I have the following code trying to solve the problem below:

Thrown n dice m times, calculate the probability of getting at least one 6.

I know that the exact probability of getting at least 1 six when throwing 2 dice is 11/36. 

My program below seems to want the probability to be 0.333, which is close, but it should be 11/36 right?

Great if the suggestions can continue on the standard code I have made, but vectorized code is also appreciated.

```
import random
from sys import argv

m = int(argv[1]) # performing the experiment with m dice n times
n = int(argv[2]) # Throwing m dice n times
s = 0            # Counts the number of times m dies shows at least one 6

print '%.g dice are thrown %.g times' % (m, n)

for i in xrange(n):
    list = []    # used to clear the list for new die count
    for q in xrange(m):
        r = random.randint(1,6)#Picks a random integer on interval [1,6]
        list.append(r)         #appends integer value
        if len(list) == m:     #when list is full, that is when m dice has been thrown
            for i in xrange(len(list)):
                #print list
                if list[i] == 6: #if the list of elements has a six add to the counter
                    s += 1
                    pass #I want the loop to exit when it finds an element = 6

print 'Number of times one of the n dice show at least one 6: %.g' % s  
print 'Probability of at least 1 six from %.g dice is = %2.3f' % (m,s/float(n))

```

I will edit the code and questions if something is unclear.

Sample on output:

```
Terminal > python one6_ndice.py 2 1000000
2 dice are thrown 1e+06 times
Number of times one of the n dice show atleast one 6: 3e+05
Probability of atleast 1 six from 2 dice is = 0.333

```