A dice is rolled until a $6$ occurs. What is the probability that the sum including the $6$ is even?

Question

A game is played where a standard six sided dice is rolled until a $6$ is rolled, and the sum of all of the rolls up to and including the $6$ is taken. What is the probability that this sum is even?

I know that this is a geometric distribution and the expected number of rolls is $\frac1{1/6} = 6$ rolls until a $6$ occurs, along with the expected value being $21$ ($6$ rolls times expected value of $3.5$ per roll), but I'm not certain how to proceed from there. Would the expected range (if it is even relevant) be from $11 = 1·5+6$ to $31 = 5·5+6$? The answer is supposedly $\frac47$. I'm also curious about how this question would change if the stopping number was anything else, say a $3$ stopping the sequence rather than a $6$. Thank you in advance!

Answer 1

Let $p$ be the desired probability, and consider the first roll. It is either a $6$, in which case we're done and the sum is even, a $2$ or $4$, in which case we want the sum of the rest of the terms to be even, or a $1,3$, or $5$, in which case we want the sum of the rest to be odd.

Thus $$p = \frac{1}{6}+ \frac{1}{3}p+\frac{1}{2}(1-p)$$ which simplifies to $p=\frac{4}{7}$.

Answer 2

We need only consider the rolls before a $6$ is obtained, because rolling an even $6$ doesn't change the parity of our total. Let $p_n$ represent the probability that the sum of $n$ rolls, not including any $6$, is even. Then we have: $p_{n+1}=\frac25p_n + \frac35(1-p_n)$, because a $2$ or a $4$ keeps a previous even total even, while a $1$, $3$ or $5$ makes a previous odd total into an even total. This simplifies to: $p_{n+1}=\frac35-\frac15p_n$. We also have $p_0=1$. We can solve this recurrence, and find that

$$p_n=\frac12\left(1+\left(-\frac15\right)^n\right)$$

Now, let $x_n$ represent the probability of rolling $n$ non-6's before the first $6$, so $x_n=\frac16\left(\frac56\right)^n$. The number we need is:

$$\begin{align} \sum\limits_{n=0}^\infty x_np_n &= \sum\limits_{n=0}^\infty \left[\frac16\left(\frac56\right)^n\cdot\frac12\left(1+\left(-\frac15\right)^n\right)\right]\\ &=\frac1{12}\sum\limits_{n=0}^\infty \left[\left(\frac56\right)^n + \left(-\frac16\right)^n\right]\\ &=\frac1{12}\left(6 + \frac67\right) = \frac47 \end{align}$$

That said, @carmichael561's answer is much, much nicer.

Answer 3

Let $p$ be the probability that the sum of the die until(and including) the first six is even.

Let $R_1$ be the roll of the first die. So partitioning on this roll, and noticing the recurance:

$$p= \underline\qquad\,\mathsf P(R_1\in\{\underline\qquad\})+\underline\qquad\,\mathsf P(R_1\in\{\underline\qquad\})+\underline {~1~}\,\mathsf P(R_1=6)$$

Fill in the blanks, evalute the probabilities, and then solve for $p$.

Answer 4

Here is an answer using exponential generating functions. This problem has the features of a basic coupon collector (six coupons drawn with replacements). Note however that there is no requirement here of seeing all coupons. Now the probability that we took $m$ rolls to see the first six is by inspection given by

$$\frac{5^{m-1}}{6^m} = \frac{1}{6}\left(\frac{5}{6}\right)^{m-1}.$$

Observe that if the sum is even the odd values must have ocurred an even number of times, which gives the marked combinatorial class (one set of slots from the $m-1$ possible ones for each of the five admissible rolls of the die)

$$\def\textsc#1{\dosc#1\csod} \def\dosc#1#2\csod{{\rm #1{\small #2}}} \textsc{SET}(\mathcal{U}\times\mathcal{Z}) \times \textsc{SET}(\mathcal{Z}) \times \textsc{SET}(\mathcal{U}\times\mathcal{Z}) \times \textsc{SET}(\mathcal{Z}) \times \textsc{SET}(\mathcal{U}\times\mathcal{Z}).$$

The corresponding EGF is

$$G(z, u) = \exp(2z)\exp(3uz).$$

Restricting to even sums we get

$$H(z) = \frac{1}{2} G(z, 1) + \frac{1}{2} G(z, -1) \\ = \frac{1}{2} \exp(5z) + \frac{1}{2} \exp(-z).$$

Extracting coefficients we find

$$(m-1)! [z^{m-1}] H(z) = \frac{1}{2} 5^{m-1} + \frac{1}{2} (-1)^{m-1}.$$

Hence the probability of an even sum given that we took $m$ draws is given by

$$\frac{1}{2} + \frac{1}{2} \left(-\frac{1}{5}\right)^{m-1}.$$

We thus get for the total probability

$$\frac{1}{12} \sum_{m\ge 1} \left(\frac{5}{6}\right)^{m-1} + \frac{1}{12} \sum_{m\ge 1} \left(-\frac{1}{6}\right)^{m-1} \\ = \frac{1}{12} \left(\frac{1}{1-5/6} + \frac{1}{1+1/6}\right) = \frac{4}{7}.$$

Answer 5

The probability of any number 1 to 6 in rolling a dice is p1= 1/6 = 0.16666, therefore probability of getting desired number with condition of even or odd sum of numbers before getting desired number should be less than p1. lets investigate with python.

this is a simple function to simulate rolling dice

def onediceprobablity(Num,numoftry):

countNum = 0
for i in range(numoftry):
    if random.randint(1,6) == Num :
        countNum = countNum +1

return countNum/numoftry

Num is a number 1 to 6 and numoftry is our tries from 100000 to 1000000

here is the function that checks condition of problem

def dicesumeven(Num,numoftry):

EvenSum = 0
L = []
for i in range(numoftry):        
    N = random.randint(1,6)       
    L.append(N)

    if N == Num :
        Sum = sum(L)             
        L = []
        if (Sum % 2)  == 0:                
            EvenSum = EvenSum + 1                 

return EvenSum/numoftry

with Num = 6 and numoftry =1000000 probability p = 0.09538

this is the function for sum equals odd number

def dicesumodd(Num,numoftry):

oddSum = 0
L = []
for i in range(numoftry):        
    N = random.randint(1,6)       
    L.append(N)

    if N == Num :
        Sum = sum(L)             
        L = []
        if (Sum % 2)  == 1:                
            oddSum = oddSum + 1                 

return oddSum/numoftry

for 1000000 tries result is p2 = 0.071222

sum of p1 and p2 is about 1/6