Question

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?

Answer 1

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.

Answer 2

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

Answer 3

he is correct though

Answer 4

you looking for this...

\[\sum_{k=1}^{\infty}k\frac{1}{6}\left(\frac{5}{6}\right)^{k-1}=6\]

Answer 5

and this

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

Answer 6

Thanks! That's really helpful

Answer 7

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

Answer 8

Nm, the K is the number of rolls.

Answer 9

correct

Answer 10

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

Answer 11

look up 'geometric sum'

Answer 12

Thanks again Zarkon!