A balanced die has six sides: $1,2,3,4,5,5$. If the die is tossed five times, what is the probability that the numbers recorded are $1,2,3,4,5$, in any order?

So for this problem, the order does matter and we have repetition so this is a permutation with repetition.

$N$ (total possible outcomes) is $5^5$ (not $6^5$, because there are two $5$ sides). and $n_A$ (number of favorable outcomes) is $\dfrac{6!}{2!}$ (number of different anagrams of $1,2,3,4,5,5$). So $P = \dfrac{n_A}{N} = \dfrac{72}{625}$.

This answer is apparently wrong, but I have no idea why. What did I do incorrectly?


2 Answers 2


Consider the balanced die with sides: $1,2,3,4,5_1,5_2$.

Then the number of ways to obtain $1,2,3,4,5_1$ or $1,2,3,4,5_2$ in any order is $2\cdot 5!$.

On the other hand the possible results are $6^5$ (in this counting we are considering $5_1$ and $5_2$ as two different values).

Hence the required probability is $$p=\frac{2\cdot 5!}{6^5}=\frac{5}{162}.$$


Your version of $n_A$ counts permutations such as $12355$ which we don't want.

Instead try $2\cdot5!$, which counts permutations of $12345$ and twice to count permutations of $12345_1$, with $5_1$ representing the 'other' $5$.

Also, this means the number of possible outcomes, $N$, is $6^5$ because the dice are effectively labeled $1,2,3,4,5,5_1$.

This gives $\dfrac{240}{7776}=\dfrac{5}{162}\approx0.0309$.

  • $\begingroup$ Sounds logical. My book gives the answer of $\dfrac{5}{162}$ though. $\endgroup$
    – Maxime
    Sep 14, 2016 at 8:59
  • $\begingroup$ I think that you have to replace $5^5$ with $6^5$. See my answer. $\endgroup$
    – Robert Z
    Sep 14, 2016 at 9:06

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