Suppose we both play a game with two fair $D6$. Suppose I roll once, and you get to roll twice and select the maximum of the two rolls. In the result of a tie, you win. If the person with the highest number wins, what is the probability I win?
Probability of tie is $\frac16$, so probability my roll is greater than any one of your given rolls is $\frac5{12}$. For me to win my roll must be strictly greater than both of your rolls so $p = \frac{25}{144}$ Is this logic valid?
On
When you roll a 1, $P(Win|\text{roll 1}) = 0$. This is because your friend can only roll a 1 or larger. If they roll a 1, you tie. If they roll anything other than a 1, you lose.
When you roll a 2, $P(Win|\text{roll 2}) = \frac{1}{36}$. You only win if your friend rolls both 1's. Any other combination results in a tie or loss.
When you roll a 3, $P(Win|\text{roll 3}) = \frac{4}{36}$. You win if your friend rolls both 1's, a 2 then a 1, or a 1 then a 2, or both 2's.
We can keep this pattern going. In fact, it can be generalized to say if you roll $X$, the probability that you win is $$P(Win|\text{roll X})=\frac{(X-1)^2}{36}$$ for that specific event.
Therefore, we have the following probabilities:
When you roll a 4, $P(Win | \text{roll 4}) = \frac{(4-1)^2}{36}=\frac{3^2}{36}=\frac{9}{36}$.
When you roll a 5, $P(Win | \text{roll 5}) = \frac{(5-1)^2}{36}=\frac{4^2}{36}=\frac{16}{36}$.
When you roll a 6, $P(Win | \text{roll 6}) = \frac{(6-1)^2}{36}=\frac{5^2}{36}=\frac{25}{36}$.
We can use conditional probability to find $P(Win)$, this is because we have $P(Win|roll X)$ for every possible event $roll X$. We can say $$P(Win)=\sum_{}P(Win|\text{roll X})P(\text{roll X})$$ $$P(Win)=P(Win | \text{roll 1})+P(Win|\text{roll 2}) + ... + P(Win|\text{roll 6})$$ $$P(Win)=(0)(\frac{1}{6})+(\frac{1}{36})(\frac{1}{6})+(\frac{4}{36})(\frac{1}{6})+...+(\frac{16}{36})(\frac{1}{6})=\frac{55}{216}$$
You win when your roll is strictly greater than either of the other player's two rolls. If you roll $x$, then the probability of this happening is $(x-1)^2/36$; and the probability of you rolling any particular $x\in\{1,2,3,4,5,6\}$ is just $1/6$. So the probability that you win is $$ \frac{0^2+1^2+2^2+3^2+4^2+5^2}{216}=\frac{55}{216} \approx 25.5\%. $$ So you're right, it's not fair... if you change the rules so you win in the case of a tie, then your win probability becomes $42.1\%$, which is an improvement.