# Probability: Biased Die

Suppose we have three 6-sided die that all share the same common bias:

For a single dice: let the probability of rolling a 2 or $P(2) = 2{\times}P(1$), let the probability of rolling a 3 or $P(3) = 3{\times}P(1)$, and so on...

Such that: $P(2) = 2P(1), P(3) = 3P(1), P(4) = 4P(1), P(5)=5P(1), P(6)=6P(1)$

Given that the sum of all the probabilities is 1, we can get that $P(1) = 1/21, P(1) {\approx} .0476$

My question are as follow:

A) Given two die are rolled what is the probability that the sum of the faces of the die is 3? How about 6?

B) Given that three die are rolled what is the probability that the sum of the faces of the die is 9? What about 10?

I made this problem myself after adapting a problem I saw that involved a single dice that had the bias described above. I have a good idea of how the probability is calculated, I just wanted to see how other people approach the problem, just to make sure I am doing it the most efficient way possible. I was also wondering if there is a frequentist method of solving the problem.

Also does anyone have Diagrams of how a PDF of would look like for 2 or 3 die rolled?

• What do you mean by "frequentest" (or frequentist?) method? – Stephen Herschkorn Jul 31 '13 at 5:54
• Meaning you look at the probabilities in terms of the frequency you would expect to see them appear in a given number of trials. So for example: randomly guessing on a 40 question multiple choice test with 4 choices per question the probability of getting a correct answer is 25%. So you would expect a frequency of about 10 correct questions on the 40 question exam. en.wikipedia.org/wiki/Frequentist_probability Thanks for the heads up on the misspell. – StudentofEuler2718 Aug 3 '13 at 2:33

For (A) and two biased dice, $P(S=3)=\frac{1\times 2 + 2\times 1}{21^2} = \frac{4}{441}$ and similarly $P(S=6)= \frac{1\times 5 + 2\times 4 + 3\times 3 + 4\times 2 + 5\times 1}{441}$ (which you can simplify).

For (B) and three biased dice, you cannot get a sum above $18$.

The probability mass functions look like this, and you can see the Central Limit Theorem starting to have an impact despite the biasedness • My mistake, I made a ridiculous error in my head when I asked part B. Regardless, your pdf answers it. Thank you very much! – StudentofEuler2718 Jul 31 '13 at 1:02

I think the best way to handle this is via the probability generating function $\pi(z) = Ez^X = \frac1{21}\sum_{k=1}^6 k z^k$ for the value $X$ from a roll of a single die. The probability generating function for the sum of $n$ independent rolls is $[\pi(z)]^n$. Multiply out the polynomilas (a CAS - e.g., Mathematica® or Maple - will help) and read off the coefficents.

• I like this approach as well. Could you please describe the variables in more detail though? I have only brief experiences with generating functions. – StudentofEuler2718 Aug 3 '13 at 2:36