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Brian plays a game in which two fair, six-sided dice are rolled simultaneously. For each die, an even number means that he wins that amount of money and an odd number means that he loses that amount of money. What is the probability that Brian makes money on his first roll?

To find the probability, do we need to find the even numbers only. There are 36 outcomes from the two dies. So is there an easy way to get the arrangements?

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Let the values of the dice in a roll be $a$ and $b$. You win money if ($1$) $a$ and $b$ are both even, ($2$) $a$ is even and $b$ is odd with $a>b$, or ($3$) $a$ is odd and $b$ is even with $a<b$. For ($1$) there are $3$ even numbers in $1,2,\dots,6$, so there are $3^2=9$ ways for $a$ and $b$ to both be positive. For ($2$) consider the three cases cases: $a=2 \implies b=1$; $a=4\implies b=1,3$; and $a=6\implies b=1,3,5$. Therefore, there are $6$ ways to get condition ($2$). Condition ($3$) is the same as ($2$) by symmetry, so the probability to win money is $$ \frac{3^2 + (2)(6)}{6^2} = \frac{21}{36} = \frac{7}{12}. $$ We can easily generalize this to any $n$-faced pair of dice when $n$ is even. Notice that the triangular numbers appear when counting conditions ($2$) and ($3$). Therefore, the general probability is $$ \frac{(n/2)^2 + 2T_{n/2}}{n^2} = \frac{n + 1}{2n}. $$

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Just fill in this table for the result of each possible die roll, and that should help you get to your answer.

$$\begin{array} {c|c|c|c|c|c|c|}&1&2&3&4&5&6\\ \hline 1&&&&&& \\ \hline 2&&&&&& \\ \hline 3&&&&&& \\ \hline 4&&&&&& \\ \hline 5&&&&&& \\ \hline 6&&&&&& \\ \hline \end{array}$$

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Nine cases have them both odd. Nine have them both even.
So you only have the other 18 to check.
And they come in pairs, as $(2,5)$ is the same as $(5,2)$, so you only have nine cases to check.

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Theres a chance of $1/2$ to get an even number on a dice, so there's a chance of $1/4$ to an even number on both dies.

Now we have to compute the chance to gain money if we get one even dice and an odd one.

There's a chance of $1/6$ to get a $6$. In that case, we can get any odd number on the other dice and we'll make money. There's a chance of $1/2$ to get an odd number so we have a chance of $1/12$ to earn money with a $6$ and an odd number.

There's a chance of $1/6$ to get a $4$ and then we need a $1$ or a $3$ (chance of $1/3$) to make money, so the chance in this case is $1/18$.

Finally, if we get a $2$ (chance of $1/6$) the only odd number that will make us make money is $1$ (chance of $1/6$) so the chance this time is $1/36$.

For the last three cases, we have two ways of obtaining that value since we have two dies. So the chances will be $1/6$, $1/9$ and $1/18$

Any other situation means losing money. The chance to make money is:

$1/4 + 1/6 + 1/9 + 1/18 = \frac{9+6+4+2}{36} = \frac{21}{36} = \frac{7}{12}$

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  • $\begingroup$ So much derp. It's fixed now. $\endgroup$ – Darth Geek Jul 24 '14 at 21:40
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It'll be 58.3% chance.

36 possible combos.

Nine are all even...

When not all even or all odd numbers: 6 being one of the two always wins, giving Six wins... 4 being one of the two wins Four... 2 being one of the two wins Twice...

9+6+4+2=21

21/36=.583^

Or use the matrix provided by @paw88789

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