How many ways can you roll a four of a kind and a two of a kind with 6 dice?

In summary, the conversation discusses the problem of rolling a four of a kind and a two of a kind with six 6-sided dice. The correct formula for finding the number of possibilities is 6C4*6*5, which equals 450. The conversation also explores the problem of two triplets out of six dice, which has a different formula of 6C3*6*5, or 600. However, it is pointed out that this method is not entirely accurate and a different approach is needed. The correct approach is to order the groups according to the earliest roll that they contain, and for the 4+2 case, this would be two cases of 5C1 + 5C3. It
  • #1
SYoungblood
64
1

Homework Statement


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?


Homework 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".



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.
 
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  • #2
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
SYoungblood said:

Homework Statement


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?


Homework 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".



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.

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
So, if I am correct, it is 6C4*6*5, or 450?
 
  • #5
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
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
SYoungblood said:
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.

Not necessarily. Think about why there might be different counting issues for (2A,4B) and (3A,3B).
 
  • #8
verty said:
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.

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
Ray Vickson said:
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.

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.
 

1. What is the probability of rolling a specific number on a single six-sided die?

The probability of rolling a specific number on a single six-sided die is 1/6 or approximately 16.67%. This is because there are six possible outcomes (numbers 1-6) and only one of those outcomes will result in the specific number being rolled.

2. What is the probability of rolling a specific number on multiple dice?

The probability of rolling a specific number on multiple dice is calculated by multiplying the individual probabilities for each die. For example, the probability of rolling a 2 on two six-sided dice is (1/6) x (1/6) = 1/36 or approximately 2.78%. This is because there are six possible outcomes for each die and only one of those outcomes will result in both dice showing the specific number.

3. What is the probability of rolling a certain sum on multiple dice?

The probability of rolling a certain sum on multiple dice can be calculated by finding all the possible combinations of numbers that can result in that sum and dividing it by the total number of possible outcomes. For example, the probability of rolling a sum of 7 on two six-sided dice is 6/36 or 1/6, as there are six possible combinations (1+6, 2+5, 3+4, 4+3, 5+2, 6+1) that can result in a sum of 7 out of 36 total outcomes.

4. How does the probability change if a die is loaded or biased?

If a die is loaded or biased, the probability of rolling a specific number will no longer be equal to 1/6. The exact change in probability will depend on how the die is loaded or biased. For example, if a die is weighted to favor one specific number, the probability of rolling that number will increase while the probabilities of rolling the other numbers will decrease.

5. How can dice probability be used in real-world applications?

Dice probability can be used in various real-world applications, such as in gambling, statistics, and decision-making. For example, understanding the probability of rolling certain numbers can help in making strategic decisions in games of chance, or in analyzing data in fields such as finance and economics. In addition, the concept of probability is important in many scientific fields and can be applied to various experiments and studies.

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