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Question 7: We roll a fair red and blue six-sided die and add them together. Keep rolling until either the red die comes up as 6 or the blue die comes up as 6, then we stop. Let $X$ be the total number of times we roll both dice. Let $D_i$ be the sum of the red and blue dice on roll $i$, and let $Y$ be the sum of all the dice rolls, that is, $Y = D_1 + D_2 + \dots + D_X$. a) Find $E(X)$. b) Find $E(Y|X = 4)$. That is, find the sum of all the dice rolls assuming that the first time you rolled a 6 on either die was the fourth dice roll. Hint: Find the conditional probability of $D_1..D_4$ first. You may consider using linearity of expectation as well as conditional expected value. c) Note that for any roll of the red and blue dice, that $D_i$ for all $1 \le i \le X$ are independent and identically distributed. What this means is that $E(D_1) = E(D_i)$ for all values $1 \le i \le X$. Wald's Identity tells us that if $X$ is the number of die rolls, and the value of $X$ depends on a stopping condition (which it does in this case), then the expected sum of all the die rolls $E(Y) = E(X) \cdot E(D_1)$. Find $E(Y)$ using Wald's Identity.

          Question 7:
We roll a fair red and blue six-sided die and add them together. Keep rolling until either
the red die comes up as 6 or the blue die comes up as 6, then we stop. Let $X$ be the total
number of times we roll both dice. Let $D_i$ be the sum of the red and blue dice on roll $i$, and
let $Y$ be the sum of all the dice rolls, that is, $Y = D_1 + D_2 + \dots + D_X$.
a) Find $E(X)$.
b) Find $E(Y|X = 4)$. That is, find the sum of all the dice rolls assuming that the first
time you rolled a 6 on either die was the fourth dice roll. Hint: Find the conditional
probability of $D_1..D_4$ first. You may consider using linearity of expectation as well as
conditional expected value.
c) Note that for any roll of the red and blue dice, that $D_i$ for all $1 \le i \le X$ are independent
and identically distributed. What this means is that $E(D_1) = E(D_i)$ for all values
$1 \le i \le X$. Wald's Identity tells us that if $X$ is the number of die rolls, and the value
of $X$ depends on a stopping condition (which it does in this case), then the expected
sum of all the die rolls $E(Y) = E(X) \cdot E(D_1)$. Find $E(Y)$ using Wald's Identity.

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Question 7:
We roll a fair red and blue six-sided die and add them together. Keep rolling until either
the red die comes up as 6 or the blue die comes up as 6, then we stop. Let X be the total
number of times we roll both dice. Let Di be the sum of the red and blue dice on roll i, and
let Y be the sum of all the dice rolls, that is, Y = D1 + D2 + … + DX.
a) Find E(X).
b) Find E(Y|X = 4). That is, find the sum of all the dice rolls assuming that the first
time you rolled a 6 on either die was the fourth dice roll. Hint: Find the conditional
probability of D1..D4 first. You may consider using linearity of expectation as well as
conditional expected value.
c) Note that for any roll of the red and blue dice, that Di for all 1 ≤ i ≤ X are independent
and identically distributed. What this means is that E(D1) = E(Di) for all values
1 ≤ i ≤ X. Wald's Identity tells us that if X is the number of die rolls, and the value
of X depends on a stopping condition (which it does in this case), then the expected
sum of all the die rolls E(Y) = E(X) · E(D1). Find E(Y) using Wald's Identity.

Added by Matthew P.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

Instant Answer

Solved by Expert Kari Hasz

10/05/2023

Step 1

Let p be the probability of rolling a 6 on either die on a single roll. Since the dice are fair, p = 1/3 (since there are 3 ways to get a 6 out of 12 possible outcomes). Let q be the probability of not rolling a 6 on either die on a single roll, so q = 2/3. The Show more…

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Question 7:[DISCRETE MATH] We roll a fair red and blue six-sided die and add them together. Keep rolling until either the red die comes up as 6 or the blue die comes up as 6 , then we stop. Let x be the total number of times we roll both dice. Let D_(i) be the sum of the red and blue dice on roll i, and let Y be the sum of all the dice rolls, that is, Y=D_(1)+D_(2)+dots+D_(x). a) Find E(x). b) Find E(Y|x=4). That is, find the sum of all the dice rolls assuming that the first time you rolled a 6 on either die was the fourth dice roll. Hint: Find the conditional probability of D_(1)..D_(4) first. You may consider using linearity of expectation as well as conditional expected value. c) Note that for any roll of the red and blue dice, that D_(i) for all 1<=i<=x are independent and identically distributed. What this means is that E(D_(1))=E(D_(i)) for all values 1<=i<=x. Wald's Identity tells us that if x is the number of die rolls, and the value of x depends on a stopping condition (which it does in this case), then the expected sum of all the die rolls E(Y)=E(x)*E(D_(1)). Find E(Y) using Wald's Identity. Question 7: We roll a fair red and blue six-sided die and add them together. Keep rolling until either the red die comes up as 6 or the blue die comes up as 6, then we stop. Let X be the total number of times we roll both dice. Let D; be the sum of the red and blue dice on roll i, and let Y be the sum of all the dice rolls, that is, Y = D+ D +...+ Dx. a) Find E(X). b) Find E(YX = 4). That is, find the sum of all the dice rolls assuming that the first time you rolled a 6 on either die was the fourth dice roll. Hint: Find the conditional probability of D..D4 first. You may consider using linearity of expectation as well as conditional expected value. c) Note that for any roll of the red and blue dice,that D; for all 1 < i < X are independent and identically distributed. What this means is that E(D)= ED) for all values 1 < i < X. Wald's Identity tells us that if X is the number of die rolls, and the value of X depends on a stopping condition (which it does in this case), then the expected sum of all the die rolls E(Y) = E(X) E(D). Find E(Y) using Wald's Identity

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00:01 So this is a problem where you roll two fair dice.

00:06 So what's the probability that the sum is seven? so the probability that the sum is seven is one out of six.

00:22 The probability, well, for part b, what's the most likely sum that is seven? because there's six possibilities or six ways to get a seven out of 36.

00:32 Part c, what's the probability the sum exceeds seven? so probability that x is greater than.

00:39 Seven, so there's 15 out of 36 ways.

00:43 The probability that the sum exceeds six is 21 out of 36.

00:53 What's the probability the sum is six or less given a die shows a two? so i'll do the probability.

01:05 I'm gonna just say x.

01:07 The probability that x is less than or equal to six given die number one shows a two.

01:27 So the intersection of a sum of less than or equal to six with die number one showing, oops, number one showing a two is four out of 36.

01:49 And the probability of die number one showing a two, there's six out of 36 ways.

01:55 So that would be four out of six or 2 3rds.

02:01 The probability that a sum is six or less given at least one die is two is seven out of 36 divided by 11 out of 36.

02:23 So that would be seven out of 11.

02:27 What's the probability that die number one shows a two given that the sum is less than or equal to six? so g is kind of a reverse of e.

02:47 So four out of 36 divided by 15 out of 36.

02:53 So that would be 4 15ths.

02:57 H is, what's the probability at least one die shows a two given the sum is six or less? so that's seven out of 36 divided by 15 out of 36, which would simplify down to 7 15ths.

03:17 I is, what is the probability that both die shows even numbers given the sum is even? so that would be nine out of 36 divided by 18 out of 36, which simplifies down to one half...

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