Odds of it happening vs When it happens on average. Okay, so I have a 6 sided dice... my odds of rolling a "6" is 1 in 6, right? (Side-question: is this "odds" or "probability", I don't know the difference between the two)
Anyways... My main question is... if my odds for rolling a "6" goes, that on average 1 in every 6 rolls will be a "6"... Then what are my odds for when this will actually happen. On average, will I roll that 1-in-6 roll after about 3 rolls? Surely if I ran this experiment 1000 times I wouldn't roll the 6 on my 6th roll most of the time right? So are my odds of rolling a "6" 1-in-6, and my typical-ness for it happening being that of the 3rd roll?
I hope this question makes sense... I'm having a hard time putting it into words haha. Essentially if I were to roll a dice until I landed on a "6"... on average, how many rolls would I have to make? And what is the formula for figuring this out? Is it just the average of the lowest amount of rolls possible vs the highest amount of rolls possible? If that were the case wouldn't it be the 3.5th roll? $(1+6)/2$
 A: Note that 6 isn't the highest number of rolls possible: it's possible (though unlikely) to roll, say, thirty 5s in a row, and then a 6.
To get a 6 for the first time on the $k$th roll means you get something else $k-1$ times (a 5-in-6 chance, $k-1$ times), and then a 6 (a 1-in-6 chance); that has probability $(\frac56)^{k-1} \frac16$.  This is a geometric distribution with $p=\frac16$.  The expected value (that is, the average number of rolls before you see a 6 for the first time) is $\frac1p$, that is, 6.  The median number of rolls, though, is about 3.8.
A: You need to think more clearly about what question you want to ask before a good answer can be given.  The distribution of probabilities for the first roll of a $6$ is $\frac 16$ for one roll, $\frac 56 \cdot \frac 16$ for two rolls (because you have to not roll a $6$ on the first roll, then roll one on the second), and $(\frac 56)^{n-1}\cdot \frac 16$ for $n$ rolls.  From that, you can ask how many rolls give you more than $50\%$ chance of getting a $6$ (four of them), the expectation value of the number of rolls to get a $6$ (six) or others.  There is no maximum-you are not guaranteed a $6$ after any finite number of rolls.
