Predict Euro Cup Matches with Simple Statistics: how to rise the ranks (pt. 4)
A football poule is not a single-player game
We have already discussed how to find the most likely outcome of a match and how to calculate the expected number of points for each predicted outcome.
Nearing the end of the tournament, we have one more thing to figure out.
How to win if you are behind?
Why Expected Value no longer works
In a real tournament, the number of matches is limited. With an infinite number of matches left to play, predicting the outcome with the highest expected value always gives the best results. There are no exceptions, this is the law of large numbers.
With a limited number of matches, this works differently. Consider the following:
You are tied with your opponent with 1 match left to play, you both have the same two available options:
— predict 2–2 as the outcome of the match, if correct, provides 10000 points
— predict NOT 2–2 as the outcome of the match, if correct, provides 1 point
The expected value of the first option will be higher. Using the neutral heatmap for equal strength teams, the chance of a 2–2 is approximately 0.04 or 4%.
The expected value for the first option:
E = 0.04 * 10000 + 0.96 * 0 = 400
Expected value for the second option:
E = 0.04 * 0 + 0.96 * 1 = 0.96
Clearly, the first option is the better one, right?
Not if you just want to beat your opponent. If our opponent picks the first option, you have a 96% chance to beat him with the second option. You will only be a single point ahead of him, but that may be all that matters.
The situation
Several factors influence the optimal strategy to overtake an opponent:
- Point difference
- Number of matches remaining
- Scoring system
- Likelihoods of possible outcomes
- The opponent’s predicted outcome
These are all five factors that we need to consider to optimize our chance of winning the poule. You may also want to consider any prizes awarded for different ending positions in the poule.
A real example
Let’s say, my opponent is 6 points ahead of me, there is only one match left to play, the same scoring system as the previous blogs is used, one country has a small advantage and our opponent uses the maximum Expected Value strategy. What should we do?
First, we have a look at the number of points our opponent will get for each outcome. The outcome with the highest EV for a small advantage is 1–0, so that will be our opponent's prediction. Below is the table for the number of points awarded for each outcome:
+---------------+---------+
| Match outcome | points |
+---------------+---------+
| 1-0 | +10 |
| 2-0 | +7 |
| 3-0 | +7 |
| 4-0 | +7 |
| 2-1 | +5 |
| 3-1 | +5 |
| 3-2 | +5 |
| 1-1 | +2 |
| 0-0 | +2 |
| 1-2 | +2 |
| 1-3 | +2 |
| 0-1 | +0 |
| 0-2 | +0 |
| 2-2 | +0 |
| 0-3 | +0 |
+---------------+---------+We need to get at least 6 more points than our opponent. We can not predict any outcome that will also give our opponent 5 or more points. After all, if our opponent gets 5 points or more, we need to get at least 11, which is not possible.
We need another column that adds our own score and compares that with our opponent’s. Let’s see what happens if we predict 1–1.
+---------------+-----------+-----+------+--------+
| Match outcome | opponent | me | net | beat? |
+---------------+-----------+-----+------+--------+
| 1-0 | +10 | +2 | -8 | |
| 2-0 | +7 | +0 | -7 | |
| 3-0 | +7 | +0 | -7 | |
| 4-0 | +7 | +0 | -7 | |
| 2-1 | +5 | +2 | -3 | |
| 3-1 | +5 | +2 | -3 | |
| 3-2 | +5 | +0 | -5 | |
| 1-1 | +2 | +10 | +8 | Yes |
| 0-0 | +2 | +7 | +5 | |
| 1-2 | +2 | +2 | +0 | |
| 1-3 | +2 | +2 | +0 | |
| 0-1 | +0 | +2 | +2 | |
| 0-2 | +0 | +0 | +0 | |
| 2-2 | +0 | +7 | +7 | Yes |
| 0-3 | +0 | +0 | +0 | |
+---------------+-----------+-----+------+--------+From the above table, we find there are only two possible outcomes that would result in taking over our opponent if we predict 1–1.
To calculate the chance of this happening, we look up the corresponding values in the heatmap above and add them.
The chances for 1–1 and 2–2 are 0.14 and 0.08, respectively.
This means we have a P(win) = 0.14 + 0.08 = 0.22 chance, or 22%, to beat our opponent with a 1–1 prediction. P stands for probability by the way.
Can we do better?
More calculations
Sure, just calculate all reasonable options and pick the best.
We already have the number of points our opponent gets for each outcome. Next, we calculate the number of points we would get for each outcome, for each prediction. This results in the following table:
Now we adjust these values for the points our opponent would receive, resulting in a heatmap with values for points gained compared to the opponent.
We still need to add likelihoods to calculate which prediction has the best chance of netting us the win. But, we can already see that some predictions never gain us enough points. For example, 1–0 always results in the same number of points as our opponent, 3–0 could at best gain 3 points more than our opponent, and 2–1 gets an additional 5 at best.
Let’s add the likelihoods and decide our optimal prediction.
The above heatmap shows the odds of beating our opponent, adding all values in a column shows the total chance of beating our opponent with that particular prediction. Green means we beat the opponent, red means we would not beat the opponent.
For example, we see for prediction 0–0:
P(win) = P(0-0) + P(2-2) = 0.13 + 0.08 = 0.21
For 0–1, 0–2 and 0–3:
P(win) = P(0-1) + P(0-2) + P(0-3)= 0.1 + 0.07 + 0.01 = 0.18
And for 1–1:
P(win) = P(1-1) + P(2-2) = 0.14 + 0.08 = 0.22
The other options have a far lower probability of beating our opponent. We were actually correct with our first guess of 1–1, although 0–0 is a very close second.
What if there is more than just one match to be played?
To calculate the odds for multiple matches, you would have to find possible combinations that score sufficient points. This will get too complicated to put in a heatmap, and you are probably better off simulating the tournament, i.e. using a Monte Carlo Simulation.
That’s for another time!
