Dice probability problem

[FfortynineE]

Here is the problem:

5. [20 point(s)] An unbiased 12-sided die has its faces numbered 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6,
and 6. This die is rolled until a 5 is observed.
(a) Find the expected number of rolls needed?
(b) A pair of such 12-sided dice is rolled. Find the expected number of rolls needed to
observe at least one 5 on at least one of the two dice.
(c) Find the smallest number of dice required so that the expected number of rolls needed
to observe at least one 5 does not exceed 5.

For part A, I figure that mu = 1/(1/6) and so six rolls will be needed.
For part B, i would think this number would be halved, since doubling the number of die doubles the chance of a 5. Part C leads me to believe that this is incorrect. Using two die also doubles the chances of getting any other number, so would the expected number of rolls be the same as in part A?
Part C, no idea, I'm hoping some illumination with regards to parts A and B will help me figure this part out.


Edit:

While I'm here I figure I might as well post another question which confuses me:

Suppose random variable X takes on only the values 10 and 12. If the expected value of X is 10.50, find the variance of X.

I'm not really sure how to approach this. E(X) = sum(x*f(x)), so I'm starting with 10.5 = (10)*c1 + (12)*c2. Var(X) = E(X[sup:tn9ku1t4]2[/sup:tn9ku1t4])-(E(X))[sup:tn9ku1t4]2[/sup:tn9ku1t4]

Thanks!

 

[soroban]

Hello, FfortynineE!

5. An unbiased 12-sided die has its faces numbered 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6.
This die is rolled until a 5 is observed.

\(\displaystyle \text{Since }\;P(1) \:=\(2)\:=\(3) \:=\(4)\:=\(5)\:=\(6)\;=\:\tfrac{1}{6}\)
. . \(\displaystyle \text{we can consider a standard 6-sided die.}\)

(a) Find the expected number of rolls needed.

Your reasoning is correct.

\(\displaystyle P(5) \,=\,\frac{1}{6} \quad\Rightarrow\quad E \:=\: \frac{1}{P(5)} \:=\:6\)

(b) A pair of such 12-sided dice is rolled. Find the expected number of rolls
needed to get at least one 5 on at least one of the dice.

Be careful . . .

\(\displaystyle \text{With a pair of dice: }\(5) \:=\:\frac{11}{36}\)

\(\displaystyle \text{Hence; }\:E \;=\;\frac{36}{11} \;\approx\;3.27\)

(c) Find the smallest number of dice required so that the expected number
of rolls needed to observe at least one 5 does not exceed 5.

In part (b) we already see that with two dice: \(\displaystyle E \leq 5.\)

Do they want a rigorous proof?


\(\displaystyle \text{With }n\text{ dice: } \(\text{no 5's}) \:=\:\left(\tfrac{5}{6}\right)^n\)

\(\displaystyle \text{Then: }\(\text{at least one 5}) \:=\:1-\left(\tfrac{5}{6}\right)^n\)

\(\displaystyle \text{Hence: }\:E \;=\;\frac{1}{1-\left(\frac{5}{6}\right)^n}\)


\(\displaystyle \text{We want: }\;\frac{1}{1-(\frac{5}{6})^n} \;\leq \;5 \quad\Rightarrow\quad 1\;\leq\;5 - 5\left(\tfrac{5}{6}\right)^n\)

. . . . . . . . . \(\displaystyle 5\left(\tfrac{5}{6}\right)^n \;\leq \;4 \quad\Rightarrow\quad \left(\tfrac{5}{6}\right)^n \;\leq \;0.8\)


\(\displaystyle \text{Take logs: }\;\ln\left(\tfrac{5}{6}\right)^n \;\leq\;\ln(0.8) \quad\Rightarrow\quad n\cdot \ln\left(\tfrac{5}{6}\right) \;\leq\;\ln(0.8)\)


\(\displaystyle \text{Hence: }\;n \;\geq \;\frac{\ln(0.8)}{\ln(\frac{5}{6})} \quad\hdots\;\text{Note: we divided by }\ln\left(\tfrac{5}{6}\right)\text{, a negative quantity.}\)


\(\displaystyle \text{Therefore: }\;n \;\geq \;1.223901086 \quad\Rightarrow\quad n \;\geq \;2\)

 

[FfortynineE]

Thanks for the help soroban! Your comment cracks me up; I'm pretty sure in any given college math/engineering/comp sci. class there's at most 3 people who do the homework honestly.

Anyone have any thoughts on the seconds problem?

 

[Denis]

Anyone have any thoughts on the seconds problem?

We'll do it for you at 10.50 PM :wink: