Rolling two or more 6s

What is the probability or rolling at least two 6s (or more) on 3, 4, 5 and 6 regular dice?

Quote: Certified

What is the probability or rolling at least two 6s (or more) on 3, 4, 5 and 6 regular dice?

For 3 dice, there are 6*6*6=216 possible ways to land.
Of those there are 4 ways to land at least 2 6s ( x66,6x6,66x and 666 where x is not a 6).
So probability for 3 dice is 4/216=1.85186%

For 4 dice, there are 6*6*6*6=1296 possible ways to land.
Of those there are 11 ways to land at least 2 6s ( xx66,x6x6,x66x,6xx6,6x6x,66xx,x666,6x66,66x6,666x, and 6666 where x is not a 6).
So probability for 4 dice is 11/1296=0.848765432%

I'll leave the rest of it to you, because I can't be bothered to do all your homework, and because I hate you for reminding me of Alan Mendleson :)
You'll need to search the forum to understand that, but let's say it's a nightmare from the past.

Quote: Certified

What is the probability or rolling at least two 6s (or more) on 3, 4, 5 and 6 regular dice?

The probability of rolling at least two 6s on N dice = 1 minus the probability of rolling zero 6s on N dice minus the probability of rolling exactly one 6 on N dice.

Quote: OnceDear

So probability for 3 dice is 4/216=1.85186%

So probability for 4 dice is 11/1296=0.848765432%

Wait, is that right? The probability of rolling a minimum of two 6s goes down with each added die?

Quote: Certified

Wait, is that right? The probability of rolling a minimum of two 6s goes down with each added die?

Hmmmmm. Seems counterintuitive. I stuffed up.

Meanwhile, consider a pair of dice.
Probability of throwing at least 2 6s is 1/36=2.77778%


Ahhhhh, I stuffed up the numerators. Apologies.

Where x is not a 6 has 5 possible values.

So, for 3 dice it should have been
166,266,366,466,566,616,626,636,646,656,661,662,663,664,665,666 : 16 ways. Therefore p=19/216=7.4074%
and for 4 dice it should have been
171/1296 =13.2%

Sorry. Been a long day.

Quote: OnceDear

So, for 3 dice it should have been
166,266,366,466,566,616,626,636,646,656,661,662,663,664,665,666 : 16 ways. Therefore p=19/216=7.4074%

thanks very much for your calculations. if you would allow me i have a follow-on question.

is the probability of have any double (from 1,1 to 6,6) on a 3-dice combination, simply 6 times your answer?

i.e. 6 x 16 ways = 96 ways out of 216 combinations = 0.444444444

thanks v much!

This is almost certainly your homework. Wow them with the following ...

With N dice there are 6^N total rolls.

There are 5^N rolls that have no 6's.

The number of ways to roll one 6 is N*5^(N-1). The 6 can appear on any one of 6 dice and the remaining dice each have N-1 choices.

So, there are 5^N + N*5^(N-1) ways of rolling 2 or more 6's.

Note that 5^N = 5*5^(N-1), so the formula above simplifies to

(5+N)*5^(N-1)

The probability of 0 or 1 6's in N dice is

p = [(5+N)*5^(N-1)]/(6^N)

The probability of 2 or more is 1 - p.

For N = 3, 4, 5, 6 respectively, these probabilities are:

0.074074074
0.131944444
0.196244856
0.263224451

Quote: OnceDear

For 3 dice, there are 6*6*6=216 possible ways to land.
Of those there are 4 ways to land at least 2 6s ( x66,6x6,66x and 666 where x is not a 6).

This is 16 ways (x can take on 5 values in each of the above).

Quote: sleepingvery

thanks very much for your calculations. if you would allow me i have a follow-on question.

is the probability of have any double (from 1,1 to 6,6) on a 3-dice combination, simply 6 times your answer?

i.e. 6 x 16 ways = 96 ways out of 216 combinations = 0.444444444

thanks v much!

Any double is easy enough.

For N dice (N <= 6), to not get a double the first dice has 6 choices, the second dice has 5 choices, the third dice has 4 choices, and so on. So, the number of ways of not getting a double is

X = 6*5*4* ... *(6-N+1).

For example, with N = 3, there are 6 choices for the first dice, 5 for the second and 4 for the third, so there are 6*5*4 = 120 ways of not getting doubles with 3 dice.

The probability of not getting doubles with N dice (N <= 6) is

p = 6*5*4* ... * (6-N+1)/6^N.

The probability of getting doubles with N dice is just 1 - p.

For N = 3, 4, 5, 6 these probabilities are

0.444444444
0.722222222
0.907407407
0.984567901

Quote: teliot

Any double is easy enough.

thanks very much for your explanations teliot!

Quote: Certified

What is the probability or rolling at least two 6s (or more) on 3, 4, 5 and 6 regular dice?

here what i got
I used combinations and permutations
one could just count the ways too (for a check)

# of dice	choose 2	choose 3	choose 4	choose 5	choose 6	6s ways	total ways	prob
2	1	0	0	0	0	1	36	0.027777778
3	15	1	0	0	0	16	216	0.074074074
4	150	20	1	0	0	171	1296	0.131944444
5	1250	250	25	1	0	1526	7776	0.196244856
6	9375	2500	375	30	1	12281	46656	0.263224451

Sally

of course, this is a binomial distribution type problem too.
a simple calculator will work
like this one

http://vassarstats.net/binomialX.html

n=# of dice
k=number of 6s wanted
p=1/6 (the prob of rolling a 6 on one die)

have fun!!

here is Excel in Google
i had it

https://goo.gl/oeHlky

easy when one knows how
Sally