I got a shoutout for last week’s solution and “nice explanation”. Definitely an improvement on the week before.

This week’s puzzle is easily stated:

This year’s World Cup has been chock full of exciting penalty shootouts. Historically, about 75 percent of soccer penalty kicks are successful. Given that number, what are the chances that a shootout goes past its fifth kick for each team and into the

even more excitingsudden-death portion of the penalty-kick period?

I see two ways of solving this one. The first is to calculate it directly, by summing the probability of each possible draw. The second is to generate every possible outcome, and find the draws among them.

I’ll try the second way first, for two reasons: firstly, I feel more confident with that method, and secondly, I can use Excel again.

The most complicated part of doing the problem is to make sure that we consider every possible result for the penalty shootout. There are up to ten penalties to be taken, and each one of the ten has two possible results, a goal or no goal.

That means there are 2 × 2 × 2 … × 2 = 2^{10} = 1024 possible combinations of goals and no goals. So one way of covering every possible combination is to look at the numbers from 0 to 1023, written in binary.

To give a couple of examples:

- zero, of course, is written as 00 0000 0000. We interpret that as all ten penalties being missed.
- 44 (decimal) is written as 00 0010 1100. We interpret that as the first four penalties being missed, the fifth being scored, the sixth being missed, the seventh and eighth being scored, and the last two being missed.

Here’s the spread sheet: riddler_express_20180713

After listing out all of the numbers from 0 to 1023 in column A, I converted them to binary using the MOD() function. That’s why the powers of two are listed from B2 to K2. Breaking the number down into columns is useful for what comes next.

Once we have all of the possible outcomes, finding the right outcomes and their total probability is relatively simple.

First we need to find the probability of this combination of goals and misses or saves. If this was a coin toss, rather than a penalty shot, all of the outcomes would be equally likely, at 0·5^{10} = 9·77 * 10^{-3}. That’s because the chance of a head is the same as the chance of a tail, 0·5. However, in a penalty shootout, the chance of a goal is not the same as a the chance of a save or miss. Any penalty shot has a 0·75 of resulting in a goal, and a 0·25 chance of not resulting in a goal. So the more goals there are in a particular result, the more likely it is overall. The probability of any particular pattern of goals and saves or misses is:

0·75^{(number of goals)} × 0·25^{(number of saves or misses)}.

For that reason, the spread sheet counts the number of goals in column L, which is just the sum of the numbers in columns B to K (this is one of the reasons why it was useful to break the binary numbers into ten columns.) The number of misses or saves in column M is just ten minus the number of goals, and then in column N the probability formula from the previous paragraph is applied.

Here also, I get to be more confident than if I was doing this with a direct calculation. If I add up the probabilities in column N, they sum to 1. That means I can be sure that I haven’t missed any possibilities out.

Now we know the chance of that particular combination of results, we can easily work out if it’s a draw or not. We just sum the results (1 for a goal, 0 for no goal) for alternating penalty shots. The team shooting any particular shot can be seen in B1 to K1, and the score for team A or team B in any combination can be seen in columns O and P.

Column Q detects whether or not a combination is a draw, by comparing the team A and B scores. Lastly, column R includes the probabilities from column N if the result is a draw, and zero if the result is not a draw. Now all we need to do is add all the values in column N, and we get the chance of a draw: 0·290 203… or 29·0%

One question you might still have is, what about the shootout ending early, because it is impossible for one team to catch up? It turns out that these don’t matter—after all, shootouts only end early if there’s no possibility of a draw. So they don’t change the values of the probabilities we have added up.

Right back at the top, I mentioned that it should be possible to do this with a direct calculation. After working through the problem in Excel, I am fairly confident that this is the formula to calculate:

This also comes to 29·0%. We can break it down a little:

- this is the sum of the probabilities of both teams scoring the same number of goals,
*i*, from zero goals to five goals. - in the brackets we can see two things:
- the probability of any particular way of scoring
*i*goals (and 5-*i*misses), multiplied by - the number of ways this can happen

- the probability of any particular way of scoring
- squaring the number in brackets reflects that we need both teams to score the same number of goals.

Truthfully, while I can see that this formula is correct now, I would not have been confident that it was correct without grinding through the problem in Excel first.