Probability that my roll on a die will be higher than yours: Why divide by 6?

I have to work out a question where on two fair, $6$ sided dice, what is the probability that the second die gives me a higher number than the first die.

So I broke the question down the long way and said "If you roll a $1$, I have a $\frac{5}{6}$ chance of beating you, if you roll a $2$, I have a $\frac{4}{6}$ chance of beating you, etc."

Then I added all these up but got the answer to be $\frac{5}{2}$.

More research showed that I was actually supposed to divide this number by $6$ to get my probability to be $\frac{15}{36} = \frac{5}{12}$, but I can't see why you divide it by $6$.

Can someone explain this to me please?

EDIT: The bit I am struggling with is the fact that why do we take the draw into consideration.

Thank you to everyone who has commented and answered, I now get that the division by 6 is because we include the probability of drawing. But in my specific question, there was nothing about a tie, I simply have to beat you, or I lose.

So why do we still include the possibility of a draw, when that's not part of the game?

Solution 1:

The probability of a tie is $\frac 16$. Given that the dice don't match, there is a $\frac 12$ chance that the higher one is the second, so the answer is $$\frac 12\times \frac 56=\frac 5{12}$$

Solution 2:

There is 1/6 chance that you roll 1, chance to win in that case is 5/6

There is 1/6 chance that you roll 2, chance to win in that case is 4/6 etc.

Your total probability to win is 1/6*5/6 + 1/6*4/6 ... That's why you divide by 6. Even you think of your probability to win, it is greater than 1 what is impossible

Solution 3:

Per Markoff Chainz’ comment, you also have to account for the probabilities of each of your possible die rolls.

Every one of the cases that you considered consists of two events: “I rolled X” and “You rolled Y.” The probabilities of these two events have to be combined to find the probability that both of them occur. Since the events are independent (loosely, my rolling some number $X$ doesn’t affect what you roll), this probability is just the product of the individual event probabilities.

The probability of your rolling any particular number is $\frac16$, so the total probability is $$\frac16\cdot\frac56+\frac16\cdot\frac46+\frac16\cdot\frac36+\frac16\cdot\frac26+\frac16\cdot\frac16+\frac16\cdot\frac06=\frac16\left(\frac56+\frac46+\frac36+\frac26+\frac16+\frac06\right).$$ That is, the actual probability is $\frac16$ times the sum you computed. This is arithmetically the same as dividing by $6$, of course, but thinking of it that way obscures what’s going on, which is that you’re multiplying by the probabilities of all of your own die rolls.