MHB Probability of Rolling Sum > 3 with Two Dice

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When rolling two fair dice, the total number of possible outcomes is 36. The sums that exceed 3 are 4 through 12. There are 34 outcomes that yield a sum greater than 3, meaning the probability of rolling a sum greater than 3 is 34 out of 36. This simplifies to a probability of 17/18. Thus, the probability of rolling a sum greater than 3 with two dice is 17/18.
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Two fair dice are rolled. What is the probability of rolling a sum that exceeds 3?
 
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skeeter said:
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is it 1/36
 
a sum that exceeds 3 is a sum $\ge$ 4
 
When rolling two dice the possible sums are 1+ 1= 2 to 6+ 6= 12. The specific possible outcomes are
1+ 1= 2
1+ 2= 3
1+ 3= 4
1+ 4= 5
1+ 5= 6
1+ 6= 7
2+ 1= 3
2+ 2= 4
2+ 3= 5
2+ 4= 6
2+ 5= 7
2+ 6= 8
3+ 1= 4
3+ 2= 5
3+ 3= 6
3+ 4= 7
3+ 5= 8
3+ 6= 9
4+ 1= 5
4+ 2= 6
4+ 3= 7
4+ 4= 8
4+ 5= 9
4+ 6= 10
5+ 1= 6
5+ 2= 7
5+ 3= 8
5+ 4= 9
5+ 5= 10
5+ 6= 11
6+ 1= 7
6+ 2= 8
6+ 3= 9
6+ 4= 10
6+ 5= 11
6+ 6= 12

A total of 6x6= 36 outcomes, not all different.

Now, how many of those "exceed 3" (i.e. are 4 or higher)? The probability of exceeding 3 is that number divided by 36.
 
You might find it simpler to count the number of rolls that give "2" or "3" and subtract that from 36.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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