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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

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    $\begingroup$ $6^2/6^4$ is not 50%. Otherwise, I think you've got it. $\endgroup$ – Will Orrick Nov 21 '15 at 21:34
  • $\begingroup$ for real, you're right.... it's $$\frac{1}{36}$$. $\endgroup$ – dc3rd Nov 21 '15 at 21:44
  • $\begingroup$ "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... $\endgroup$ – vonbrand Nov 21 '15 at 21:48
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    $\begingroup$ @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 $\endgroup$ – dc3rd Nov 21 '15 at 21:50
  • $\begingroup$ In that case, the first roll is completely irrelevant. $\endgroup$ – vonbrand Nov 21 '15 at 21:55
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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?

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  • $\begingroup$ Be careful. $\frac{6^2}{6^4} = \frac{1}{6^2}$. $\endgroup$ – N. F. Taussig Nov 21 '15 at 22:29
  • $\begingroup$ 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 $\frac{1}{36 * 36}$? $\endgroup$ – dc3rd Nov 21 '15 at 22:30
  • $\begingroup$ 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. $\endgroup$ – user19405892 Nov 22 '15 at 0:17
  • $\begingroup$ actually I just worked it out. but the only way I was able to work it out was through writing out all of the cases. is there a way of avoiding that or in this case the only way to solve it is by writing the cases. This one was symmetric so the cases were able to be reduced, but that won't always be the case right? $\endgroup$ – dc3rd Nov 22 '15 at 0:45
  • $\begingroup$ 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 $6\dfrac{1}{6^4}+30\dfrac{2}{6^4} = \dfrac{11}{216}$. $\endgroup$ – user19405892 Nov 22 '15 at 1:04

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