The Pólya urn

1. Step 1 · One order with 6 goals and 4 misses

   $\displaystyle (1 \times 2 \times \dots \times 6) \times (1 \times 2 \times 3 \times 4) \div (2 \times 3 \times \dots \times 11) = \frac{1}{2310}$

   Each goal’s chance has the goals so far on top, each miss’s the misses so far, and the bottoms run 2, 3, …, 11 in any order.

2. Step 2 · Orders with 6 goals

   $\displaystyle \binom{10}{6} = 210$

   Choose which 6 of the 10 kicks are goals.

3. Step 3 · Chance

   $\displaystyle 210 \times \frac{1}{2310} = \frac{1}{11}$

   Every order has the same chance, so multiply. The answer is $1 \div 11$ for any number of goals from 0 to 10.

Answer$\displaystyle \frac{1}{11} \approx 9.09\%$