# Probability of All Distinct Faces When Six Dice Are Rolled

If six fair dice are rolled what is probability that each of the six numbers will appear exactly once?

Imagine you throw one after the other. You consider a throw as a success if the number is different from all previous numbers. You start with one. This is always a succes so $P(\text{first}) = 1 = \frac{6}{6}$. Your second throw is a success if one of the remaining $5$ numbers shows, so $P(\text{second}) = \frac{5}{6}$. And so on. Since all the throws are independent, the total probability is the product of all separate probabilities:

$P(\text{all numbers are different}) = \frac{6}{6} \cdot \frac{5}{6} \cdot \frac{4}{6} \cdot \frac{3}{6} \cdot \frac{2}{6} \cdot \frac{1}{6}$

Pretend the dice are distinguishable. There are $6!$ ways the desired numbers can appear (any permutation of $\{1, 2, 3, 4, 5, 6\}$ ). There are $6^6$ total possible rolls, so the probability is $$\frac{6!}{6^6} \approx 0.015.$$

The probability of getting each number is $1/6$.

$1$ dice: The number have $6$ choices to fill, therefore $6\cdot 1/6$.

$2$ dice: Now the number has already come on the first dice so the second number has only $5$ choices, $5\cdot 1/6$

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Similarly finally the equation is $6/6\cdot 5/6\cdot 4/6\cdot 3/6\cdot 2/6\cdot 1/6$.

## protected by Community♦Apr 16 '13 at 17:31

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