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I found the probability of having the same number when throwing 3 dice to be $1\times\left(\frac16\right)^2$.

In addition, I don't understand how do people get the equation $\left(\frac16\right)^3\times6=\frac1{36}$, like why do we have to multiply $\left(\frac16\right)^3$ by six?

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up vote 4 down vote accepted

Just a small supplement to the nice answers addressing your question, since I didn't found it mentioned explicitly.

We can calculate the probability of an event by calculating the ratio of the number of favorable choices or number of successes of an event with the number of all choices of an event.

This way we obtain \begin{align*} \frac{\text{number of successes}}{\text{number of all choices}}=\frac{6}{6^3}=\frac{1}{6^2}=\frac{1}{36} \end{align*}

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The probability that the first die is a 1 is $\frac16$. The probability that the second die is a 1 is $\frac16$. The probability that the third die is a 1 is $\frac16$. Multiplying this together gives $\frac1{6^3}$ for the probability that all dice are a 1.

We have the same chance $\left(\frac1{6^3}\right)$ of rolling three 2s, three 3s, three 4s, three 5s, or three 6s, so the probability of rolling any number on all three dice at the same time is $6\cdot\frac{1}{6^3}$, or $\frac1{36}$.

Also note that both your answer $\frac{1}{6^2}$ and $6\cdot\frac{1}{6^3}$ are equal to $\frac1{36}$.

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By counting: There are $6^3$ possible outcomes: 1-1-1, 1-1-2, ..., 6-6-5, 6-6-6. There are $6$ outcomes with all the numbers the same: 1-1-1, 2-2-2, ..., 5-5-5, 6-6-6. So the probability of all the numbers being the same is $6/6^3$.

Alternatively: Throw the first die. It doesn't matter what number you get.

Throw the second die. The probability of getting the same number is $1/6$.

Throw the third die. The probability of getting the same number is again $1/6$.

So the probability of three numbers the same is $1/6 \times 1/6$.

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FYI: you can use $\frac{num}{den}$ to render a fraction vertically. – Mark Reed Feb 27 at 21:20

Are you're asking for the probability of all three dice having the same number, regardless of which number it is?

Imagine rolling the dice one at a time. We don't care which number we see, so we don't care about the outcome of the roll of the first die; it just tells us what we're looking for.

Whatever that number is, the chances that the second die comes up on it is $\frac16$.

The chances that the third die comes up with it is also $\frac16$.

So the chances that both the second and third dice come up with the same number that's on the first is $\frac16$ × $\frac16$ = $\frac1{36}$.

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