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!

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.
