# Dice probability question

1. Aug 11, 2014

### SYoungblood

1. The problem statement, all variables and given/known data
Hello, thank you in advance for help here. I am just trying to figure this out for my own edification. If I took 6 dice, each a regular 6 sided cube, how many possibilities are there of me rolling a four of a kind and a two of a kind? In other words, rolling a 646646, 222112, 333344, any abaaba series of numbers?

2. Relevant equations
A four of a kind is nothing more than 6C4, then you multiply that by 6 for each possibility, then 5 for each roll that is not the same as the four of a kind, and then multiply that by four, to account for the rolls that are not the same as the four of a kind or the first "odd dice out".

3. The attempt at a solution
I got that problem right, 6C4*6*5*4, or 1800. If you account for the number of possibilities of any roll, 6^6, the odds are 1800/46,656, but I cannot figure out how to go from that answer to getting a roll of a four and two. All help is appreciated, thank you.

Last edited: Aug 11, 2014
2. Aug 11, 2014

### verty

Hint 1: How many ways are there to roll four 6's and two 5's in any possible order?
Hint 2: Suppose the dice are numbered 1 to 6.

3. Aug 11, 2014

### Ray Vickson

When you multiply C(6,4) by 6*5*4 you are counting for things that do not occur in in your events of interest. You should count only the things that DO occur. There are 6 ways to choose the pair, and for each such way there are then 5 ways to choose the quartet.

4. Aug 14, 2014

### SYoungblood

So, if I am correct, it is 6C4*6*5, or 450?

5. Aug 14, 2014

### SYoungblood

Another question -- how many possibilities are there for two triplets out of 6 dice? Using the same logic, I have 6C3*6*5, or 600.

6. Aug 14, 2014

### verty

Sorry, my hint was rather cryptic because I had made a mistake. Yes, 450 is correct. And 600 must also be correct because it's the same calculation with different numbers.

7. Aug 14, 2014

### Ray Vickson

Not necessarily. Think about why there might be different counting issues for (2A,4B) and (3A,3B).

8. Aug 14, 2014

### Ray Vickson

No, the calculations are different. Look at the smaller case of 3-sided dice, where we can enumerate everything.

For (4,2) the different outcomes are (4A,2B), (4A,2C), (4B,2A), (4B,2C), (4C,2A), (4C,2B), so 6 altogether. For (3,3) the different outcomes are (3A,3B), (3A,3C), (3B,3C), so 3 altogether. Note that (3A,3B) and (3B,3A) are the same.

9. Aug 14, 2014

### verty

When this happens, it means the method is wrong. The right method should scale to analogous problems. I know from before (doing all the combinatorics problems in a well-known textbook) that binomial coefficients are not all that safe to use, I was trying to avoid them and usually I can but here, there is no easy way around them.

The formula for the 4+2 case looks deceptively general: 6c4 * 6*5. 6*5 is a number that we would expect to find in any "split the dice into two groups" problem and 6c4 looks like a very high-level operation that should scale to other problems. What could go wrong?

The fact is, binomial coefficients are the wrong decomposition for combinatorics problems.

Right, what is the correct way to approach both the 3+3 and the 4+2 cases, that works for both without any technicalities? It is this: order the groups according to the earliest roll that they contain. In the 3+3, roll 1 appears in group 1, the number 5c2 is obviously appropriate. In the 4+2, roll 1 appears in one of the two groups, two cases: 5c1 + 5c3.

I apologise for the bad advice, I do try very hard with these combinatorics problems (since I got the helper badge) to answer them in a very principled way. The lesson for me here is that a principled approach may still not scale to analogous problems, if binomial coefficients are involved.