Coin Toss Game with differential probability

Question

This problem is from QuantGuide(namely Tricky Bob 1):
Bob proposes a game where Alice and Bob are each given a coin. Each is allowed decide the probability of heads for their coin. Afterwards, they both flip their coins. If it comes up HH, Alice gives Bob 6 dollar. If it comes up TT, Alice gives Bob 4dolla. Otherwise, Bob gives Alice 5 dollar. However, Alice must tell Bob the success probability $p_1$ ​ they have chosen before Bob decides his $p_2$. The interval of values for $p_1$​ such that regardless of what Bob selects, Bob has non-positive expected value on the game is in the form [a,b], where a and b are rational numbers.
My Approach:
The expectation value equation will be: \begin{equation} 6p_1p_2+4(1-p_1)(1-p_2)-5p1(1-p_2)-5p2(1-p_1)=20p_1p_2-9(p_1+p_2)+4 \le 0 \end{equation}. Hence the conditions will be : \begin{equation} p_1 \le \frac{9p_2-4}{20p_2-9} \end{equation} In the range from 0 to 1 for $p_2$ it assumes a negative value as well as a value greater than 1. Hence according to me, there can't be any value for $p_1$ such that Bob will get a non-positive expected value.

Answer 1

Your calculation of the expected value is correct. However, I think you've misinterpreted the nature of the question. Let $$E(p_1, p_2) = 20 p_1 p_2 - 9 (p_1 + p_2) + 4 = 20 \left( p_1 - \frac{9}{20} \right)\left(p_2 - \frac{9}{20}\right) - \frac{1}{20}.$$ be the expected value. The question is asking, for what $p_1 \in [0,1]$ is is true that $E(p_1, p_2) \le 0$ for all $p_2 \in [0,1]$? Well, it's clear that $E(9/20, p_2) = -1/20 \le 0$ no matter what choice of $p_2$ we take. So this immediately demonstrates that there is at least one choice of $p_1$ that meets the problem's criteria.

Moreover, we can intuitively see that if $p_1$ is "sufficiently close" to $9/20$, that we could get a non-positive expectation for any choice of $p_2$. This suggests letting $a = p_1 - 9/20$ and $b = p_2 - 9/20$, hence $$E(p_1, p_2) \propto ab - \frac{1}{400}.$$ And now it becomes very clear that if $$b \le \frac{1}{400 a}$$ if $a > 0$, or if $$b \ge \frac{1}{400 a}$$ if $a < 0$, then $E(p_1, p_2) \le 0$. What we need to do is find out what value(s) of $a$ exist such that the respective inequality is true for all permissible $b$. In other words, if $a > 0$, we require $$p_2 - \frac{9}{20} = b \le \frac{1}{400a}$$ or $$\frac{1}{400(p_1 - 9/20)} + \frac{9}{20} \ge 1, \quad \text{and} \quad p_1 > 9/20.$$ This implies $$\frac{9}{20} < p_1 \le \frac{5}{11}.$$ Similarly, if $a < 0$, we require that $1/(400a) \le 0$, or $$\frac{1}{400(p_1 - 9/20)} + \frac{9}{20} \le 0, \quad \text{and} \quad p_1 < 9/20.$$ This implies $$\frac{4}{9} \le p_1 < \frac{9}{20}.$$ Putting all of this together, we find that whenever $$\frac{4}{9} \le p_1 \le \frac{5}{11},$$ then $E(p_1, p_2) \le 0$ for any choice of $p_2 \in [0,1]$.