# Math Help - Dice problem

1. ## Dice problem

5 regular 6 sided dice are thrown. What is the probability to get at least 4 dice with 6:es?

The "at least" part is a bit tricky.

Solved it.

Found this video that helped me solve it.

2. ## Re: Dice problem

Originally Posted by TriForce
5 regular 6 sided dice are thrown. What is the probability to get at least 4 dice with 6:es?
Do not over complicate it. Either four or five:
$5\left( {\dfrac{5}{6}} \right){\left( {\dfrac{1}{6}} \right)^4} + {\left( {\dfrac{1}{6}} \right)^5}$

3. ## Re: Dice problem

"At least four" means "four or five". I don't see why that would be "tricky".

Write "S" to mean a six, "N" to mean "not a six". Then one way of getting 4 sixes in five rolls would be "SSSSN", another would be "SSSNS", etc. It should be easy to see that there are five ways (NOT 4) to write those: the single "N" in each of the 5 places. And the probability of each of those is $\left(\frac{5}{6}\right)\left(\frac{1}{6}\right)^4$ so the probability of "four sixes in five rolls" is $5\left(\frac{5}{6}\right)\left(\frac{1}{6}\right)^ 4$.

The probability of 5 sixes in 5 rolls is, of course, $\left(\frac{1}{6}\right)^5$. The probability of "at least four sixes= either 4 sixes or 5 sixes" is the sum of those:
$5\left(\frac{5}{6}\right)\left(\frac{1}{6}\right)^ 4+ \left(\frac{1}{6}\right)^5$.

(Again, "5", not "4". That could also be calculated as $\frac{5!}{4! 1!}$.)

4. ## Re: Dice problem

Yes, thanks for the clarifications, it's a lot easier to recognize the path to solution when the technique is clarified.

And the technique is basically:
First step is identifying where the at least bit is, in my case it was at 4 out of 5.
Next step is to sum all remaining bits, including the staring one, with the following formula:
P = n C r * p^r * q^(n-r) ~ as explained in the video.