Sometimes a single question in the early evening program (in this case from the esteemed presenter Kai Pflaume) is enough to turn a harmless quiz show finale into a minor optimization problem. That's exactly what happens on "Who Knows What?" Master question: The category is known, the answer not yet – but the stakes already determine which outcomes are still good.
Let's take two teams \(A\) and \(B\) . Before the final question, team \(A\) had won the amount \(x_a\) , and team \(B\) had won the amount \(x_b\) . We consider the case
$$
x_a > x_b > 0.
$$
The teams are now betting whole numbers.
$$
1 \leq y_a \leq x_a,\qquad 1 \leq y_b \leq x_b.
$$
If the answer is correct, the staked amount is added; if the answer is incorrect, it is subtracted. For the four possible outcomes, the following final scores result.:
$$
\begin{array}{c|c|c}
\text{Fall} & A & B\\
\hline
A \text{ richtig}, B \text{ richtig} & x_a+y_a & x_b+y_b\\
A \text{ richtig}, B \text{ falsch} & x_a+y_a & x_b-y_b\\
A \text{ falsch}, B \text{ richtig} & x_a-y_a & x_b+y_b\\
A \text{ falsch}, B \text{ falsch} & x_a-y_a & x_b-y_b
\end{array}
$$
In the show, a tie leads to an estimation question. Therefore, it remains visible in the matrix as a separate case "=". For the percentage values of the winning probability, we count direct wins in the four equally likely sub-cases. A pattern like \(|A|B|A|A|\) yields three direct wins for Team \(A\) and one for Team \(B\) . A tie does not count as a direct win for either team. This more precise counting is crucial; it is not sufficient to simply count an entire cell as "blue" or "red".
This model contains an assumption: We treat the four answer combinations as equally likely. Therefore, it's not about whether Team \(A\) or Team \(B\) knows the category better, but only about the strategy employed before answering.
Let's
$$
d=x_a-x_b.
$$
Then \(d>0\) is the advantage of team \(A\) . The question now is: What is the optimal stake?
The complete matrix of possible bets can be calculated dynamically.:
Team A's perspective
We first examine when Team \(A\) wins with fixed stakes \(y_a\) and \(y_b\) .
The case
$$
A \text{ richtig}, B \text{ falsch}
$$
It always goes to Team \(A\) , because
$$
x_a+y_a > x_b-y_b
$$
This applies automatically because \(x_a>x_b\) and \(y_a,y_b>0\) .
For the other three cases, we receive:
$$
\begin{array}{c|c}
\text{Fall} & A \text{ gewinnt genau dann}\\
\hline
A \text{ richtig}, B \text{ richtig} & x_a+y_a>x_b+y_b\\
A \text{ falsch}, B \text{ richtig} & x_a-y_a>x_b+y_b\\
A \text{ falsch}, B \text{ falsch} & x_a-y_a>x_b-y_b
\end{array}
$$
With \(x_a=x_b+d\) this becomes:
$$
\begin{array}{c|c}
\text{Fall} & A \text{ gewinnt genau dann}\\
\hline
A \text{ richtig}, B \text{ richtig} & d+y_a>y_b\\
A \text{ falsch}, B \text{ richtig} & d-y_a>y_b\\
A \text{ falsch}, B \text{ falsch} & d-y_a>-y_b
\end{array}
$$
So:
$$
\begin{array}{c|c}
\text{Fall} & A \text{ gewinnt genau dann}\\
\hline
A \text{ richtig}, B \text{ richtig} & y_b<y_a+d\\
A \text{ falsch}, B \text{ richtig} & y_b<d-y_a\\
A \text{ falsch}, B \text{ falsch} & y_b>y_a-d
\end{array}
$$
The counting method is now crucial. Previously, one might be tempted to evaluate each cell solely based on whether it contains more \(A\) cases than \(B\) . However, this is too simplistic for calculating the probability of winning. The four sub-cases are themselves equally probable events. Therefore, \(|A|B|A|A|\) does not count as a single win for \(A\) , but rather as three won sub-cases for \(A\) .
For a fixed stake \(y_a\) of team \(A\) we therefore sum the individual \(A\) entries in the four cases over all possible stakes \(y_b=1,2,\ldots,x_b\) .
The case " \(A\) correct, \(B\) incorrect" always goes to team \(A\) . This already yields \(x_b\) won subcases.
The following numbers result for the three other cases.:
$$
\begin{aligned}
N_1(y_a)&=\min(x_b,d+y_a-1),\\
N_3(y_a)&=\min(x_b,\max(0,d-y_a-1)),\\
N_4(y_a)&=\begin{cases}
x_b, & y_a\leq d,\\
\max(0,x_b-y_a+d), & y_a>d.
\end{cases}
\end{aligned}
$$
This means the number of subcases won by Team \(A\) is
$$
N_A(y_a)=x_b+N_1(y_a)+N_3(y_a)+N_4(y_a).
$$
The corresponding probability of winning is
$$
P_A(y_a)=\frac{N_A(y_a)}{4x_b}.
$$
Since a tie is not included here, \(P_A\) is exactly the probability of winning the master question directly (without an estimation question).
This more precise counting slightly shifts the optimum compared to the simple majority of cells. For Team \(A\) the following optimal application areas result.:
$$
\boxed{
\begin{cases}
1\leq y_a\leq2, & x_b=1,\ d=2,\\
d\leq y_a\leq x_b-d+1, & 2d\leq x_b+1,\\
1\leq y_a\leq d, & 2d=x_b+2,\\
1\leq y_a\leq \max(1,x_b-d+1,d-x_b-1), & 2d>x_b+2.
\end{cases}
}
$$
All bets in this area maximize Team \(A\) 's probability of winning. If you want to bet the largest possible amount among the equally good bets, you should always use the right edge of the area.
An example:
$$
x_a=30,\qquad x_b=22.
$$
Then
$$
d=x_a-x_b=8.
$$
There
$$
2d=16\leq 23=x_b+1
$$
The optimal range applies.
$$
8\leq y_a\leq 15.
$$
The greatest optimal use is therefore
$$
\boxed{y_a=15}.
$$
The old approach of considering whole cells would have suggested the range \(9\leq y_a\leq 14\) . The partial case count shows more precisely that the two boundary values \(8\) and \(15\) are also optimal.
Team B's perspective
Now we consider the same situation from the perspective of the trailing team \(B\) . Here too, we no longer count only entire cells, but the individual \(B\) entries in the four subcases.
The case
$$
A \text{ richtig}, B \text{ falsch}
$$
Team \(B\) always loses. In the three remaining cases, Team \(B\) receives the following number of won sub-cases for a fixed stake \(y_b\) when summing over all possible \(y_a=1,2,\ldots,x_a\):
$$
\begin{aligned}
M_1(y_b)&=\max(0,y_b-d-1),\\
M_3(y_b)&=x_a-\max(0,d-y_b),\\
M_4(y_b)&=\max(0,x_b-y_b).
\end{aligned}
$$
So is
$$
N_B(y_b)=M_1(y_b)+M_3(y_b)+M_4(y_b)
$$
and the associated probability of winning
$$
P_B(y_b)=\frac{N_B(y_b)}{4x_a}.
$$
For \(y_b\leq d\) this simplifies to
$$
N_B(y_b)=2x_b.
$$
For \(y_b>d\) one only gets
$$
N_B(y_b)=2x_b-1.
$$
This is the optimal area for Team \(B\)
$$
\boxed{
1\leq y_b\leq \min(d,x_b).
}
$$
This is an important correction compared to the cruder cell majority approach: The optimal bet for Team \(B\) is not necessarily uniquely \(1\) For example, if Team \(A\) is ahead with \(d=8\) and Team \(B\) can bet at most \(22\) , then all bets for \(B\) are optimal with respect to the probability of winning.:
$$
1\leq y_b\leq 8.
$$
The intuition remains similar: The trailing team should not bet unnecessarily high. While excessively high bets improve individual scenarios, they worsen others. As soon as \(y_b\) exceeds the deficit \(d\) , team \(B\) loses a scenario overall. Therefore, the leading team bets in such a way that, across all possible bets by the opponent, they win as many individual scenarios as possible.
The pursuer doesn't necessarily bet exactly one euro, but at most as much as the deficit. The master question is thus a good example of how much game theory is contained in a seemingly simple quiz rule: what matters is not only which cell ends up blue or red, but how many of the four sub-cases within that cell are actually won.