One Of Us Has To Be Bigger!

The possible outcomes are 36 :

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

One of the "diagonals" shows there are six draws , which means there is no winner (1-1, 2-2, 3-3, 4-4, 5-5, 6-6).

There are 30 possibilities left, 15 of which have you as a winner (the second number is greater than the first one). Therefore, in the 15 left cases, I am the winner.

The probability of you getting a grater number than I is 15/36 = 5/12.

Note that $P(\text{your number } > \text{ my number}) = P(\text{your number } < \text{ my number})$ , since in any situation where your number is bigger than mine, swapping the two numbers would yield a situation where your number is smaller than mine.

Furthermore, $2P(\text{your number } > \text{ my number})$ $= P(\text{your number } > \text{ my number}) + P(\text{your number } < \text{ my number})$ $= 1 - P(\text{both numbers same})$ since the three events are mutually disjoint. Thus, to compute $P(\text{your number } > \text{ my number})$ , it will suffice to compute $P(\text{both numbers same})$ .

But $P(\text{both numbers same})$ is just $1/6$ ; whatever you roll, the probability that I roll the same thing is $1/6$ . Thus, we conclude $P(\text{your number } > \text{ my number}) = \frac{1}{2}\cdot \left( 1-\frac{1}{6} \right) = \frac{5}{12}.$

For the first dice, probability is 1/6. Now if the number is 1, we have 5 possible outcomes for second dice, if number is 2, we have 4 possible outcomes for second dice and so on. So the probability is 1/6(5/6+4/6+3/6+2/6+1/6)

The probability of me rolling any $j^{th}$ number is obviously $\frac{1}{n}$ , where $n$ is the number of sides my die has. The probability of rolling another, $k^{th}$ , number from $j+1$ to $n$ , given $j$ , is:

$P(k > j | j) = \frac{1}{n}$ $\frac{n - j}{n} = \frac{n-j}{n^2}$

, since the rolls do not depend on each other. We then do the sum from $j = 1$ to $j = n-1$ (because if we went to $n$ , the final numerator would just be $0$ anyway):

$P(k > j) = \sum_{j = 1}^{j = n-1} \frac{n-j}{n^2} = \frac{n-1}{2n}$

Assuming this is a 6 sided die (although not explicitly stated in the question), we plug in $n = 6$ and attain $\boxed{\frac{5}{12}}$ .

Probability is $\frac{15}{36}$ or $\frac{5}{12}$

I'm curious...what if we were each throwing two dice, recording their respective sums, and finding the probability of one person having a greater sum?