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!

  • $\begingroup$ 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

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 $1/6^6$.


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.