Probability of rolling the same number twice in more than one round If two people roll a die, the probability of rolling the same number is $1/6$.
But what if they try it for example $3$ more times? How high is the probability that they will roll the same number at least once in $4$ rounds?
I think that after the first round the probability is $1/6$ that the numbers are the same and $5/6$ that they are not.
In a second round this continues like in the tree below:
    ___________
   /           \
  1/6          5/6
 /   \        /   \
1/6  5/6     5/6  1/6

Now to calculate the probability after the second round, that they have at least in one round the same answer is:
$$(1/6 \cdot 1/6) + (1/6 \cdot 5/6) + (5/6 \cdot 1/6) =  11/36 \approx 30\%$$
Is that right?
 A: Your reasoning is correct, however there is usually a nicer way to think about it.  Whenever you encounter a question that says "at least one time" or something of the sort a good intuition is to try to calculate the probability of the event occuring no times and subtracting from $1$.
In your example you would attempt to calculate the probability that in 2 rolls both players would roll different results.  This is, in my opinion, an easier calculation. The chance of rolling differently a single time is, as you mentioned, $5/6$, thus the chance of rolling differently both times is $5/6 \times 5/6$ or $25/36$.
Since players must either roll differently each time or the same at least once (and can't do both), the probabilities must sum to one.  Thus to get the probability that both players will roll the same thing more than once we subtract $25/36$ from $1$ to get $11/36$, the same result you arrived at.
This method is also pretty easy to calculate for larger values where as the mothod you used, while perfectly valid, can become confusing or arduous to calculate for larger values.  For example if we wanted to know the probability after 3 rounds we would just do $1 - (5/6\times 5/6\times 5/6)$ resulting in $91/216$.  In general you can calculate the probability after $n$ rounds as $1 - (5/6)^{n}$.
