Probability distribution of rolling 6 dice and getting 6, 5, 4, 3, 2 and 1 unique results

Question

I understand that getting all 6 to be unqiue is the formula:

1 x 5/6 x 4/6 x 3/6 x 2/6 x 1/6

...as the first number does not matter, but as we continue to roll dice numbers will be 'banned' from appearing otherwise the result of 6 uniques will not occur. Therefore the answer is 5/324.

I cannot get how to calculate the other questions though. Please explain with working, thanks!

Answer 1

There are many ways to approach this problem. I will use inclusion/exclusion for 2-4.

The easiest one, which you didn't provide, is the number of ways of getting one different result.

To have all be the same, we have 6 ways (all of each value), so

P(1) = 1/6^5 = 1/7776

P(2) We have C(6,2) possible pairs. Then, the number of ways with 2 numbers is 2^6 - 2*1 (all numbers the same), so P(2) = C(6,2)*(2^6-2)/6^6 = 15*62/6^6 = 155/7776

P(3) = C(6,3) * (3^6 - 3*2^6 + 3*1^6)/6^6 = 20 * 540/6^6 = 25/108

P(4) = C(6,4) *(4^6-4*3^6+6*2^6-4)/6^6 = 23400/6^6 = 325/648

P(5) = you have to have 1 number twice and 1 number not at all. Then, the number selected twice can be in C(6,2) positions, and the other 4 can be in 4!, so the probability is 6*5*C(6,2)*4!/6^6 =

10800/6^6 = 25/108

P(6) = 5/324 as given (you could have derived this as 6!/6^6)

Just for fun, I also programmed the solution in R.

The code is easy to understand. You could put it into your language of choice

counts <- rep(0,6)

for(a in 1:6){

for(b in 1:6){

for(c in 1:6){

for(d in 1:6){

for(e in 1:6){

for(f in 1:6){

flags <- rep(0,6)

flags[a]<- 1

flags[b] <- 1

flags[c] <- 1

flags[d]<- 1

flags[e] <- 1

flags[f] <- 1

count <- sum(flags)

counts[count] <- counts[count] + 1

}}}}}}

prob <- counts/6^6

counts