Where should the penalty taker aim?
Tonight's idea
A penalty taker shoots left or right. The keeper must dive as the ball is struck, so she is choosing too. Suppose the chance of a goal is (invented numbers; left and right as the kicker sees them):
| Keeper dives left | Keeper dives right | |
|---|---|---|
| Kick left | 50% | 90% |
| Kick right | 80% | 60% |
No single choice is safe. Always shoot left and she always dives left: 50%. Always right, and she dives right: 60%. So the kicker shoots left at random, with probability \(p\): a mixed strategy, as in matching pennies in grade 10.
If she dives left, the goal chance is \(50p + 80(1 - p) = 80 - 30p\). If she dives right, it is \(90p + 60(1 - p) = 60 + 30p\): two straight lines in \(p\). A keeper who knows \(p\) picks the lower one, so the kicker wants the lower line as high as possible. That is where the lines cross:
\[80 - 30p = 60 + 30p, \qquad p = \tfrac13, \qquad \text{goal chance } 70\%.\]
Shoot left one time in three, at random, and you score 70% whatever the keeper does. This equalising mix makes her choice not matter.
Watch one: the keeper's side
The keeper dives left with probability \(q\). Against a kick left the goal chance is \(50q + 90(1 - q) = 90 - 40q\). Against a kick right it is \(80q + 60(1 - q) = 60 + 20q\).
The kicker picks the higher line, so she wants it as low as possible: the crossing again.
\[90 - 40q = 60 + 20q, \qquad q = \tfrac12, \qquad \text{goal chance } 70\%.\]
The same 70%, the value of the game. He can guarantee at least 70 and she can hold him to at most 70, so neither gains by changing plan alone: a Nash equilibrium. Von Neumann proved the two numbers agree in every zero-sum game, where one side's gain is the other's loss.
Your turn
- The kicker ignores the advice and shoots left half the time. Find the goal chance against each dive. Which way does the keeper dive, and how much has the kicker given away?
- A different kicker (numbers invented again): kick left scores 20% against a left dive and 80% against a right dive; kick right scores 90% and 50%. Find his \(p\), the keeper's \(q\), and the value.
- Back to the first table. The kicker trains until his left shot scores 60% against a left dive, not 50%. Find the new \(p\), \(q\) and value. The keeper is no better or worse than before. Why does her mix change?
- A third kicker: kick left scores 70% and 90%; kick right scores 50% and 60%. Write the two lines in \(p\). Where do they cross? What should each player do, and what is the value?
Stretch: one formula
Call the table \(a, b\) on the top row and \(c, d\) on the bottom. Suppose the lines cross between 0 and 1, with one rising and the other falling. Show that \(p = \dfrac{d - c}{a - b - c + d}\) and that the value is \(\dfrac{ad - bc}{a - b - c + d}\). Check both on the first table. Then try \(a, b, c, d = 90, 70, 50, 60\). The lines still cross between 0 and 1, but the formula gives the wrong value. Why?
Answers
- Left dive: \(80 - 30 \times \tfrac12 = 65\%\). Right dive: \(60 + 30 \times \tfrac12 = 75\%\). The keeper dives left every time and the kicker scores 65%, giving away 5 points out of 70.
- Left dive: \(20p + 90(1 - p) = 90 - 70p\). Right dive: \(80p + 50(1 - p) = 50 + 30p\). Equal when \(p = 0.4\), value \(62\%\). Keeper: kick left \(20q + 80(1 - q) = 80 - 60q\), kick right \(90q + 50(1 - q) = 50 + 40q\). Equal when \(q = 0.3\), value \(62\%\) again.
- Left dive: \(60p + 80(1 - p) = 80 - 20p\). Right dive: \(60 + 30p\) as before. So \(p = 0.4\) and the value is \(72\%\). Keeper: \(60q + 90(1 - q) = 90 - 30q\) and \(60 + 20q\), so \(q = 0.6\). Her mix is chosen to make the kicker's two shots equally good, and his numbers changed. If she stayed at \(q = \tfrac12\), a kick left would score 75% and a kick right 70%, so he would shoot left every time.
- Left dive: \(70p + 50(1 - p) = 50 + 20p\). Right dive: \(90p + 60(1 - p) = 60 + 30p\). They cross at \(p = -1\), which is not a probability. Between 0 and 1 the left-dive line is always the lower one, and it is highest at \(p = 1\). So the kicker always shoots left, the keeper always dives left, and the value is 70%. Kicking left beats kicking right against either dive (70 against 50, 90 against 60), so there is nothing to mix.
Stretch. Left dive: \(ap + c(1 - p)\). Right dive: \(bp + d(1 - p)\). Setting them equal, \(p(a - c - b + d) = d - c\), so \(p = \dfrac{d - c}{a - b - c + d}\). Then \(1 - p = \dfrac{a - b}{a - b - c + d}\), and the value is \(ap + c(1 - p) = \dfrac{a(d - c) + c(a - b)}{a - b - c + d} = \dfrac{ad - bc}{a - b - c + d}\). First table: \(p = \dfrac{60 - 80}{50 - 90 - 80 + 60} = \dfrac{-20}{-60} = \tfrac13\), and the value is \(\dfrac{3000 - 7200}{-60} = 70\). With 90, 70, 50, 60 the lines are \(50 + 40p\) and \(60 + 10p\). They cross at \(p = \tfrac13\), at about 63%, but both rise, so the lower line is highest at \(p = 1\). Always shooting left guarantees 70%, and the keeper holds him to 70% by always diving right. The value is 70%, not 63%. The keeper's formula \(q = \dfrac{d - b}{a - b - c + d} = -\tfrac13\) gives the warning: it is not a probability. The crossing gives the value only when it lies between 0 and 1 and one line rises while the other falls, which is the same as both \(p\) and \(q\) lying between 0 and 1. Otherwise compare the ends of the lower line.