There is a die that, when rolled, gives a random real number between 0 and 1. We play the following game:
- Player 1 rolls the die
- Player 1 may either tell Player 2 the number, or lie and report any number between 0 and 1
- Player 2 can either roll the die to try and beat the number they were told, or accuse Player 1 of lying
- Player 2 wins if they beat the number they were told, or they accuse Player 1 of lying and is right
- Player 1 wins otherwise
What’s the best strategy for Player 1?
The best strategy for Player 1 to tell the truth if they roll higher than \(\frac{1}{e}\), or otherwise lie, reporting in the range \(\left[\frac{1}{e}, 1\right]\) according to the probability density function:
\[p(x) = e \left( \frac{1}{x} - 1 \right)\]Proof
Suppose Player 1 has chosen a strategy, and has reported a value in the range \([x, x+dx]\). This is referred to as the range “near” \(x\) .
If Player 2 chooses to trust, then they will win with probability:
\[P_{trust} = 1 - x\]If Player 2 chooses to accuse, then their probability of winning depends on how often Player 1 lies “near” \(x\). Let \(L(x) dx\) be the probability that Player 1 lies and reports a value in the range \([x, x+dx]\). The probability that the roll really was in the range \([x, x+dx]\) is \(dx\). Therefore, Player 2 will win when accusing with probability:
\[\begin{align} P_{accuse} & = \frac{L(x) dx}{L(x) dx + dx} \\ & = 1 - \frac{1}{L(x) + 1} \end{align}\]Player 2 should only accuse if:
\[\begin{align} P_{accuse} & > P_{trust} \\ 1 - \frac{1}{L(x) + 1} & > 1 - x \\ L(x) & > \frac{1}{x} - 1 \end{align}\]Now if Player 1 ever wants to report “near” \(x\) they should make sure that there are enough lies “near” \(x\) so that Player 2 is indifferent to trusting and accusing:
- If Player 1 doesn’t lie “near” \(x\) often enough, then Player 2’s best choice will be to trust. Knowing this, Player 1 can lie a little more often “near” \(x\) rather than reporting “near” \(y\) for any \(y<x\).
- If Player 1 lies too much “near” \(x\) , then Player 2’s best choice will be to accuse. Knowing this, Player 1 can lie a little less often “near” \(x\) and instead report “near” \(z\) for the largest \(z\) that doesn’t get lied “near” often enough (even if \(z < x\)).
Therefore, for any \(x\), Player 1’s best strategy will involve either never reporting “near” \(x\), or lying “near” \(x\) with probability:
\[L(x) = \frac{1}{x} - 1\]We can see that this means that \(P_{accuse} = P_{trust}\), and thus Player 2 will be indifferent to trusting or accusing. Thus we can assume that Player 2 will trust.
Given that Player 2 will trust always in this strategy, Player 1 should report high values as often as possible, and never report low values. Thus values less that some \(N\) will never be reported, and values greater than \(N\) will be lied “near” with a probability \(L(x)\).
We require that the probability of lying for \(x > N\) be the same as the probability for rolling \(x < N\):
\[\begin{align} N & = \int_N^1 \left( \frac{1}{x} - 1 \right) dx \\ & = \left[ \log x - x \right]_N^1 \\ & = \log 1 - 1 - \log N + N \\ \log N & = -1 \\ N & = \frac{1}{e} \end{align}\]Now we require that for rolls where \(x < N\), Player 1 will lie, reporting “near” \(y\) with probability \(cL(y)\), and that the sum of all probabilities of \(y\) in the range \([N, 1] = 1\):
\[\begin{align} \int_N^1 c L(x) dx & = 1 \\ c \int_N^1 \left( \frac{1}{x} - 1 \right) dx & = 1 \\ c \frac{1}{e} &= 1 \\ c & = e \end{align}\]Giving the required probability distribution in the equation: \(p(x) = e \left( \frac{1}{x} - 1 \right)\)