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?

Karsyn Beltran 2022-07-22 Answered
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)?
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Answers (1)

akademiks1989rz
Answered 2022-07-23 Author has 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 X = x is geometrically distributed with probability mass function P ( N = k X = x ) = ( 1 p ) k 1 p ( k 1 ) where p = 6 x + 1 6 .
Step 2
The law of total expectation yields that
E N = E ( E N X ) = E 6 6 X + 1 = 6 E 1 6 X + 1 = 6 × 1 6 ( 1 + 1 2 1 6 ) .
So E N = k = 1 6 1 k
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