Suppose that we roll a fair die until a 6

Anne Wacker 2021-12-15 Answered
Suppose that we roll a fair die until a 6 comes up.
a) What is the probability that we roll the die n times?
b) What is the expected number of times we roll the die?
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Expert Answer

Wendy Boykin
Answered 2021-12-16 Author has 35 answers
Step 1
a) If we roll a fair die, then we have 1 chance in 6 of rolling a six and 5 chances in 6 of not rolling a 6.
P(6)=16
P(Not 6)=56
We roll the die until a 6 is rolled.
Let X be the number of rolls of the die.
When we roll the die n times, then the first n1 rolls cannot be a 6, while the n-th roll has to be a 6:
P(X=n)=(P(Not 6))n1P(6)=(56)n1×16
Note: The variable X has a geometric distribution with p=16.
Step 2
b) The expected value of a random variable X with a geometric distribution is the reciprocal of the constant probability of success p=P(6)
E(x)=1p=1P(6)=116=6

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Heather Fulton
Answered 2021-12-17 Author has 31 answers
a) We would have to roll n1 numbers that are not 6 followed by a 6, so the probability is (56)n1(16)
b) E(N)=n=1n(56)n1(16)=(16)(11(56))2=6.
This is a nice answer since after 6 rolls we would expect to have rolled exactly one 6.

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nick1337
Answered 2021-12-27 Author has 575 answers

a. If we roll the die n times (assuming n1 and the dice is 6-sided and fair), then we must roll n-1 "not 6" rolls followed by 1 "6" roll. The probability of that is:
(56)n1(16)1
b. We could roll the die any number of times from 1 to infinite. Consider a random variable R which is the number of rolls to roll a 6. We could determine the expected number in two ways:
i. We could say that R is a geometric RV with a chance of success of 16, so E(R)=1(16)=6
ii. We could show i. explicitly, using the formula for expected value:
E(R)=sSR(s)p(s)
=i=1(i)((56)i1(16)1)
=16(i=1(i)((56)i1)
This summation takes the form
i=1(i)(ri1), where |r|<1
i=1(i)(ri1)=i=1ddr(ri)
=ddri=1(ri)
=ddr11r (since |r|<1)
=1(1r)2
Let r=56(|r|=56<1.) Then,
E(R)=16(1(156)2)
E(R)=6
Either way, we get the same answer: 6 rolls.

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