Dice probabilities 6 dice

The Wizard has answered dice probabilities for 5 Dice with this question: What are the probabilities for a 5 of a kind, 4 of a kind, 3 of a kind, full house, 2 pair, pair, straight, and nothing with the roll of five dice? in the "Probability - Dice" section.

I thought I would try to do the same for a 6 Dice section (of course would also have to calculate 6 of a kind then). i was able to figure out:
6 of a kind
5 of a kind
4 of a kind
1 of a kind

But got lost trying to figure out 3 of a kind and 4 of a kind and full house and 2 pair and 3 pair (which is now possible with 6 dice).

Anyone could with probability that could show me how to calculate this?

Quote: Bibbs1000

The Wizard has answered dice probabilities for 5 Dice with this question: What are the probabilities for a 5 of a kind, 4 of a kind, 3 of a kind, full house, 2 pair, pair, straight, and nothing with the roll of five dice? in the "Probability - Dice" section.

I thought I would try to do the same for a 6 Dice section (of course would also have to calculate 6 of a kind then). i was able to figure out:
6 of a kind
5 of a kind
4 of a kind
1 of a kind

But got lost trying to figure out 3 of a kind and 4 of a kind and full house and 2 pair and 3 pair (which is now possible with 6 dice).

Anyone could with probability that could show me how to calculate this?

And don't forget 2 trips (AAABBB.) For this, I would do: ways = [6*5/2] * [6*5*4 / (3*2)] = 300. So, the probability of rolling 2 trips is 300(1/6)⁶.

A full house would be this pattern: AAABBC. Ways = [6*5*4] * [6 * (5*4/2)] = 7200. Probability = 7200(1/6)⁶.

3 pair would be the pattern AABBCC. Ways = [6*5*4 / (3*2)] * [(6*5/2) * (4*3)/2] = 1800.

2 pair would be AABBCD. Ways = [(6*5/2) * (4*3)/2] * [(6*5/2) * (4*3)] = 16200.

4-of-a-kind would be AAAABC. Ways = [6 * (5*4/2)] * [6*5] = 1800.

4-of-a-kind with pair would be AAAABB. Ways = [6*5] * [6*5/2] = 450.

Thank you very much for the reply. I like the approach as it is very structured and logical.

However, I'm sorry to be so dense as I can't quite tell where the numbers come from - particularly those answers that have divided by 2 in them.

Take for example the 2 trips answer at the top of your response. I can see from simulation that your response is correct. But I can't figure out the logic. I was trying to solve using combinations so for AAABBB I had 6 ways for the A and 5 ways for the B. There are 6C3 ways for the AAA to occur and 3C3 ways for the BBB to occur (since there are only 3 dice left after considering the AAA). If I multiply 6 x 5 x 6C3 x 3C3 I get 600 which is exactly double what the answer should be. So I should divide by 2 as you've done - but for the life of me I can't figure out why this is needed.

My approach gets even worse with the 3 pair or AABBCC. I had 6 ways for the A, 5 ways for the B and 4 ways for the C. Then I said there are 6C2 ways for the A's to be grouped and 4C2 ways for the Bs to be grouped (since there are 4 dice left after considering the As) and 2C2 ways for the Cs to be grouped after considering the As and Bs. This works out to 6 x 5 x 4 x 6C2 x 4C2 x 2C2 = 10,800. No where near the correct answer of 1,800. So clearly my approach is very flawed. But I can't figure out why this doesn't work.

If you could explain to me where all those numbers come from in your answer that would help me figure it out. Your approach is much simpler than doing combinations and factorials, but I'm struggling to understand the logic.

Much appreciated.

Quote: Bibbs1000

Thank you very much for the reply. I like the approach as it is very structured and logical.

However, I'm sorry to be so dense as I can't quite tell where the numbers come from - particularly those answers that have divided by 2 in them.

Take for example the 2 trips answer at the top of your response. I can see from simulation that your response is correct. But I can't figure out the logic. I was trying to solve using combinations so for AAABBB I had 6 ways for the A and 5 ways for the B. There are 6C3 ways for the AAA to occur and 3C3 ways for the BBB to occur (since there are only 3 dice left after considering the AAA). If I multiply 6 x 5 x 6C3 x 3C3 I get 600 which is exactly double what the answer should be. So I should divide by 2 as you've done - but for the life of me I can't figure out why this is needed....

I do the 2 trips in two parts. First, the two trips can be 1s and 2s, 1s and 3s, 1s and 4s, 1s and 5s, 1s and 6s, 2s and 3s, 2s and 4s, 2s and 5s, 2s and 6s, 3s and 4s, 3s and 5s, 3s and 6s, 4s and 5s, 4s and 6s, or 5s and 6s. I count 15 ways to choose those two numbers. The quick way to do that is: combin(6,2) = 6*5/2 = 15. (Whenever you see that I divided by 2, you know I had used the combination function.)

And second, the numbers can occur on various dice. For example, for three 1s and three 2s, the 1s can occur on the 1st, 2nd, and 3rd dice; 1st, 2nd, and 4th dice; 1st, 2nd, and 5th dice; 1st, 2nd, and 6th dice; etc. Rather than write out all the possibilities to count them, you know that there are combin(6,3) = 6*5*4/(3*2) = 20 ways the three 1s can occur. (We don't have to count where the three 2s occur because they can only happen where the 1s didn't.)

So, the number of ways for two trips is: combin(6,2) * combin(6,3) = 15*20 = 300.


In each of my other calculations, the first part is counting the number choices, and the second part is counting on which dice the numbers happen on.

Thanks for this. I think I (mostly) get it now.

I think if you have two trips then you can make a Full House, e.g. 666555 can make 66655, so it should be added into that column. Similarly 665544 creates Two Pairs.

It can be difficult to work out when you need to divide by two (or six sometimes). For instance with three pairs you might think it's 6*5*4, but this counts all ways to get 654 (654 645,564 546 465 456), so you need to divide by 6.