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If I roll one $6$ 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 $2.7\% (1/6 \times 1/6)$.

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 $6$'s 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).

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This is very much not about the order in which you roll dice or whether you roll them at the same time or one after the other. It is about whether you examine the conditional probability or the overall probability.

Given die one shows a 6, die two has a 6-probability of $1/6$.

Not given that any one of them has a 6, the probability of them both getting 6 is $1/6\times1/6=1/36$

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Dice rolls are independent. The die doesn't "remember" if you just rolled a six (or any other number).

What this means is that the probability of a 6 on a single roll is always $\frac{1}{6}$.

The apparent contradiction in your question arises from conditional probability.

Conditional probability expresses the event "given that I just rolled a six, what's the probability that the next roll is a six?" We just established that there's no "memory" to this process, so the information "I just rolled a six" is irrelevant, so we can re-state the problem as "what's the probability that the next roll is a six?" without any loss of information. Therefore, the probability of rolling a six is (still) $\frac{1}{6}$.

But if you're rolling two dice at the same time, each die is independent of the other (the dice can't "see" what each other is doing), so the joint probability is $\frac{1}{6} \cdot \frac{1}{6} = \frac{1}{36}$.

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  • $\begingroup$ No I know that they are inherently independent and they can't "see" what the other is doing. But I'm asking if they did and if they were dependent. Let me rephrase the situation: When it comes to the numbers of rolls that can happen, it can be either one or two, I'll distinguish them as the first and second roll. I want a 6 on the second roll, but for there to be a second roll, I need to roll a 6 on the first roll. If I get any other number on the first roll, the second roll will not happen. Me getting a 6 on the second roll entirely depends if I get a 6 on the first die. $\endgroup$ – MarcLikesMath May 18 '18 at 16:44
  • $\begingroup$ This is the same as two dice rolled at the same time yielding two sixes. The probability that there is a second roll at all is dependent on a six, so 1/6. The probability of a six on the second roll is 1/6. The rolls are independent so we have 1/6 * 1/6 = 1/36. $\endgroup$ – Sycorax May 18 '18 at 17:59
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Here's a way to think about it that clearly shows that the results of rolling two simultaneously have the same probability as one after the other:

Since you asked specifically about the chance of failure, lets make that explicit: You correctly noted that the chance of success when rolling simultaneously is 1/36, so the chance of failure is 35/36.

When rolling the first die separately, there are 6 possible outcomes that could lead to failure. If you haven't failed already, then when you roll the second die, there are five possible outcomes that lead to failure. So your probability of failure is 6/6 * 5/6 which is also 35/36.

[Conversely, when you roll 1st, there are 1/6 possible results that could lead to success, and the same when you roll the second, so the chance of success is 1/6 * 1/6 = 1/36.]

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