Probability problem driving me crazy

[DyslexicHobo]

So I'm going to the casino this weekend and it got me thinking about probability.

If I have an n-sided die and roll it n times, what are the chances of me hitting a specific number, x?

For example, if I flip a coin (n=2) twice, what are the chances of me hitting heads at least once?

Answer (obvious for n=2): 1-.5^2 = .75

It's obvious for most n's until they get large. I've simplified it to this equation:

1-((n-1)^n)/(n^n)

If I go back to n=2, we see that we arrive at the same answer of .75.

My TI-89 can calculate up to an n of 100, which gives a value of .6340.

I'm very interested in this value as n increases, though. Unfortunately I don't have access to math software that could solve the problem as a limit. How can I solve this, or more importantly... WHAT'S THE ANSWER?

It's driving me crazy! My best guess is that it tends towards 1/2.

Thanks for any help!

 

[Tedjn]

Close to 1/2. The actual answer is 1/e, which is slightly less. Details: http://en.wikipedia.org/wiki/E_(mathematical_constant)#Bernoulli_trials

 

[uart]

Close to 1/2. The actual answer is 1/e, which is slightly less. Details: http://en.wikipedia.org/wiki/E_(mathematical_constant)#Bernoulli_trials

Yes the \left(\frac{n-1}{n}\right)^n term aproaches 1/e so the limiting probability DyslexicHobo seeks is 1 - 1/e, which is about 0.632