thanks to all who have answered, let me explain the actual reason for the question ;
whilst in a pub at weekend they had a little quirk whereby after buying a round of drinks
you were offered the opportunity of rolling two dice, a free round if you throw an odd total,
nothing if you threw an even total. the bar staff weren`t bothered the order the two dice landed
and only the total. i reasoned that although a bit of fun the odds were slightly in favour of the pub
and a long argument ensued with arguments both ways.
the combinations for a win,
to me were 1-2,1-4,1-6,2-3,2-5,4-3,4-5,5-6,6-3 = 9ways
and the combinations for a loss,
to me were 1-1,2-2,3-1,3-3,4-2,4-4,5-1,5-3,5-5,6-2,6-4,6-6= 12ways
are there any other combinations ?
sadly i`m not as clever as you guys and only wondered whether my reasoning was correct,
thanks for the replies
Think about a red die and green die.
Calculate probability that sum is even. That means both dice are even or both are odd.
\(\displaystyle \dfrac{1}{36}\ for\ red\ 1\ and\ green\ 1.\) Sum is 2
\(\displaystyle \dfrac{1}{36}\ for\ red\ 1\ and\ green\ 3.\) Sum is 4
\(\displaystyle \dfrac{1}{36}\ for\ red\ 1\ and\ green\ 5.\) Sum is 6
\(\displaystyle \dfrac{1}{36}\ for\ red\ 2\ and\ green\ 2.\) Sum is 4
\(\displaystyle \dfrac{1}{36}\ for\ red\ 2\ and\ green\ 4.\) Sum is 6
\(\displaystyle \dfrac{1}{36}\ for\ red\ 2\ and\ green\ 6.\) Sum is 8
\(\displaystyle \dfrac{1}{36}\ for\ red\ 3\ and\ green\ 1.\) Sum is 4
\(\displaystyle \dfrac{1}{36}\ for\ red\ 3\ and\ green\ 3.\) Sum is 6
\(\displaystyle \dfrac{1}{36}\ for\ red\ 3\ and\ green\ 5.\) Sum is 8
\(\displaystyle \dfrac{1}{36}\ for\ red\ 4\ and\ green\ 2.\) Sum is 6
\(\displaystyle \dfrac{1}{36}\ for\ red\ 4\ and\ green\ 4.\) Sum is 8
\(\displaystyle \dfrac{1}{36}\ for\ red\ 4\ and\ green\ 6.\) Sum is 10
\(\displaystyle \dfrac{1}{36}\ for\ red\ 5\ and\ green\ 1.\) Sum is 6
\(\displaystyle \dfrac{1}{36}\ for\ red\ 5\ and\ green\ 3.\) Sum is 8
\(\displaystyle \dfrac{1}{36}\ for\ red\ 5\ and\ green\ 5.\) Sum is 10
\(\displaystyle \dfrac{1}{36}\ for\ red\ 6\ and\ green\ 2.\) Sum is 8
\(\displaystyle \dfrac{1}{36}\ for\ red\ 6\ and\ green\ 4.\) Sum is 10
\(\displaystyle \dfrac{1}{36}\ for\ red\ 6\ and\ green\ 6.\) Sum is 12
Which adds up (at least until almost closing time) to \(\displaystyle \dfrac{18}{36} = 50\%.\)
You can use those lines to achieve the same result a different way, but it will answer the true but irrelevant objection that the odds of rolling every sum are NOT equal.
Probability of the sum being two = \(\displaystyle \dfrac{1}{36}.\)
Probability of the sum being four = \(\displaystyle \dfrac{3}{36}.\)
Probability of the sum being six = \(\displaystyle \dfrac{5}{36}.\)
Probability of the sum being eight = \(\displaystyle \dfrac{5}{36}.\)
Probability of the sum being ten = \(\displaystyle \dfrac{3}{36}.\)
Probability of the sum being twelve = \(\displaystyle \dfrac{1}{36}.\)
And \(\displaystyle \dfrac{1}{36} + \dfrac{3}{36} + \dfrac{5}{36} + \dfrac{5}{36} +\dfrac{3}{36} + \dfrac{1}{36} =\dfrac{18}{36} = 50\%.\)