Consider the experiment of rolling a fair die independently until the same number/face occurs 2 successive times and let X be the trial on which the repeat occurs, e.g. if the rolls are 2,3,4,5,1,2,4,5,5, then X=9. a. find the probability function f(x)=P(X=x). b. compute EX

bergvolk0k 2022-10-20 Answered
MathJax(?): Can't find handler for document MathJax(?): Can't find handler for document Consider the experiment of rolling a fair die independently until the same number/face occurs 2 successive times and let X be the trial on which the repeat occurs, e.g. if the rolls are 2,3,4,5,1,2,4,5,5, then X = 9.
a. find the probability function f ( x ) = P ( X = x )
b. compute EX
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Answers (2)

ehedem26
Answered 2022-10-21 Author has 13 answers
Step 1
This is basically a geometric distribution. At each step, independent of whatever happened previously, you have a 1/6 chance of rolling the same as you did on the previous try. You keep going until you get a success, and you are counting the number of tries until a success. Let's call this random variable Y.
Step 2
Now here comes the only twist: You start not on the first try (on which it is impossible to duplicate a previous roll), but on the second. So X is one more than Y, hence f X ( t ) = f Y ( t 1 ), and E ( X ) = E ( Y + 1 )
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Marilyn Cameron
Answered 2022-10-22 Author has 1 answers
Step 1
You could solve this directly, expressing your answer as a sum of the form n = 2 n P [ X = n ]. And, with a bit of work, you could evaluate the sum explicitly (using the formula for the sum of a differentiated geometric series; see, e.g., the ideas here)).
The expected value is n = 2 n ( 5 / 6 ) n 2 ( 1 / 6 ). One can show this sum has value 7.
Step 2
Or, you could find the expected value by conditioning on the result of the second roll:
Let X be the number of rolls to obtain the same result twice in a row. Let Y be the event that the second roll is the same as the first. Then
E ( X ) = E ( X Y ) P ( Y ) + E ( X Y C ) P ( Y C ) = E ( X Y ) 1 6 + E ( X Y C ) 5 6 . .
Step 3
But E ( X Y ) = 2 and E ( X Y C ) = E ( X ) + 1 (here, it's as if you started playing the game at the start again on the second roll). So
E ( X ) = 2 1 6 + ( E ( X ) + 1 ) 5 6 .
Solving the above for E(X) gives E ( X ) = 7.
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