Rolling one die after another

Question

If I roll 1 6 sided die, the chances of getting one number, say 6, is 1/6 or 16.6%. If I roll 2 dice simultaneously, then the probability of rolling that number twice is 1/36 or 2.7% (1/6 x 1/6).

What if I have this situation: I roll a die, and if I get a 6, then I'll roll again. What are the chances of me getting a 6 the second time? I think it's also 1/36, but the difference is that I make the second roll if I get a 6 the first time. If I roll 2 dice simultaneously, let's say I differentiate them as the first and second die, the first die can be any number other than 6 while the second die still rolls. Is it still the same?

Is not getting the 6 the first roll, and not rolling the second die, and rolling two dice simultaneously and not getting two 6s the same level of failure in terms of probability? (even if 1 die rolls a 6 the other doesn't? Say I get a 6 on the second die but not the first).

Answer 1

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:

1. A six-sided die is rolled.
2. 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.