In the realm of statistical analysis I'm about as green as they come, so forgive me if my question isn't stated perfectly, I'm trying my best to explain. There's a table-top game my friends and I play called Warhammer (40k for those who know the difference) that involves a lot of 6-sided die rolling. While trying to plan out a strategy I became interested in the probability of certain outcomes, and got sucked into what players of this game refer to as the dreaded "Math-hammer" and decided to build a spreadsheet of the statistics involved. So far I've muddled my way through it to come up with some fairly accurate numbers, but now I'm to a challenging section that I could use some advice on. Here's the scenario...
Part 1: Suppose you have n 6-sided dice (n being between 1 and 6). You roll them all at once with the goal of getting them all to be equal to or higher than a target number x (between 2 and 6). What are the chances of meeting that goal at least once for each possible combination of x and n? Bonus: How many times will you meet that goal for each of those possible outcomes?
I've established the possible ranges for meeting the goal of x at least once, but am having trouble filling in the blanks and figuring out how to express the bonus part of the question. Here's an example of what I have so far (rounded for simplicity):
For n = 6
1/6 2/6 3/6 4/6 5/6 6/6
x = 6 99+% ? ? ? ? ~0%
x = 5 99+% ? ? ? ? 0.14%
x = 4 99+% ? ? ? ? 1.56%
x = 3 99+% ? ? ? ? 8.78%
x = 2 99+% ? ? ? ? 33.49%
Part 2: Certain conditions allow you to re-roll once any dice that did not meet or exceed x on the initial roll. I have no idea how to express this. Thoughts?
Edit: Using the example provided by @marc I've attempted to solve the table above but have run into a couple questions that I can't seem to wrap my head around. The thought process I used was "Each die has a 1/6 chance of rolling a 6, but this should be multiplied by the number of dice you are throwing at once because they each have that same chance. So 6(1/6) = 100% chance of getting at least one 6." I know from experience this is definitely not the case, but I have no idea why or how to express this. Surely there's a small chance that this will not happen, but I don't see how to adjust the equation for this. The equation $n(\frac{7-x}{6})^n$ makes sense to me but I know after working out the table that it's wrong.
Edit 2: Using $P($"at least one 6"$)=1−P($"not any 6"$)=1−(\frac{5}{6})^n$ (as provided by @marc again) I can see the actual chance of getting at least one 6 when rolling $n$ dice. How do I adjust this to account for rolling two 6's when $2 < n < 7$?
Given $n$ fair $6$-sided dices, and fixed $2 \leq x \leq 6$, if you roll them at once the probability to get a number not less than $x$ on each of them is $\left(\frac{7-x}{6}\right)^n$. If you want to know how many times, on average, you'll meet a certain event, that depends on the number of times you roll the dices. For example, if you'd like to expect to see, on average, at least $t = 5$ times the event "throw 4 dices and get a number $\geq 3$ on each of them" you should solve $\frac{5}{t} \leq \left(\frac{7-3}{6}\right)^4$