Six-Sided Dice Probability - Getting a 5 or 6

What are the chances of four of X (variable) rolled six-sided dice producing a result of “5” or “6” in one roll?

4 dice = 1.2%

5 dice = ???

6 dice = ???

etc.

I know that each die has a 1 out of 3 (or 1/3, or 33.4%) chance of producing the desired result of 5 or 6 (also called a "hit"). I know that the chances of all four dice of a four-die roll "hitting" is equal to the chance of one of them hitting times the chance of each other one hitting, i.e.

(1/3)*(1/3)*(1/3)*(1/3)

Thank you for your time and your help!

Re: Six-Sided Dice Probability - Getting a 5 or 6

Quote:

Originally Posted by

**jpmadigan1** What are the chances of four of X (variable) rolled six-sided dice producing a result of “5” or “6” in one roll?

4 dice = 1.2%

5 dice = ???

6 dice = ???

etc.!

I simply do not understand.

Are you asking: If we roll seven dice, what is the probably getting at **least one** 5 or 6?

OR, Are you asking: If we roll seven dice, what is the probably getting at **exactly four ** 5's or 6's?

Or is it some else all together?

Re: Six-Sided Dice Probability - Getting a 5 or 6

If we roll seven dice, what is the probability of getting exactly four 5's or 6's.

Sorry for the ambiguity.

Re: Six-Sided Dice Probability - Getting a 5 or 6

Quote:

Originally Posted by

**jpmadigan1** If we roll seven dice, what is the probability of getting exactly four 5's or 6's.

Sorry for the ambiguity.

If $\displaystyle N\ge 4$ is the number are dice rolled the probability of getting exactly four 5's or 6's is:

$\displaystyle \binom{N}{4}\left(\frac{2}{3}\right)^4\left(\frac{ 1}{3}\right)^{N-4}$

where $\displaystyle \binom{N}{4}=\frac{N!}{4!(N-4)!}$