# Probability question with similar/distinct dice.

The question:

We're throwing $$3$$ normal dice twice. (Suppose that every vector with length of $$6$$ has an equal probability). What is the probability to get two similar results from the two throws of the $$3$$ dice if:
a) The dice are different.
b) The dice are similar.

My Work:
for (a):
$$\Omega = \{(a_1,a_2,...,a_6) \mid a_i \in \{1,...,6\} ,(i=1,2,...,6)\}$$
if all the dice are different then: $$|\Omega| = 6^6$$
Let $$A\subset\Omega$$, ($$A$$- we get similar results from the two throws), then $$|A|=6^3$$ (first dice throw can be whatever, but the second one must be the same).
So the final answer I got is $$P(A) = \frac {6^3}{6^6}$$.

for (b): (and here comes the confusion):
Things I tried to do and didn't get the right answer:

1. I tried to choose $$3$$ of the dice (since they're all similar) throw them, and sort them, so what I got is $$\binom{6}{3} \cdot 6^3 \cdot 3!$$.
2. I tried to choose $$1$$ of the two throws, and find how many options there are and sort them (same as before, each dice has $$6$$ sides so $$6^3$$ for all), and got $$\binom{2}{1} \cdot 6^3 \cdot 3!$$.

Both of the two ways led me to wrong answers, I'm trying where are my mistakes and why my thinking led me to wrong answers, which can lead me to reach the true answer.

The Final answer is $$\frac {996}{6^6}$$

EDIT:
After deeply thinking about the problem I obtained a new approach (I couldn't reach the answer with it, but I'm not sure if it's my weak combinatorics or if it's a wrong approach).
3. By writing down a couple of examples, I noticed that there's something that I can also use to calculate the probability, we can divide this into $$3$$ possible cases:
First: Getting same number on all of the six dice. $$(2,2,2, 2,2,2)$$
Second: Getting $$2$$ different numbers. $$(2,4,2, 2,4,2)$$
Third: Getting $$3$$ different numbers. $$(1,2,3, 1,2,3)$$
So I could calculate the number of options each case has and divide by the total options $$|\Omega|=6^6$$.
I felt like I've gone too far but I can't see why it wouldn't be true. Any Feedback is really appreciated.
• The word dice is plural; the singular is die. The word dices means to cut into small cubes. – N. F. Taussig Mar 21 at 10:19
• Formatting tip: type $\binom{n}{k}$ to obtain $\binom{n}{k}$. – N. F. Taussig Mar 21 at 10:23
• Your new approach is the right one. However, I don't see how the author obtained $996$ favorable cases. – N. F. Taussig Mar 21 at 10:40
• @N.F.Taussig Thanks alot I'm happy to hear that, could you recognize why the first two approaches reached wrong answers? I still can't understand why they led me to wrong answers and trying to see what problems I have there if it's using binomials the wrong way or just my logic wasn't true. Like for example in approach 1, I chose the $3$ dice of the first throw, and they could be whatever, so the cases that I added in my new approach (according to my understanding) are included in the first approach. seems like I have a big misunderstanding somewhere. – Pwaol Mar 21 at 10:48
• Your first two approaches fail since if your original outcome was, say $(5, 5, 4)$, there are only three ways the second roll could match the first, depending on which die matched the $4$, not $3!$. – N. F. Taussig Mar 21 at 10:52

With the help of N. F. Taussig in the comments, I think I have reached the right solution, So I decided to answer my own question, and would love to hear feedback.

Starting with why my first two approaches are wrong, The number of ways that the second roll could match the first is different whenever there's $$3$$ different numbers or $$2$$ different or all of them are the same, but in my first two approaches, I didn't pay attention to that.

Solution: (Approach 3):
The number of ways to get the same number on all dice, is just simply choosing a number which is $$\binom{6}{1}=6$$.
The number of ways to get two different numbers in the first roll is:
$$\binom{6}{2}$$ choosing $$2$$ numbers, $$\binom{2}{1}$$ choosing the number that will appear twice, $$\binom{3}{1}$$ choosing which place is the single number in the first roll, $$\binom{3}{1}$$ choosing in which place is the single number in the second roll. So in total by multiplying all of them we get $$\binom{6}{2}*2*3*3 = 270$$ ways.

The number of ways to get three different numbers in the first roll is:
$$\binom{6}{3}$$ choosing $$3$$ numbers, $$3!$$ ordering the first roll, $$3!$$ ordering the second roll. in total $$720$$.

Hence we reach the answer by adding them up and dividing by $$|\Omega|$$.

• This is correct. The reason I did not initially obtain $996$ was that I forgot that there were only three dice. – N. F. Taussig Mar 21 at 11:14