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I want to calculate the chance for the following situation: You throw a die 5 times. How big is the chance to get the numbers "1,2,3,3,5" if the order does not matter (i.e. 12335 = 21335 =31235 etc.)?

I have 4 different solutions here, so I won't include them to make it less confusing. I'm thankful for suggestions!

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I would recommend that you do show your solutions. Also, keep in mind that a problem can have more than one way of solving it (correctly) and getting the (correct) answer, and so four solutions need not mean four different answers! –  Dilip Sarwate Jan 6 '12 at 21:41

2 Answers 2

up vote 3 down vote accepted

There are $5$ options for placing the $1$, then $4$ for placing the $2$, and then $3$ for placing the $5$, for a total of $5\cdot4\cdot3=60$. Alternatively, there are $5!=120$ permutations in all, and pairs of these are identical because you can exchange the $3$s, which also gives $120/2=60$. The total number of combinations you can roll is $6^5=7776$, so the chance is $60/7776=5/648\approx0.77\%$.

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If you already have 4 solutions, what is it that you seek ?

There is more than one way to skin a cat, and i would suggest that you use the one that strikes you as simplest.

To add to the list, here's one (hopefully not already there in exact form)

favorable ways = 5!/2! = A

total ways = 6^5 = B

Pr = A/B

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