# Probability : Maximize the expected payoff

Given $2$ random variables $X, Y$ that take integer values with uniform distribution from $0$ to $100$. You play a game in which a random value of $x$ comes first & you have to decide if the random value of $Y$ will be greater than $x$ or less than $x$. You will get reward of $y$ if your guess is correct and else $0$. Find the value you may like to pay for the game.(That is maximize the expectation payoff by finding the value of threshold $c$ if $x\ge c$ --> you guess $y < x$ and if $x < c$ --> you guess $y > x$)

• Equality of X and Y can occur with positive probability. So, I think that your decision should include greater or equal than x or less than x or vise versa. – Jimmy R. Nov 20 '14 at 10:13

You have two strategies say $l$ for guessing that $Y$ will be less than the given $x$ and $g$ for guessing that $Y$ will be greater than or equal to the given $x$ (since the case of equal is missing from the exercise, I make the assumption that you can put it with greater than). Now the expected reward $R$ of each choice is equal to $$R_l=\sum_{y=0}^{x-1}P(Y=y)\cdot y+P(Y\ge x)\cdot 0=\sum_{y=0}^{x-1}\frac{1}{101}\cdot y=\frac{1}{101}\cdot\frac{x(x-1)}{2}$$ and similarly \begin{align*}R_g&=P(Y< x)\cdot 0+\sum_{y=x}^{100}P(Y=y)\cdot y=\sum_{y=x}^{100}\frac{1}{101}\cdot y=\frac{1}{101}\cdot\frac{(x+100)(100-(x-1))}{2}\\\\&=\frac{1}{101}\cdot\frac{(100+x)(101-x)}{2}\end{align*} Now choice $g$ has a greater (or equal) expected reward than choice $l$ iff $$\frac{1}{101}\cdot\frac{(100+x)(101-x)}{2}\ge \frac{1}{101}\cdot\frac{x(x-1)}{2}$$ which is equivalent to $$-2x^2+2x+10100\ge0 \implies -(x+70.565)(x-71.565)\ge 0$$ Thus for every $x$ in $\{0,1,\ldots 71\}$ $g$ has a greater expected payoff and the opposite for $x \in \{72, 73, \ldots, 100\}$. In total the optimal strategy of the player is $$\text{Optimal choice }=\begin{cases}\text{greater or equal than }, & 0\le x\le 71 \\ \text{less }, & 72\le x \le 100 \end{cases}$$ (The threshold $c$ is equal to $71.565$).