5

i have to calculate how probably it is, that the sum of 2 dice rolls is even.

From my course i know that i have N^n Possibilities = 6^2 = 36. 6 = sides of dice and 2 = number of dice rolls

Know i calculated the possibilities for all even values. P = Probability and x = value:

P(x=2)=1/36

P(x=4)=3/36

P(x=6)=5/36

P(x=8)=5/36

P(x=10)=3/36

P(x=12)=1/36

Sums of all P should be the probability i am searching for:

P(x=2)+...+P(x=12)=18/36=1/2=50%

Is this correct? Is there any way to solve this "easier" or by using a formula. Because the next task is for 3 dice rolls :-)

Thanks a lot.

N. F. Taussig
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hukachaka
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5 Answers5

6

Yes, there is a simpler way. It is called principle of deferred decisions. It goes like this. Condition on the parity of the first die throw. It is either 0 or 1. Now, the parity determines uniquely what the parity of the second die throw must be to have the resultant sum have parity 0. Thus,

\begin{align} P(\text{even sum}) &= P(\text{even sum, first was even} + P(\text{even sum, first was odd}) \\ &= P(\text{even|first was even})P(\text{first was even}) + P(\text{even|first was odd})P(\text{first was odd}) \end{align}

Now, note that $P(\text{even|first was even}) = P(\text{second is even}) = 1/2$. Similarly, $P(\text{even|first was odd}) = P(\text{second is odd}) = 1/2$. Thus, we have $P(\text{even sum}) = 1/2\left(P(\text{first was even}) + P(\text{first was odd})\right) = 1/2\left(1\right) = 1/2$.

ADDENDUM: The principle of deferred decisions helps solve much harder similar problems. Consider this problem - you throw a die 10 times. What is the probability that the sum is a multiple of 6?

\begin{align} P(\text{sum $\equiv 0 \mod 6$}) &= \sum_{i=0}^5 P(\text{sum $\equiv 0 \mod 6$}|\text{previous sum was $\equiv i \mod 6$}) \cdot P(\text{previous sum was $\equiv i \mod 6$}) \end{align}

Note that $P(\text{sum $\equiv 0 \mod 6$}|\text{previous sum was $\equiv i \mod 6$}) = 1/6$ because there is only one particular die value that will give us $(6-i) \mod 6$. Thus, $P(\text{sum $\equiv 0 \mod 6$}) = 1/6\left(\sum_{i=0}^5 P(\text{previous sum was $\equiv i \mod 6$})\right) = 1/6\left(1\right) = 1/6.$

Abhiram Natarajan
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  • Thank you very much :-) Solved the Problem for me – hukachaka Oct 22 '17 at 13:14
  • I still have one question. For the next task i need to do it for 3 dice rolls and if the sum is divisible by 3.You answer with 10 times should be helping me solving this. But i dont understand the "5" above the sigma. doesnt this have to be 10 since you said you throw the dice 10 times? – hukachaka Oct 22 '17 at 13:29
  • @AlVed it can be done more simple: every set ${n+1,\dots,n+6}$ contains exactly $2$ elements that are divible by $3$. – drhab Oct 22 '17 at 13:35
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    @drhab, yes but aren't there different compositions possible. Like first 2,2,2 ? Doesnt the probability change then? – hukachaka Oct 22 '17 at 13:52
  • @Al Ved The 0 to 5 are the remainders when divided by 6. So for divisible by 3, you sum over 0, 1, 2. The beauty of 'principle of deferred decisions' is that 10 does not appear anywhere in the calculation. That is what makes it so powerful. The same method would work whether it is 10 or 1000 or GOOGOL; specifically, the answer would be the same. – Abhiram Natarajan Oct 22 '17 at 13:54
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    The fact that we are rolling 10 times is encapsulated in "previous sum was ...". The point is we do not care about the probabilities of the previous sum because in our calculation we are summing over all probabilities, which by definition is 1. Let me know if this needs more explanation. – Abhiram Natarajan Oct 22 '17 at 13:57
  • so i just need to sum from 0 to 1 the formula ? i think i am gonna try this on paper and post this again in another thread. Thanks to all :-) – hukachaka Oct 22 '17 at 14:03
  • If I understand right, your question is that you have to throw three dice and have the sum be a multiple of three right? Then you have to sum from 0 to 2. $P(\text{total sum $\equiv 0 \mod 3$}) = \sum_{i=0}^2 P(\text{total sum $\equiv 0 \mod 3$} | \text{penultimate sum $\equiv i \mod 3$}) \cdot P(\text{penultimate sum $\equiv i \mod 3$}) = P(\text{current die throw $\equiv 3-i \mod 3$})\cdot P(\text{penultimate sum $\equiv i \mod 3$}) = \frac{1}{3} \left(\sum_{i=0}^2 P(\text{penultimate sum $\equiv i \mod 3$})\right) = \frac{1}{3}\left(1\right) = \frac{1}{3}.$ – Abhiram Natarajan Oct 22 '17 at 14:08
  • Intuitionally you can view the total number of outcomes in a given throw of $n$ dice is always even (as $6^n$ will always be even $\forall n \in \mathbb{N}$), exactly half of the outcomes will be even. – Bensstats Oct 22 '17 at 19:43
  • Hi. Isnt it the same method to find even sum when 1 dice throw 2 times? – Dini Nov 16 '19 at 06:36
  • Yes. The die throws must be independent, that's all. – Abhiram Natarajan Nov 17 '19 at 14:58
5

The sum of two dice rolls is even if both dice display even numbers or both display odd numbers. Hence, the probability that the sum of the two dice rolls is even is $$P(\text{sum is even}) = P(\text{second is even} \mid \text{first is even})P(\text{first is even}) + P(\text{second is odd} \mid \text{first is odd})P(\text{first is odd}) = \frac{1}{2} \cdot \frac{1}{2} + \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{2}$$

For the case of three dice rolls, we obtain an even sum if the outcome of the roll is even given that the sum of the first two rolls is even or if the outcome of the third roll is odd given that the sum of the first two rolls is odd.

N. F. Taussig
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4

Correct.

New question (as introduction for an easyer way of working):

Let it be that $n$ is some integer. Now throw a die and let $D$ denote the outcome.

What is the probability that $n+D$ is even?

Well, for every integer $n$ there are exactly $3$ even numbers in the set of possible outcomes: $\{n+1,n+2,n+3,n+4,n+5,n+6\}$ and all outcomes are equiprobable.

So the answer is $\frac36=0.5$.

For $n$ you can take anything you like.

Also a random $n$ will do, like the sum of one, two (or more) dice.

Let $X$ be a random variable with $\mathsf P(X\in\mathbb Z)=1$, let $D$ be the outcome by throwing one die and let $X$ and $D$ be independent.

Then:

$$\begin{aligned}\mathsf{P}\left(X+D\text{ is even}\right) & =\sum_{n\in\mathbb Z}\mathsf{P}(X+D\text{ is even}\mid X=n)\mathsf{P}(X=n)\\ & =\sum_{n\in\mathbb Z}\mathsf{P}(n+D\text{ is even})\mathsf{P}(X=n)\\ & =\sum_{n\in\mathbb Z}0.5\cdot\mathsf{P}(X=n)\\ & =0.5\sum_{n\in\mathbb Z}\mathsf{P}(X=n)\\ & =0.5 \end{aligned} $$

drhab
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2

you have two dices and you want even sum there are only two ways in which you can get even sum either $even+even$ or $odd+odd$

now if you want even on both dices you have 3 choice(2,4,6) for one dice and three choices for another dice thus $total 3*3=9 choices$ similarly for $odd+odd$ $3*3=9 choices $ thus $total 18 choices$ are in your favour $$probability =18/36=1/2$$

now for 3 dices $even+even+even$ , $even+odd+odd $, $odd+even+odd$ and $odd+odd+even $there are total 4 possibilities thus $total 4*27 ways$

probabalitiy =$\frac{4*27}{6^3}$=1/2

Deepesh Meena
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1

An easy way to answer the OP's question is by drawing a picture and counting.

Make a $6 \times 6$ square showing every possible combination of numbers, like this:

$$\begin{matrix} 1,1 & 1,2 & \cdots & 1,6 \\ 2,1 & 2,2 & \cdots & 2,6 \\ \vdots & \vdots & \ddots & \vdots \\ 6,1 & 6,2 & \dots & 6,6 \end{matrix}$$

It should be obvious that the odd and even sums form a checkerboard pattern, with three odd and three even in each row (and each column).

So there are the same number of odd and even sums - i.e. the probabilities of getting an odd or an even sum are both $1/2$.

There are two out of the six sums in each row (and column) that are divisible by $3$, so the probability of getting a sum divisible by $3$ is $1/3$.

alephzero
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