The probability of rolling any particular number with one die is 1 in 6, so if you've managed to throw a 3 on your first throw the chance remains 1 in 6 that you'll throw a 3 on the next throw of the die

The 1/6 probability is correct but consecutive probabilities require multiplication as these probabilities are not considered independent events.

1/6 * 1/6

**= 1/36**

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Before I explain the steps, if you'd like, you may get some review on how to deal with problems like these through this video here:

For my personal explanation, read on.

Let's analyze that question piece by piece. A "fair die" is a cube with six faces - each one labeled as either 1, 2, 3, 4, 5, or 6, depending on the number of dots located on that face.

"Consecutive" in this case means to roll the die once, get whatever number it lands on, then re-roll the same die and get the second number it lands on. And in this case, they both landed on 3 and it's asking for the probability of that.

So let's ask ourselves, what is the probability of the die to land on 3 if the die were to be rolled only once? There are six possible sides, and it lands on only one side. Landing on a 3 is considered __ one__ side out of the

__. So the odds are 1/6.__

**possible six**Now we're going to do the same thing, but for the second roll. What are the chances of the die landing on any side, even 3, out of the possible six? 1/6 again, because we're using the SAME die that has 6 sides, and whatever number it lands on, there will always be a 1/6 chance that it lands on that number.

The die lands on 3 on the first roll - 1/6 chance

The die also lands on 3 on the second roll - 1/6 chance

All you have to do here is multiply the two fractions, because the probability of rolling a 3 twice is (1/6) * (1/6).

(1/6) * (1/6) = 1/36

**Therefore, the probability of rolling a 3 on two consecutive rolls of a fair die is 1/36.**

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