# What is the probability two die show the same values on their second rolls?

What is the probability two die, a red and a white one, show the same values on their second rolls as on their first rolls?

so i first computed the total number of ways of just getting outcomes from two sets of rolls:

$$6^4$$

now the next step I wanted to perform was to obtain the number of outcomes in which the value obtained on the first roll is obtained on their second rolls:

$$6^2$$

my reasoning for the $6^2$ is that after we get whatever number for the first set of rolls, the second set doesn't matter. But I feel this just may be too high an outcome because the probability would work out to $0.50$

I am not concerned about finding the probability, I am concerned on performing the combinatorial idea so any feedback there would be helpful. Thanks

• $6^2/6^4$ is not 50%. Otherwise, I think you've got it. – Will Orrick Nov 21 '15 at 21:34
• for real, you're right.... it's $$\frac{1}{36}$$. – dc3rd Nov 21 '15 at 21:44
• "Same value on each die on each of two rolls" or "roll two dice twice, the faces are the same on the second roll"? Quite different... – vonbrand Nov 21 '15 at 21:48
• @vonbrand it is the latter. That is how they phrased it in the textbook and it took me a moment to decipher what they meant – dc3rd Nov 21 '15 at 21:50
• In that case, the first roll is completely irrelevant. – vonbrand Nov 21 '15 at 21:55

The answer is simply $P(\text{both dice show same number on the first and second rolls}) = \dfrac{6^2}{6^4} = \dfrac{1}{36}$.
Sidenote: say the order didn't matter in which you got the same pairs. For example, $(4,3)$ and $(3,4)$ are the same. What would be the answer in that case?
• Be careful. $\frac{6^2}{6^4} = \frac{1}{6^2}$. – N. F. Taussig Nov 21 '15 at 22:29
• If order didn't matter the result could occur in $\frac{1}{36}$ ine way and the exact same the other way, would it be $\frac{1}{36 * 36}$? – dc3rd Nov 21 '15 at 22:30
• But some would have only $1$ copy, i.e. $(1,1)$ while others such as $(4,3)$ would have two copies, i.e. $(4,3),(3,4)$. So you must account for that. – user19405892 Nov 22 '15 at 0:17
• There are only $6$ cases where we have only $1$ copy: $(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)$. So the probability of matching for each of those is $\dfrac{1}{6^4}$. On the other hand, the probability of matching for each of the other $6^2 - 6 = 30$ cases where there are $2$ copies is $\dfrac{2}{6^4}$. Therefore, the probability here is $6\dfrac{1}{6^4}+30\dfrac{2}{6^4} = \dfrac{11}{216}$. – user19405892 Nov 22 '15 at 1:04