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This is probably an extremely simple question and my lack of sleep is preventing me from understanding it completely, but I figured I would ask.

Suppose you have one fair die. You roll it 3 times. What is the probability that the first and second roll are the same number? My guess is that it is 1/6; however, I was also thinking about how to solve the problem with the frequentist approach.

We know that there are 6^3 = 216 possible combinations of a 6 sided die in 3 rolls. Given that there are only six possible combinations where the numbers of the first and second roll can be the same (1-1,2-2,3-3,4-4,5-5,6-6), I would think the probability (using this approach) would be 6/216. However, I think this is clearly wrong, since by increasing the number of rolls to say 5 (so the total number of combinations is 6^5 = 7776), the same method would say the probability of the first two numbers being the same is now 6/7776 (clearly does not make sense!).

What am I overlooking? Are there underlying assumptions of the frequentist approach that I am not considering? Clearly, the probability of the same number of the first and second roll should be independent of how many rolls we perform (beyond 2 rolls).

Thanks

Edit: Err...title should read "One Die" *embarrassed*

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# Rolling One Die 3 times

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