# Probability of rolling identical numbers with polyhedral dice. [closed]

Is there a way to calculate the number of identical rolls in polyhedral dice rolls?

A specific example: I roll $3$ times a $12$-sided dice.

What are the odds of rolling some number twice with 3 dice? E.g. $(4,4,7)$ , $(4,2,4)$...

AND: What are the odds of rolling some number $3$ times with $3$ dice? E.g. $(4,4,4)$, $(11,11,11)$...

• What have you tried? Do you know how to calculate the probability of this when rolling 2 dice? – Matthew Conroy Mar 24 '17 at 21:46

For rolling three $n$-sided dice and getting exactly one pair of dice which match (counting results like (4,2,4) but not counting results like (6,6,6)) we count how many possible triples of numbers have this property via multiplication principle.

• Pick the number used for the pair
• Pick the number used for the singleton
• Pick the location in the triple used by the singleton

There are $n$ choices for the number used for the pair. Given that selection, there are $n-1$ remaining choices to use for the singleton. Regardless of these choices, there are $3$ positions available for the singleton to be placed in in our triple. Applying multiplication principle, there are then $3n(n-1)=3n^2-3n$.

In the case of a $d6$ that would be $6\cdot 5\cdot 3 = 90$ possibilities.

If we were to include also the results where all three dice show the same result, that would be an additional $n$ possibilities, bringing the new total to $3n(n-1)+n=3n^2-2n$, in the case of a $d6$ a total of $90+6=96$ possibilities.

To find the probability, we divide by the total number of triples, which would be $n^3$, seen also by multiplication principle. In the example of a $d6$ that would be $6^3=216$.

The correct probability for the six-sided dice example would then be $\frac{90}{216}$ if we are concerned with exactly one pair, and would be $\frac{96}{216}$ if we are concerned with at least one pear.

The term "odds" is related in concept to probability, but technically different.

While the probability for exactly one pair in the six-sided dice example is $\frac{90}{216}$ the odds will instead be $90~::~126$, in reference to how there are $90$ favorable outcomes versus $126$ unfavorable outcomes.

This sounds awfully like a homework question. Please include the homework tag.

If you have three 6 sided dies then the probability of rolling the same number two times is $96/216$. The probability of all the dies having the same value is $6/216$.

Why 216? That is the total amount of possible outcomes from the dies.

Then you can count the amount of times a specific outcome would occur.

For a 12 sided dice it follows the same logic. The probability of all the 3 dies having the same value is $12/1728$. The probability of 2 of them being the same is $408/1728$.

• Why is $66$ the numerator? Why is $276$? – JMoravitz Mar 24 '17 at 22:03
• I think it's wise to let students do some thinking on their own and do their own homework. This is a hint in the right direction. It's for him to understand the problem. – JahKnows Mar 24 '17 at 23:34
• Do you think its wise to also give incorrect numbers which might serve to only confuse them further? – JMoravitz Mar 24 '17 at 23:35
• Fixed. Thanks for pointing it out. – JahKnows Mar 24 '17 at 23:48