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

Question

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?

Answer 1

Probability to get $1$ in first dice is $\dfrac{1}{6}$ and then to get higher on second dice is $\dfrac{5}{6}$. so, probability for first case is $\dfrac{1}{6}\dfrac{5}{6}$. So are the other cases.

So, probability
$\dfrac{1}{6}\dfrac{5}{6}+\dfrac{1}{6}\dfrac{4}{6}+\dfrac{1}{6}\dfrac{3}{6}+\dfrac{1}{6}\dfrac{2}{6}+\dfrac{1}{6}\dfrac{1}{6}$

Answer 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

Answer 3

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 $$\frac 12\times \frac 56=\frac 5{12}$$

Answer 4

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.

Answer 5

When rolling two dice there are $36$ possible outcomes. Six of these outcomes are draws (1 followed by 1, 2 followed by 2 etc), which leaves 30 outcomes that aren't draws.

Of these 30 remaining outcomes, 15 have the first roll higher and 15 have the second roll higher because of the symmetry.

Thus the probability of the first roll being higher is $\frac{15}{36}$.

Answer 6

Okay:

If the first die is a $1$ the probability of the second die being higher is $5/6$. But that is the probability only if the first die is $1$.

If the second die is $2$ the probability of the second die being higher is $2/3$. But again that is only the probability if the first die is $2$.

And so one. You have six different probabilities, form $5/6$ to $0$ depending on the what the first die is.

As the first die is equally likely to be any one of those, and as it must be one and only one and never more than one of those, then the total probability of the second die being higher than the first is THE AVERAGE of all the probabilities.

To find the average probability add up the probabilities and divide by the equally likely possibilities of the first die. i.e. $\frac 16[5/6 + 2/3 + 1/2 + 1/3 + 16 + 0] = 5/12$.

or..

(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)

(2,1)(2,2)(2,3)(2,4)(2,5)(2,6)

(3,1)(3,2)(3,3)(3,4)(3,5)(3,6)

(4,1)(4,2)(4,3)(4,4)(4,5)(4,6)

(5,1)(5,2)(5,3)(5,4)(5,5)(5,6)

(6,1)(6,2)(6,3)(6,4)(6,5)(6,6)

Probability is: $P = 15/36$

$P = \frac {5/6+2/3+1/2 + 1/3 + 1/6 + 0}6$

$P = \frac {5+4+3+2+1+0}{36}$

$P = \frac 16*\frac 56 + \frac 16*\frac 23+\frac 16*\frac 12+\frac 16*\frac 13+\frac 16*\frac 16 + \frac 16*0$

They are all equal ways of thinking of it.

Answer 7

what is the probability that the second die gives me a higher number than the first die , i.e. $P(R_{2} < R_{1}) = ? $

$P(R_{2} < R_{1}) + P(R_{2} > R_{1}) + P(R_{2} = R_{1}) = 1 $

By symmetry ,

$P(R_{2} < R_{1}) = P(R_{2} > R_{1}) = x(say) $

Then, $ 2x + P(R_{2} = R_{1}) = 1 => x = P(R_{2} < R_{1}) = \frac{1 - P(R_{1} = R_{2}) }{2} = \frac{1 - \frac{1}{6}}{2} = \frac{5}{12}(Ans) $

Answer 8

Just offering a simple visual representation here;

There are 6x6 possible combinations of pairs that can be arranged in a 6x6 grid. Only the upper triangular portion of the grid corresponds to the second dice having a higher number than the first in this case.

The prob is the ratio of the area of this triangular section to the total area of the grid (36) = 15/36 = 5/12.