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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 $2 < n < 7$?

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Probably all the answers you are going to see here have something like "Assuming fair rolls" or something similar. If you are using GW dice you should correct for the deviation ;P… – Achifaifa Feb 13 '14 at 22:28
Wow I had no idea there was such a huge difference... My friends and I were all stunned at the sheer number of 1's we roll sometimes... guess that explains it! – thanby Feb 17 '14 at 13:20
@thanby the numbers do not make sense to you because the probability to get at least one six is surely less than 1: $P(\mbox{"at least one 6"}) = 1 - P(\mbox{"not any 6"}) = 1 - (\frac{5}{6})^n$, for $n$ dices. This explains that small chance ;) – marc Feb 17 '14 at 23:48
Oops my last comment made no sense so I had to delete it >.< Okay this works now, I just calc'd the table again using that formula and it looks a lot better. Though I would have expected there to be a higher chance of rolling a 6 with 6 dice, it works out to be 66.51% I suppose that's right though, n'est pas? – thanby Feb 18 '14 at 15:29
up vote 2 down vote accepted

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 $\frac{5}{t} \leq \left(\frac{7-3}{6}\right)^4$

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Wow the formula I used to determine the ranges is actually right, I just never knew how to write it out. For n=6 I did (1/6)^6 but for some reason it never occurred to me that the power I was raising it to should actually just be n, for each step. Like I said, I'm pretty green when it comes to math :/ The time function you explain makes perfect sense as well, I think I can use that to finish the rest of the chart. Thanks for the help! – thanby Feb 17 '14 at 13:31

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