Probability of an event happening more times than at least one other event I'm having trouble formulating this problem: How to find out the probability of an event A happening more times than at least one of events B, C or D? Assuming the events may have different chances of happening.
For example, if I throw a 6-sided die 10 times, what are the odds that 2 will appear more often than either 3, 4 or 5?
Context: I was discussing the Settlers of Catan boardgame with some friends. In the game you throw 2 dice and add-up the results. I'm trying to figure out, after throwing the pair of dice 10 times, what the odds are that the sum 4 will appear more times than 6, 7 or 8. Thanks in advance.

For your example , total outcomes =$6^{10}$ ,each equally likely.
Now the hard part is to calculate the no. of favorable cases.  Take the cases where 3 occurs $x$ times with $x$ ranging from 1 to 10 and no. of time 2 appears goes from 0 to $x-1$ for each case.(do this by counting ways to select req. no. of positions from 10 positions  ).
Similarly Take the cases for 4 . But consider only  the cases where appearance of 3 and 5 are less than or equal to the appearance of 2.
Similarly Take the cases for 5. But consider only  the cases where appearance of 4 and 3 are less than or equal to the appearance of 2.
