This probability tree represents the game and guides the calculations:
The blue node at the left represents the start. At this point there is a 5% chance of success (leading to the up and left). If success is achieved now, only one attempt is made, as indicated in the orange circle.
Lacking success, we progress down and to the right to the next blue node. This time the chance of success is 10% and the chance of failure 90%. And so on, up to the 20th blue node where the chance of success (after 20 tries) equals 100% (and so there is no possibility of failure).
By axiomatic laws of probability, the chance of arriving at any terminal orange node is the product of the chances along the edges leading to that node. For instance, the chance of success in exactly two attempts is 95% times 10%, equal to $19/200,$ and the chance of success in exactly three attempts equals 95% times 90% times 15%, equal to $513/4000.$
By carrying out all the multiplications we compute the chances of success after exactly $k$ attempts for $k=1, 2, \ldots, 20$:
$$\frac{1}{20},\frac{19}{200},\frac{513}{4000},\frac{2907}{20000},\frac{2907}{20000},\frac{26163}{200000},\ldots,\frac{14849255421}{640000000000000000}.$$
From these values all quantities of interest may be calculated. For instance, the chance of success after four attempts is $\frac{1}{20}+\frac{19}{200}+\frac{513}{4000}+\frac{2907}{20000} = \frac{2093}{5000} = 0.4186$ and the chance of success after five attempts is greater by $\frac{2907}{20000}$ again, giving $\frac{11279}{20000} = 0.56395.$ Therefore the median number of attempts lies between four and five.
The expected (mean) number of attempts is (by definition) obtained by multiplying the number of attempts by its chance and summing over all possible numbers. The value is $\frac{3387894135040576041}{640000000000000000}$, approximately equal to $5.29358$. Its reciprocal, approximately $18.8908$%, answers the last question: the fixed chance of success with the same expected number of attempts.
A similar probability tree can help answer any similar questions where the value of $5$% may differ and can even change from one attempt to the next: just write in the appropriate probabilities and do comparable calculations.
As other people have pointed out in comments, the correct answer to the question "what is the probability of rolling another 6 given that I have rolled a 6 prior to it?" is indeed $\frac{1}{6}$. This is because the die rolls are assumed (very reasonably so) to be independent of each other. This means that past rolls of the die does not affect future die rolls.
Expressed mathematically, independence of two variables $X$ and $Y$ imply that $Pr(Y=y | X = x) = Pr(Y = y)$.
Letting $X$ be a variable denoting the outcome of the first die roll and $Y$ be a variable for the second die roll, we can use the definition of independence to arrive to the conclusion that $Pr(Y=6 | X = 6) = Pr(Y = 6)=1/6$.
The reason that the answer is not 1/36 is due to the fact that we are making a conditional statement. We are saying "given that we already have rolled a six in the first roll". This means that we are not interested in the likelihood of that first roll occuring. We are only interested in what happens next.
It might be helpful to enumerate all possible outcomes here. I have done this below in the form {x, y}, where x is the outcome in the first roll and y in the second.
{1, 1} {1, 2} {1, 3} {1, 4} {1, 5} {1, 6}
{2, 1} {2, 2} {2, 3} {2, 4} {2, 5} {2, 6}
{3, 1} {3, 2} {3, 3} {3, 4} {3, 5} {3, 6}
{4, 1} {4, 2} {4, 3} {4, 4} {4, 5} {4, 6}
{5, 1} {5, 2} {5, 3} {5, 4} {5, 5} {5, 6}
{6, 1} {6, 2} {6, 3} {6, 4} {6, 5} {6, 6}
Now, the probability you are interested in is the event {6, 6}. If you give the information that you are in the last row (which corresponds to having rolled a 6 in the first roll), you only have six possibilities of outcomes. Only one of them is a "success", so the probability of that event is 1/6.
Edit:
After re-reading the OP's question, it appears that I have missed part of the question. The question there seems to be regarding the following scenario:
- A six-sided die is rolled.
- If the die rolled a 6, roll a second die. Otherwise, do not roll a second die.
The question is there: What is the probability that this procedure results in two sixes having been rolled? Equivalently: What is the probability that this procedure results in us rolling a six in step 2?
The answer to this question is indeed 1/36. Heuristically, the reason for this is that we now are not conditioning on something that has happened anymore. We are instead asking for the probability of an event that can occur after we go through a procedure.
Let us now prove that the probability is 1/36. Letting once again $X$ be the result of the first roll and $Y$ the result of the second roll. We are interested in $Pr(Y=6)$. Note that if $X\neq 6$ then the probability that $Y=6$ is zero since the second die won't be rolled. Thus $Pr(Y=6\mid X\neq6)=0$. We use the law of total probability to note that $Pr(Y=6)=\underset{x=1}{\overset{6}{\sum}}Pr(Y=6 \mid X=x) \cdot Pr(X=x)$.
Now since $Pr(Y=6 \mid X=x)=0$ $\forall x\neq 6$, we see that
$Pr(Y=6) = 0+0+0+0+0+Pr(Y=6\mid X=6)\cdot Pr(X=6)$.
This simplifies to $Pr(Y=6) = \frac{1}{6}\cdot\frac{1}{6}=\frac{1}{36}$ which completes the proof.
Best Answer
You're looking for the binomial distribution, which has two parameters: n and p. The first is the number of rolls, the second is the probability that the desired event happens. So we have p locked to 0.1 (10%) and we can vary n.
The mean is np and the variance is np(1-p), and the standard deviation is the square root of the variance.
So there are two conflicting influences: as we increase the number of rolls, it becomes less and less likely that we'll get 10% exactly, because there are more and more possibilities. 1 out of 10 (39%) is more likely than 2 out of 20 (29%), which is likelier than 10 out of 100 (13%).
But as we increase the number of rolls, the square root (as a fraction of the mean) gets lower, so percentage of successes gets closer and closer to 95%; that is, if you build an interval around the mean that 95% of sample means will fall into, the lower and upper bound of that interval get closer and closer to 10%.
So get it exactly, the best shot you have is rolling 10 dice, and you're capped at 39%; if you want to get close, the more dice you roll the closer you'll get. (If you want to get at most a particular distance from the proportion with some probability, you can calculate how many dice you need to roll using the cumulative distribution function.)