Here's the question:
6 Dice are rolled. Which is more likely, that you get exactly one 6, or that you get 6 different numbers?
Here's what I've done:
The number of possible outcomes is $6^6 = 46656$.
The probability of rolling exactly one 6 = $\frac{1}{6}\times(\frac{5}{6})^5 = \frac{3125}{46656}$
The probability of getting six different numbers is:
$C(6,1)\times C(5,1)\times C(4,1)\times C(3,1)\times C(2,1)\times C(1,1) = \frac{720}{46656}$
Therefore if everything I've said above is true, then it is more likely that you will roll exactly one six. However I'm really not sure about the last part. Is this correct way to solve this type of problem and can the combinations part be simplified?
EDIT: Since I'm also looking for a better idea of how to solve this type of problem, rather than just this specific case, so could you please include how to solve this problem for rolling 5 dice, as well as/or instead of 6 dice in your answer, so that I can see the pattern of what is happening? Many thanks.
Best Answer
The question is very easy to answer without computing probabilities. Every combination with six different numbers contains exactly one six. There are then additional combinations which contain exactly one six - e.g. $111116$. So the probability of exactly one six is greater.