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

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3 Answers

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:

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

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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 $$ \frac{6!}{6^6} \approx .015. $$

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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 $6/6\cdot 5/6\cdot 4/6\cdot 3/6\cdot 2/6\cdot 1/6$.

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protected by Community Apr 16 '13 at 17:31

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