# Probability of dependent combined events

I have a problem that might look trivial at first but it seems to me it is not. Let´s see if somebody can help. I have to roll a dice. If a roll a 6 I get 2 points. If I roll a 5 I get 1 point. What´s the probability of getting a certain amount of points in a certain amount of rolls? Let´s say I roll the dice 10 times and I want to know the probability of getting at least 4 points. If I only consider the probability of one event, for example, rolling a 5, my chances of getting 4 points can be calculated with the binomial, being the probability of getting exactly 4 successes out of 10: $${10 \choose 4} *(1/6)*(5/6)^5=0.0542$$ The probability of getting at least 4 successes (4 points or more) can be calculated too, using the binomial, and is 0.06972. If I consider the other event (rolling a 6), I can calculate the probability of getting at least 2 successes (4 points) in a similar way, it is 0.5154. But how do I calculate the probability of getting at least 4 points? I cannot add the probabilities because there is a nonzero chance of both events occurring at the same time, and there is also a probability of no one of them happening, and getting 4 points anyway (if I roll one 6 and three 5s, for example). Of course I could do a computer simulation and calculate it by brute force, but this is just a simplification of the actual problem I´m facing, so I´m wondering if somebody could come up with an algebraic solution to it. Thank you all in advance!

• HINT: Calculate the probability of getting less than 4 and subtract that from 1. Commented Oct 25, 2013 at 11:27

The probability of getting at least $4$ points is one minus the probability of getting at most $3$ points. We can list all ways to get no more than $3$ points:

$\bullet$ no $5$s and $6$s

$\bullet$ one $5$, no $6$s

$\bullet$ two $5$s, no $6$s

$\bullet$ three $5$s, no $6$s

$\bullet$ no $5$s, one $6$

$\bullet$ one $5$, one $6$

The probabilities of these events can easily be calculated. There isn't a "nice" algebraic solution though, if the list of possibilities becomes too big you will probably have to resort to simulation.

• Just to be sure, how would you calculate, for example, the probability of getting three 5s and no 6s? Is this correct?:$$(1/6)^3*(4/6)^7*{10 \choose 3}$$ Commented Nov 21, 2013 at 9:41
• Yes, that is correct. Commented Nov 21, 2013 at 14:43

Assuming that the dice is the usual dice, and fair, and that the rolling procedure is also fair, then each roll $i$ can be mapped to suit your purposes to a discrete random variable $X_i$ taking values $\{0,1,2\}$ with probability mass function

\begin{align}P(X_i=0) =& 4/6 \\ P(X_i=1) =&1/6 \\ P(X_i=2) =& 1/6 \\ \end{align}

Then you want to consider the random variable

$$S_n = \sum_{i=1}^{n}X_i$$

that has support $\{0,1,...,2n\}$

Assuming further that each roll is independent for all others, then the expression of the distribution of the sum of i.i.d. discrete random variables having a general distribution, like your variables do, has been derived here . It indeed gets complicated as $n$ increases, but in principle it will give you the probability $p(S_n=s)$ of any desired pair $\{(n, s)\}$.

ADDENDUM: USING THE PROBABILITY GENERATING FUNCTION

The probability generating function (PGF), denote it $G(z)$, of each $X_i$ is

$$G_{X_i}(z) = \sum_{i=0}^{2}p_{X_i}(x_i)z^{x_i} = \frac 46 + \frac 14 z + \frac 14 z^2$$

Then the PGF of $S_n$ is (due to the i.i.d assumption)

$$G_{S_n}(z) = \left(\frac 46 + \frac 14 z + \frac 14 z^2\right)^n$$

The probability mass function of $S_n$ relates to its PGF by

$$P(S_n=s) = \frac {1}{s!}\frac {d^sG_{S_n}(z)}{dz^s}|_{z=0}$$

This means that if for example you want to consider the probability of getting $s=20$ points, you will have to calculate the 20th derivative of the PGF. You will end up a nerve-wreck, but it can be done (or you can use software). In the end, the expression will be greatly simplified because you will set all $z$ present equal to zero. The result will be a function of $n$, and then by varying the value of $n$, you will obtain the probability of any pair $(s=20,n)$. Etc