# Multiple dice probability

Suppose I have 6 dice. One 20-sided die, one 12-sided, one 10-sided, one 8-sided, one 6-sided, and one 4-sided. The dice are in a bag and you pick two of them at random (each of them equally likely to be picked) and roll them. The first questions is what is the sample size of that experiment. The second question is what is the probability that the sum of the two dice is even? And lastly what is the probability that you roll an 8 or higher on both die.

For the first question I believe there should $20\cdot 12\cdot 10\cdot 8\cdot 6\cdot 4\cdot 30$ total outcomes. The $30$ in the last part being there to account for order, $6$ ways to choose the first $5$ for the second dice.

For the second question, would it be possible to say that since there are $31$ different sum totals, and that $16$ of those are even the probability would be $\frac{16}{31}$?

For the third question, would the number of ways of rolling an 8 or higher on both be $13\cdot 5\cdot 3 \cdot 30$? Then the probability would be that over the value in question 1?

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Do you mean sample space? Have you tried drawing a probability tree? – ncmathsadist Feb 22 '12 at 2:55
"what is the sample size of that experiment" - do you mean to simply record the sum of their faces, as in $12$? Do you mean to keep track of what pairs of numbers lead to that sum, as in $(10,2)\neq(4,8)$? Or do you mean to keep track of everything, as in $(10_{20},2_8)\neq(10_{20},2_4)$? – alex.jordan Feb 22 '12 at 3:13

For the second question, each die has an equal probability to give even or odd, so the total has an equal probability to be even or odd. Among the 31 possible totals, not all are equally probable, so you cannot just say 16/31 because 16 of them are odd.

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How many outcomes there are depends on what you’re counting as different outcomes. If each possible choice of dice and each possible way the two chosen dice can come up is counted as a separate outcome, there are $2840$ possible outcomes; here’s the reasoning.

If you pick the $20$- and $12$-sided dice, there are $20\cdot12=240$ possible outcomes; if you pick the $20$-and $10$-sided dice, there are $20\cdot10=200$ possible outcomes; and so on. The total possible number of outcomes is therefore the sum of all possible products of two of the numbers $20,12,10,8,6$, and $4$. You can compute this by brute force, or you can take advantage of the fact that \begin{align*}&(a_1+a_2+a_3+a_4+a_5+a_6)^2=\\ &a_1^2+a_2^2+a_3^2+a_4^2+a_5^2+a_6^2+2(\text{the sum that you want})\;, \end{align*} so that \begin{align*}&\text{the sum that you want}=\\ &(20+12+10+8+6+4)^2-(20^2+12^2+10^2+8^2+6^2+4^2)=\\ &60^2-(400+144+100+64+36+16)=\\ &3600-760=\\ &2840\;. \end{align*}

These $2840$ outcomes are not all equally likely, however. Since there are $\binom62=15$ possible pairs of dice, and all pairs are equally likely, the probability of drawing any particular pair is $\frac1{15}$. If you draw the $20$- and $12$-sided dice, each of the $240$ possible rolls is equally likely, so the overall probability of drawing those two dice and getting a particular roll is $\frac1{15}\cdot\frac1{240}=\frac1{3600}$. If you draw the $6$- and $4$-sided dice, on the other hand, there are only $24$ possible rolls, so the probability that you draw those two dice and get a particular roll is $\frac1{15}\cdot\frac1{24}=\frac1{360}$, or ten times as large.

However, you might decide to count as different outcomes only the different possible totals on the two dice. The minimum possible total is $1+1=2$, the maximum is $20+12=32$, and every integer total between these extremes is possible, so there are $32-1=31$ possible totals, as you said. Of course these $31$ outcomes also have very different probabilities: to take the easiest example, there is only one way to get a total of $32$, but there are $15$ ways to get a total of $2$, one for every possible pair of dice.

This is relevant to the second question: because the different totals have different probabilities, you can’t simply say that there are $16$ even and $15$ odd totals, and therefore the probability of getting an even total is $\frac{16}{31}$. You’re going to have to analyze the possibilities in more detail.

Let’s look at a particular pair of dice, say the $10$- and the $6$-sided dice. The $10$-sided die has $5$ even and $5$ odd numbers, and the $6$-sided die has $3$ even and $3$ odd numbers. The $10\cdot 6=60$ possible rolls of these two dice can therefore be broken down as follows: $5\cdot 3=15$ rolls with an even number on both dice; $5\cdot 3=15$ rolls with an even number on the $10$-sided die and an odd number on the $6$-sided die; $5\cdot 3=15$ rolls with an odd number on the $10$-sided die and an even number on the $6$-sided die; and $5\cdot 3=15$ rolls with an odd number on both dice. When do you get an even total? In the first and last cases, which amount to $15+15=30$ out of the $60$, or half of them. Thus, if you happen to draw the $10$- and the $6$-sided dice, your probability of rolling an even total is $\frac12$.

We could have got this with much less work by realizing right away that in order to get an even total, we must get either two even or two odd numbers. Each die has half its faces even and half of them odd, so no matter which two dice we pick, the probability of getting even + even is $\frac12\cdot\frac12=\frac14$, and so is the probability of getting odd + odd, which means that the probability of getting an even total is $\frac14+\frac14=\frac12$. Since the probability of getting an even total is $\frac12$ no matter which pair of dice we draw, the overall probability of getting an even total must also be $\frac12$.

For the last question you must also look a bit more closely at the possible outcomes. First, you can’t roll an $8$ or higher on both dice unless both dice have at least $8$ faces, so you must begin by drawing two of the four dice with the most faces; there are $\binom42=6$ ways to do this, so $\frac9{15}=\frac35$ of the time you won’t even draw dice on which it’s possible to get $8$ or higher on both. What you should do is look at these six pairs individually, counting the number of rolls that give you two numbers that are at least $8$. For instance, if you’re rolling the $12$- and $10$-sided dice, you need to get one of the $5$ largest numbers on the $12$-sided die and one of the $3$ largest on the $10$-sided die; the probability of doing so is $\frac5{12}\cdot\frac3{10}=\frac18$. Of course the probability of drawing these two dice in the first place is only $\frac1{15}$, so the probability of drawing them and getting a successful roll is $\frac1{15}\cdot\frac18=\frac1{120}$. Similarly, the probability of success when you roll the $12$- and the $8$-sided dice is $\frac5{12}\cdot\frac18=\frac5{96}$, and the probability of getting those two dice in the first place is $\frac1{15}$, so the probability of drawing them and getting a successful roll is $\frac1{15}\cdot\frac5{96}=\frac1{288}$. If you perform similar calculations for each of the other four pairs of dice that can give you successful outcomes, you can then add the results to get the final probability of rolling two numbers greater than or equal to $8$.

You can’t simply count successful outcomes and divide by $2840$, since the $2840$ possible rolls aren’t equally likely, as noted at the beginning.

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Thank you. This was very clear and helpful. – Tom Davis Feb 22 '12 at 3:50