Number of ways a penalty kick shootout can be decided in the first round

Solution 1:

I know this question was asked about 2 years ago, but it is never too late if the outcome is good!

Compute the total amount of 10-penalty shoot-out combinations

The way I solved this problem was to compute all possible combinations for the first-to-shoot team (team A) in order to win, lose or draw with respect to the other team (team B) in the 10-penalty shoot-out. The green ticks and the red crosses indicate whether a penalty is scored or not, respectively. I used tuples for counting the number of goals and misses for each scenario.

Notice that some of the shoots are shaded, indicating the relevant shots to permute for getting new combinations without duplicates. The $C(n, r)$ operation stands for a $r$-combination with $n$ elements, where $$ C(n, r) = \frac{n!}{r!(n-r)!} = \binom{n}{r}. $$

Team A wins

Team B does not score goals

Team B scores 1 goal

Team B scores 2 goals

Team B scores 3 goals

Team B scores 4 goals

If we add the total amount of combinations for each scenario, team A has $19 + 79 + 87 + 24 + 1 = 210$ ways of winning.

Team B wins

Team A does not score goals

Team A scores 1 goal

Team A scores 2 goals

Team A scores 3 goals

Team A scores 4 goals

If we add the total amount of combinations for each scenario, team B has $15 + 59 + 87 + 44 + 5 = 210$ ways of winning.

Draw

For this case, both teams must score the same amount of penalties at the end of the shoot-out. Therefore, the number of possible combinations will be $$ \sum_{r=0}^5C(5, r)^2 = 1^2 + 5^2 + 10^2 + 10^2 + 5^2 + 1^2 = 252.\tag{1}\label{eq:draw} $$

Solution

1. As we see, there are $210 + 210 = 420$ ways of settling the match in the first 10-penalty shoot-out.
2. Following the product rule and \eqref{eq:draw}, we need to multiply the number of ways to tie by the number of ways of winning or losing. Thus, there are $252 \cdot 420 = 105840$ ways of settling the match in the second 10-penalty shoot-out.
3. We have to add both previous answers including the rest of the combinations given by the sudden death. In order to continue in a sudden death, both teams must score or miss. The sudden death ends when team A scores and team B misses or viceversa. For a $n$ pair of penalty shots, this situation can be visualized as filling a bitstring with $n-1$ score-score and/or miss-miss results and a last one with a score-miss or miss-score result. This gives us $2^{n-1} \cdot 2 = 2^n$ possible combinations. Since $n$ must be less than $6$, a sudden death with $n<6$ pairs of penalty shots can be settled in $$ \sum_{n=1}^52^n = 2 + 4 + 8 + 16 + 32 = 62\text{ ways}. $$ Following the product rule, we will have $252 \cdot 252 \cdot 62 = 3937248$ ways of ending a sudden death with $n<6$ pairs of penalty shots. Therefore, the final answer is $420 + 105840 + 3937248 = 4043508$.