'How much would you pay to play this game' question

Question

'We play a game: I pick a number $n$ from 1 to 100. If you guess correctly, I pay you $n^2$ dollars and zero dollars otherwise. How much would you pay to play this game?'

I know the answer for the case where you are paid $n$ dollars if you guess correctly, and just want to see if I am correct for this other case.

The idea was to pick $k$ with probability $\frac{C}{k}$ where $C = \left(\sum_{j=1}^{100}\frac{1}{j}\right)^{-1}$. So then $\mathbb{E}[\text{winnings}] = \sum_{j=1}^{100}j^2 \mathbb{P}(\text{winnings} = j) = \sum_{j=1}^{100} j^2 \frac{1}{100}\frac{C}{j} = \frac{101}{2}C$. So I'd be happy to pay any amount less than or equal to $\frac{101}{2}C$. Is this correct? I'm worried it's a little too large as a value.

Answer 1

I think your $\frac{101}{2}C \approx 9.735$ looks far too large if the bank can just choose $1$. It rather depends on whether the bank has to choose uniformly at random or has a deliberate choice. Let's assume it is a deliberate choice or strategy chosen by the bank but unknown to the player. I suspect

• the game-theoretic optimal strategy by the player is then, for each $k\in \{1,2,\ldots,100\}$, to bet on $k$ with probability $\dfrac1{k^2\, H_{100,2}}$ where $H_{100,2} = \sum\limits_{n=1}^{100} \frac1{n^2}\approx 1.635$, a generalised harmonic number
• if the banker has chosen a particular $n$ then the expected amount paid by the banker to the player is then $\frac{n^2}{n^2\, H_{100,2}}= \frac1{H_{100,2}} \approx 0.6116$ and is the same for all $n$
• so $\frac1{H_{100,2}} \approx 0.6116$ is the most the player should be willing to pay to play the game.

If the banker changed the rules to remove the $100$ maximum so there was no limit on the positive integer chosen, then this amount would decrease slightly to $\frac6{\pi^2} \approx 0.6079$.

If the banker had to choose uniformly at random from $\{1,2,\ldots,100\}$, then the best strategy for the player would be to always bet on $100$, with an expected amount paid by the banker to the player of $\frac1{100}100^2 =100$. A better game-theoretic strategy for the banker would be the same as the optimal strategy as the player, picking $n$ with probability $\dfrac1{n^2\, H_{100,2}}$.