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If I have 6 regular dice, (each numbered 1-6):

  1. What is the probability that when rolled that each will be a different number.(each individual di is a different number from 1-6, but a random order)

  2. What is the probability that of the six rolls, each will be an increasing number starting with 1. (First roll is 1, second is 2, etc.)

I came up with this problem after watching an episode of numberphile. Thanks!

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  • $\begingroup$ Hint: the second part is fairly easy! (each desired event is independent of the others and each occurs with probability $\frac{1}{6}$). How does the first part differ from the second? $\endgroup$ – lulu Jul 7 '15 at 1:00
  • $\begingroup$ The first part is a random, but still different numbers, and the second is a specific, increasing, order. $\endgroup$ – K. W. Cooper Jul 7 '15 at 1:06
  • $\begingroup$ have you tried to count the favourable cases against the possible ones? in many circumstances (as in this one) counting is a way to go $\endgroup$ – Conrado Costa Jul 7 '15 at 1:07
  • $\begingroup$ The probability of throwing {1,2,3,4,5,6} in that order is the same as the probability of throwing exactly {2, 1, 3, 4, 5, 6}, say. So all you need is the probability of getting one particular ordering and the number of orderings. $\endgroup$ – lulu Jul 7 '15 at 1:10
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There are $6^6$ ways of throwing the dice in total. There are $6!$ ways of throwing all the numbers from 1 to 6 in any order. There is only one way of throwing $1,2,3,4,5,6$.

Hence your first answer is $6!/6^6$, and your second is $1/6^6$.

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For a hand-waving solution to Part 1:

Roll the first die; any number is acceptable; probability is $\frac66$ or $1$

Roll the second die; only five acceptable numbers left; probability is $\frac56$

Roll the third die; only four acceptable numbers left: probability is $\frac46$

More of the same for the last three dice; multiply each probability together, giving$$P_{different}=\frac{6\times 5\times 4\times 3\times 2\times 1}{6\times 6\times 6\times 6\times 6\times 6}$$which of course matches the correct answer of @Dr Xorile

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