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.
 A: 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.
A: Suppose $A_j$, $j=1\ldots k$ are mutually exclusive and exhaustive events (i.e. in each trial exactly one of these will occur) with probabilities
$p_1, \ldots, p_k$ respectively.  Take $n$ independent trials with these 
probabilities, and let $N_i$ be the number of times $A_i$ occurs in these $n$ trials.  Then $(N_1, \ldots, N_k)$ have a multinomial distribution with parameters
$n, p_1, \ldots, p_k$: 
$$P(N_1 = x_1, \ldots, N_k = x_k) = \dfrac{n!}{x_1! \ldots x_k!} p_1^{x_1} \ldots p_k^{x_k}$$
where $x_i$ are nonnegative integers with $x_1 + \ldots + x_k = n$.
If we're interested in the probability that $N_1$ is greater than at least one of $N_2, N_3, N_4$, we may as well lump all possibilities other than $A_1, \ldots, A_4$ together as $A_5$.  We then must add $P(N_1 = x_1, \ldots, N_5 = x_5)$ for all cases where $x_1 > \min(x_2, x_3, x_4)$.  Note that
$x_5 = n - x_1 - \ldots - x_4$, and we must have $x_5 \ge 0$ so $x_1 \le n - x_2 - x_3 - x_4$.
$$ \sum_{x_1 = 1+\min(x_2,x_3,x_4)}^{n - x_2 - x_3 - x_4} \sum_{x_2 = 0}^{n - x_3 - x_4} \sum_{x_3 = 0}^{n - x_4} \sum_{x_4 = 0}^n P(N_1 = x_1, \ldots, N_4 = x_4, N_5 = n - x_1 - x_2 - x_3 - x_4)$$
