?

haha

1 seems to fit all the rules for being a prime number. A search (on google) continualy says the definition of prime numbers includes only those greater than one, however. I'm not sure why, but 1 isn't prime.

It was a good analogy. I wasn't aware he was hit in the neck, that's all. If he wasn't, it wouldn't make sense, but then again I haven't seen too much about it to know.

Oh wow, I think my IQ has just dropped below 140.

I apologize Blike. That was typed as a joke without my knowledge or consent. Just the friend who knew the engineer was a litttle loco . Seriously though, what is zetalk?