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# What are the odds and probability of rolling 5 dice, and all 5 dice are the same number?

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16
May 7, 2018

$\frac{1}{6} ^ 4 = \frac{1}{1296}$

#### Explanation:

There are six faxes with six distinct numbers. For each dice, the
probability is (number of favorable cases)/(the total number of all possible cases).
Because any first roll is favorable, this probability can be modeled by:
=6*1/(6^(n) or $\frac{1}{{6}^{n - 1}}$

For all showing the same number, it is the compound probability
= product of the separate probabilities.
$= \frac{1}{{6}^{5 - 1}}$
$= \frac{1}{6} ^ 4$

Then teach the underlying concepts
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2
May 7, 2018

$P \left(\text{all 5 dice show the same number}\right) = \frac{1}{1296}$

Odds are $1 : 625$

#### Explanation:

Let's look at the probability first.

$P \left(5 \text{ dice give the same number}\right)$

They must be all $1 s$ or all $2 s$ or all $3 s$ ... and so on

$= P \left(1 , 1 , 1 , 1 , 1\right) + P \left(2 , 2 , 2 , 2 , 2\right) + \ldots \ldots \ldots + P \left(6 , 6 , 6 , 6 , 6\right)$

$\left(\frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} \times \frac{1}{6}\right) + \ldots . + \left(\frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} \times \frac{1}{6}\right)$

$= \frac{1}{6} ^ 5 + \frac{1}{6} ^ 5 + \ldots \ldots \ldots . + \frac{1}{6} ^ 5$

$= 6 \times \frac{1}{6} ^ 5$

$= \frac{1}{6} ^ 4 = \frac{1}{1296}$

The odds are given as a ratio of the number of ways this will happen compared to the number of ways it doesn't.

So if the first die is $1$, the other four will not be,

$6 \times \left(\frac{1}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6}\right) = \frac{625}{1296}$

The odds for this happening are $1 : 625$

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