There is such a game with chips or cards. There are an even number of chips, and there are N different types of chips among them. Chips are randomly shuffled and divided equally between two players. The winner is the one with the most different types of chips. And we need to find the probability of a draw in the game, with a certain number of different chips.
For example:
black - 3; white - 2; red - 1;
Permutations:
rww vs bbb : probability - 1/10 (not draw)
rwb vs wbb : probability - 3/5 (not draw)
rbb vs wwb : probability - 3/10 (DRAW!)
So we need to find out the number of permutations with a draw.
I'd be appreciated for any ideas. Thanks.
P.S. (the problem is practical)
The number of chips of the same color <= 10
The number of colors <= 50
Suppose there are $t_i$ chips of type $i$, for $i = 1 \ldots k$, with $\sum_i t_i = 2 n$. Let $X_i$ be the number of chips of type $i$ given to player $1$. Thus if $x_i$, $i=1\ldots k$, are integers with $\sum_i x_i = n$ and $0 \le x_i \le t_i$, $$ \mathbb P(X_1 = x_1, \ldots, X_k = x_k) = \frac{\displaystyle \prod_i {t_i \choose x_i}}{\displaystyle {2n \choose n}} $$
To get the probability of a draw, you must sum these up for all $(x_1, \ldots, x_n)$ such that the number of $i$ for which $x_i = 0$ is equal to the number of $i$ for which $x_i = t_i$. So for each pair of disjoint subsets $A, B$ of $\{1, \ldots, k\}$ with equal cardinality, you want to consider $$ \mathbb P(X_i = 0\ \text{for}\ i \in A,\; X_i = t_i\ \text{for}\ i \in B,\; 1 \le X_i \le t_i-1\ \text{otherwise}) = \sum_x \frac{\displaystyle \prod_{i \notin A \cup B} {t_i \choose x_i}}{\displaystyle {2n \choose n}}$$ where the sum is over $x_i = 1 \ldots t_i-1$ for $i \notin A \cup B$, with $\sum_{i \notin A \cup B} x_i = n-\sum_{i\in B} t_i$. Then add those up for all such $A$ and $B$. I have no reason to think this will simplify in any nice way.
Thus in your example, you have $k=3$, $t = [3,2,1]$, $n=3$. Since $t_3 = 1$, $1 \le x_3 \le t_3 - 1$ is impossible, so $3$ must be in $A$ or $B$. For $A = \{1\},\; B = \{3\}$ we get $ \mathbb P(X_1 = 0,\; 1 \le X_2 \le 1,\; X_3 = 1) = 0$ because that would make $\sum X_i=2$, not $3$. Similarly for $A = \{3\},\; B = \{1\}$. For $A = \{2\},\; B = \{3\}$ we have $$\mathbb P(1 \le X_1 \le 2,\; X_2 = 0,\; X_3 = 1) = \mathbb P(X_1 = 2,\; X_2 = 0,\; X_3 = 1) = \frac{\displaystyle{3 \choose 2}}{\displaystyle{6 \choose 3}}= \frac{3}{20}$$ and similarly for $A=\{3\},\; B = \{2\}$. So the probability of a draw is $3/20 + 3/20 = 3/10$.