Regarding combinatorics on dice...

Question

The problem:

If you are to roll a die five times, what is the probability the fifth die is not a repetition?

Apparently the answer is $$\frac{6\cdot5^4}{6^5}$$ but I don't really see why. What that answer says to me is that "We have 6 options for the first roll and then for each roll after that we have 5 options for each." That means, to me, that there can indeed be repetition on the 5th roll, which is not what we're going for! My original thought for this was to just opt out of repetition completely, but that leaves only one of the cases. Can anyone help me understand the logic behind this problem?

Answer 1

There are $6^5$ possible sequences of die rolls, which explains the denominator. Now, given a number from 1 to 6, there are $5^4$ ways we can roll the die 5 times and get this number last and ONLY last (each of the first 4 rolls has 6-1=5 possibilities). So the number of rolls where the last is not a repeat is $6\cdot5^4$, explaining the numerator.

Answer 2

I think this question can be interpreted as: If you roll a die 5 times, what is the probability that the first roll is different from the other 4 rolls? Now, imagine rolling the first die. You will get a value from 1 to 6, call it $a$. Then, the probability that the second roll isn't $a$ is $\frac56$. Assuming each roll is independent, the probability that the third roll isn't $a$ is also $\frac56$. Continuing this way we see that the probability we wish to calculate is equal to $(\frac56)^4$. Can anyone correct me if I'm wrong?