how much would you be willing to pay for this card game?

Question

suppose there are 12 cards with number 1 - 12 on each card. All cards are faced down and are listed one by one in front you.

You get to pick one card, if the card has 1 on it, you get $1.

If it's 2, you get $2, so on and so forth. Now you have the chance to put the first card back, re-shuffle it, and re-pick another card. If the second card is higher than the first card, you will get paid by an amount equivalent to the number of the second card. if the second card is less, you will get paid by whatever amount printed on the first card. How much would you be willing to pay for this game?

Answer 1

For each $k\in \{1,\cdots, 12\}$ let $\psi(k)$ denote the probability that both of your choices are $≤k$. Since the draws are independent we have $$\psi(k)=\left(\frac k{12}\right)^2$$

Now let $P(k)$ denote the probability that the maximum of the two draws is exactly $k$. Clearly we have $$P(1)=\psi(1)\quad \& \quad P(k)=\psi(k)-\psi(k-1)\quad \text {for}\,k>1$$

It is now easy to compute the expect value of the max, $$E=\sum kP(k)=8.486$$

Answer 2

For what it's worth, I just made a $12\times12$ array to represent the payouts for every possible pair of (1st draw, 2nd draw). The payout being the max of the two draws.

The average of the values in the table is the expected payout of the game. It is approximately $8.49$. "How much would you be willing to pay?" is a bit of a subjective question, but if I was running a casino with this game, I would certainly charge more than that for people to play it.