OpenStudy (anonymous):

Suppose "r" is 1, 2, 3, 4, 5, or 6 You roll single 6 sided dice. What is the expected number of times one would roll the dice before getting r. Also, suppose r = 5. We now are dealing with a loaded dice with the values 1, 2, 3, 4, 5, 5. What is the expected number of rolls to get "r" in this case?

5 years ago
OpenStudy (anonymous):

well the answer to the first question is 6 times as you have 6 options. well you have 5 options two of them are r so probability of r is 2/5. you cant simplify r 2/5 any further but keep in mind that 2/5 > 1/3 but less then 1/2 so i would say 3 rolls.

5 years ago
OpenStudy (anonymous):

That's not concrete enough :/ Thanks anyway :)

5 years ago
OpenStudy (zarkon):

he is correct though

5 years ago
OpenStudy (zarkon):

you looking for this... \[\sum_{k=1}^{\infty}k\frac{1}{6}\left(\frac{5}{6}\right)^{k-1}=6\]

5 years ago
OpenStudy (zarkon):

and this \[\sum_{k=1}^{\infty}k\frac{2}{6}\left(\frac{4}{6}\right)^{k-1}=3\]

5 years ago
OpenStudy (anonymous):

Thanks! That's really helpful

5 years ago
OpenStudy (anonymous):

Zarkon: Do you mind if I ask what the k represents?

5 years ago
OpenStudy (anonymous):

Nm, the K is the number of rolls.

5 years ago
OpenStudy (zarkon):

correct

5 years ago
OpenStudy (anonymous):

I'm sorry to bug you again, but how can a summation that goes on into infinity be equal to anything?

5 years ago
OpenStudy (zarkon):

look up 'geometric sum'

5 years ago
OpenStudy (anonymous):

Thanks again Zarkon!

5 years ago
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