Sorry missed your reply. The mathematical formulation is as follow:

Let \( \mathbf{X} = (X_1, X_2, \ldots, X_k) \sim \text{Multinomial}\left(n; \frac {1} {k}, \frac {1} {k}, \ldots, \frac {1} {k}\right)

\)

be the counts of each ball picked by person 1, where \( n \) is the number of picks and \( k \) is the number of balls in the bag. In particular, \( n = 6 \) and \( k = 11 \) in your question.

Similarly we can let \( \mathbf{Y} = (Y_1, Y_2, \ldots, Y_k) \) be another multinomial vector representing the picks from person 2, which has the identical distribution as \( \mathbf{X} \) and they are independent.

The number of matches, by definition, is

\( Z = \sum_{i=1}^k \min\{X_i, Y_i\} \)

and you want to find out the distribution of this discrete random variable.

I have not find a nice method to tackle this problem yet; just use R to simulate 1 million times to obtain a numerical solution first.

Code:

```
m<-1000000
k<-11
n<-6
pr<-rep(1/k,k)
x<-rmultinom(m,n,pr)
y<-rmultinom(m,n,pr)
count<-rep(0,7)
for (i in 1:m) {
j<-sum(pmin(x[,i],y[,i]))
count[j+1]<-count[j+1]+1
}
count/m
[1] 0.041412 0.209483 0.371843 0.280376 0.087191 0.009471 0.000224
```

where the last line gives you the estimate of probability mass function \( \Pr\{Z = z\}, z = 0, 1, \ldots, 6 \) from left to right.