Rolling a Fair Die: Probability of Getting At Least One Six After Multiple Rolls

Amannequin
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Homework Statement
How many times do we need to roll a fair die to get a better than evens chance of at least one six?

The attempt at a solution
Let n be the number of rolls. Am I trying to find how large n must be so P(At least one six)≥ 1/2?
I'm working on the basis that I need to solve the probability of the union of events Ai, where Ai denotes the event of rolling i sixes, i =1,2,3..., being greater than 1/2.
Any help or nudge in the right direction will be appreciated.
 
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I'd attack it by finding the odds of NOT getting a six with one roll, then 2 rolls, then 3 rolls, and so on. As soon as that probability drops below 50% you've got the answer.
 
Amannequin said:
Homework Statement
How many times do we need to roll a fair die to get a better than evens chance of at least one six?

The attempt at a solution
Let n be the number of rolls. Am I trying to find how large n must be so P(At least one six)≥ 1/2?
I'm working on the basis that I need to solve the probability of the union of events Ai, where Ai denotes the event of rolling i sixes, i =1,2,3..., being greater than 1/2.
Any help or nudge in the right direction will be appreciated.

P(at least one six) = 1 - P(no sixes)
 
pasmith said:
P(at least one six) = 1 - P(no sixes)

Ah ... isn't that what I just said?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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