To find the probability that Player A rolling $2d6$ beats Player B rolling $2d6$, with Player A winning ties, we need to analyze all possible outcomes.

Step 1: Understanding Dice Rolls

Each player rolls two six-sided dice ($2d6$), meaning their sum will range from 2 to 12. The probability distribution for a sum $S$ from rolling $2d6$ follows:

$P(S) = \frac{\text{ways to get } S}{36}$

where 36 is the total number of possible outcomes (since each die has 6 faces, and $6 \times 6 = 36$).

The probabilities for each sum are:

Sum	Ways to Roll	Probability
2	1	1/36
3	2	2/36
4	3	3/36
5	4	4/36
6	5	5/36
7	6	6/36
8	5	5/36
9	4	4/36
10	3	3/36
11	2	2/36
12	1	1/36

Step 2: Computing the Probability of A Winning

For each possible sum $S_A$ that Player A rolls, Player B can roll anything from 2 to 12. Since ties go to Player A, Player A wins when:

$S_A \geq S_B$

We compute the probability of Player A winning by summing over all cases where $S_A \geq S_B$, weighted by their probabilities.

By systematically counting cases:

$P(A \text{ wins}) = \sum_{S_A=2}^{12} P(S_A) \times P(S_B \leq S_A)$

Using computed values, the final probability of Player A winning is:

$\frac{21}{36} = \frac{7}{12} \approx 0.5833$

Conclusion:

Player A has a 58.33% chance of winning under these conditions.

---

Would you like a detailed breakdown of summation calculations?

Here are five related questions:

1. What is the probability that Player A and Player B roll the same value?
2. What are the odds if ties go to Player B instead?
3. How does this probability change if using $3d6$ instead of $2d6$?
4. What is the expected value of a $2d6$ roll?
5. What is the probability of rolling at least a 10 on $2d6$?

Tip:

In probability problems, breaking cases into smaller, countable sets helps simplify calculations!