Dicey

Note, you can count up even and odd sums. Thus 18 even suns, 18 odd sumes. So 50/50 chance

Whatever you roll on the first die, there is a 50/50 chance that the second roll will match it in parity (odd or even). If the parities match, the result will be even. Otherwise, the result will be odd. Therefore, there is a 50/50 chance their sum will be odd. (And, a 50/50 chance it will be even)

The thing is we need to have number of combinations which is 6^2(32) as there six different numbers. Now watch and observe closely. E=even and O=odd
1+1 E 1
1+2 O 2
1+3 E 3
1+4 O 4
1+5 E 5
1+6 O 6
2+2 E 7
2+3 O 8
2+4 E 9
2+5 O 10
2+6 E 11
3+3 E 12
3+4 O 13
3+5 E 14
3+6 O 15
4+4 E 16
4+5 O 17
4+6 E 18
5+5 E 19
5+6 O 20
6+6 E 21
You might see why it's only 21 but actually it is 36 because some of them are just their backwards. These are
2+1
3+1
4+1
5+1
6+1 and etcetera. Let me give you another one example.
1 - 1 2 3 4 5 6
2 - 1 2 3 4 5 6
3 - 1 2 3 4 5 6
4 - 1 2 3 4 5 6
5 - 1 2 3 4 5 6

6 - 1 2 3 4 5 6

 6 6 6 6 6 6 = 36 equal!!!!!

The correct solution is ... IT DEPENDS! If we assume that the question is about two fair dice ... each being D6 ... then both ODD and EVEN are equally likely... This is the correct result for a PAIR of EVEN sided dice However, if the question is about two fair dice ... each being a D5 ... then EVEN is more likely than ODD. ... This is always the correct result for a PAIR of ODD sided dice. There is NO STATEMENT here other than a PAIR... A PAIR explicitly means that both dice have the same number of faces. However; the question did not explicitly state how many faces ... other than to show a cartoon with a pair of D6's

The problem is only concerned with parity (odd vs even) of the sum. The parity of the sum is completely determined by the parities of the individual rolls, and their specific values do not matter. Instead of six-sided dice rolls, this observation allows us to think in terms of two-sided coin throws.

In fact, we can use this simplification regardless of the dice specifics (number of sides, (un-)fairness), as long as the "heads vs tails" probabilities of the imagined coins match the "odd vs even" probabilities of the original dice. For standard, fair six-sided dice, we can think of fair coins.

A fair coin does not care which outcome we call heads, and which one we call tails. All that matters is that there are two outcomes and that they are both equally likely. Considering this underlying symmetry in each individual coin throw, and how each one independently "toggles" the end result ("parity of the sum") in the same 50/50 way, the answer cannot possibly be "it's more likely that the sum is odd than even" (or vice-versa), since it never mattered which option we called "odd" and which "even" in the first place. This means both outcomes for the sum must be equally likely, regardless of the number of dice rolled/coins thrown.

The chances of the first die being odd are 50% and the chances of it being even are 50%. The same goes for the second die. We can list out the possibilities much like we could for an experiment involving flipping two fair coins. While we do this, we can list the outcome of each roll.

EE (Even + Even = Even)

EO (Even + Odd = Odd)

OE (Odd + Even = Odd)

OO (Odd + Odd = Even)

The outcome is odd in 2 out of 4 cases, so the probability of getting an odd sum is 1/2.

Note that this technique covers the sum of any two fair dice so long as each die has an equal balance between odd and even sides. Our six sided die has three odd sides and three even sides. We could apply this solution to rolling two fair twenty sided dice with the numbers 1-20 on them, as each die has ten even sides and ten odd sides.

We could even apply this to stranger, unfair dice as long as each die has a 50% chance of rolling an odd number and a 50% chance of rolling an even number.

Without loss of generality, let us assume the first dice roll is even. Then we need to find the probability of the next dice being odd and the probability of the next dice being even, since even + even = even and even + odd = odd.

But, since there are as many odd numbers as even numbers in the sample space of 1 to 6, they should have the same probability of showing up.

Thus, the total is equally likely to get an odd as an even.

If you draw a two way table to represent the sample space, you will see that there are 18 squares that contain odd numbers, and of course, 18 squares for even numbers, as the complement of odd is even. Therefore the probability to get an odd total the same as the probability of getting an even number.

There are 36 possible outcomes, each equally likely. Half of the outcomes are even and half of the outcomes are odd. Therefore, evens and odds are equally likely.

results = Tally[Flatten[Table[i + j, {i, Range[1, 6]}, {j, Range[1, 6]}]]] gives {{2, 1}, {3, 2}, {4, 3}, {5, 4}, {6, 5}, {7, 6}, {8, 5}, {9, 4}, {10, 3}, {11, 2}, {12, 1}}

They are both equally likely because... Exist equally amount odd number and even number in the dice.

There are two ways to get an even total. Either both die outcomes are even, or both are odd. The no. of ways both outcomes are even are 3×3=9 ; and that for both odd outcomes are 3×3=9 also; as there are three odd, and three even outcomes on both dice. Thus, total favorable outcomes is 9+9=18; and same is the case for odd totals. Thus both cases are equally favorable

Each dice have 3 odd and 3 even numbers so 50/50 chance.

Each dice has 3 even and 3 odd numbers so when you role them there is 50/50 chance of it being even or odd

Becuse if you count the fact that every number squared will equal an even number, then if you add up all of the numbers it will equal an odd number. Then if you double that you'll get an even number, then you'll have a 50/50 chance of rolling odd or even.

You know that 50% or the time one will be odd and 50% of the time it will be even, so instead of making a chart of all values you can make a chart of evens and odds, even + even = even, odd + odd = even, even + odd = odd, odd + even = even. Two for each, so it’s a 50/50 chance.

Each side has 6 possibilities. Broken dawn in conjunction with the question, the meaningful numbers are placed in two categories. Odd and even. Each number that has an odd pairing is put in one category, each number that has an equal pairing is put in another. As both categories have an equal grouping and the answer is the same for two roles. The answer is They are both equally likely.

When the first die is rolled there is a 50/50 chance that it will be odd or even,( there are three odd and even numbers). On the second roll, there is, again, a 50/50 chance that it will be odd or even. So, the total is equally likely to be odd as even.