Karsyn Beltran

2022-07-22

I chose a random number from 1 to 6. Afterwards, I roll a die till I get a result that is even or higher than my chosen number. What is the E(x) of the number of times I throw the die?

So I thought it's Geometric distribution will "success" where "success" is to get my number. So first the probability of choosing a random number in the die is 1/6. Now I can't configure what is the probability to get something even or higher? Since I chose the number randomly. and afterthat, I just need to divide 1 by the probability I get (like the geometric E(x) formula)?

So I thought it's Geometric distribution will "success" where "success" is to get my number. So first the probability of choosing a random number in the die is 1/6. Now I can't configure what is the probability to get something even or higher? Since I chose the number randomly. and afterthat, I just need to divide 1 by the probability I get (like the geometric E(x) formula)?

akademiks1989rz

Beginner2022-07-23Added 16 answers

Step 1

Let X be uniformly distributed on {1,…,6} and let N be the number of times that you need to roll the die. Note that $N\mid X=x$ is geometrically distributed with probability mass function $P(N=k\mid X=x)=(1-p{)}^{k-1}p\phantom{\rule{1em}{0ex}}(k\ge 1)$ where $p=\frac{6-x+1}{6}$.

Step 2

The law of total expectation yields that

$EN=E(EN\mid X)=E\frac{6}{6-X+1}=6E\frac{1}{6-X+1}=6\times \frac{1}{6}(1+\frac{1}{2}\cdots \frac{1}{6}).$

So $EN=\sum _{k=1}^{6}\frac{1}{k}$

Let X be uniformly distributed on {1,…,6} and let N be the number of times that you need to roll the die. Note that $N\mid X=x$ is geometrically distributed with probability mass function $P(N=k\mid X=x)=(1-p{)}^{k-1}p\phantom{\rule{1em}{0ex}}(k\ge 1)$ where $p=\frac{6-x+1}{6}$.

Step 2

The law of total expectation yields that

$EN=E(EN\mid X)=E\frac{6}{6-X+1}=6E\frac{1}{6-X+1}=6\times \frac{1}{6}(1+\frac{1}{2}\cdots \frac{1}{6}).$

So $EN=\sum _{k=1}^{6}\frac{1}{k}$

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