This week, we return to the brilliant and ageless game show, “The Price is Right.” In a modified version of the bidding round, you and two (not three) other contestants must guess the price of an item, one at a time.
Assume the true price of this item is a randomly selected value between 0 and 100. (Note: The value is a real number and does not have to be an integer.) Among the three contestants, the winner is whoever guesses the closest price without going over. For example, if the true price is 29 and I guess 30, while another contestant guesses 20, then they would be the winner even though my guess was technically closer.
In the event all three guesses exceed the actual price, the contestant who made the lowest (and therefore closest) guess is declared the winner. I mean, someone has to win, right?
If you are the first to guess, and all contestants play optimally (taking full advantage of the guesses of those who went before them), what are your chances of winning?
Playing optimally is not just taking advantage of the guesses of those who went before. It is also contemplating what every player after will choose as well.
Let us define a block in this game. A block occurs when some future player guesses a real number immediately before or immediately after the guess of another player.
It is in every players' best interest (except the last) to not be blocked by any future player. In order to accomplish that, each player must allow every future player to have a chance greater than or equal their own.
In order to recieve $\frac{1}{3}$ chance of winning, choose $\$66 \frac{2}{3}$. Player 2 will also incorporate the same strategy of avoiding being blocked. It is in their best interest to choose $\$33\frac{1}{3}$. Player 3 will choose $\$0$.
Note that Player 2 has no interest choosing a number greater than yours, because that would yield a chance of winning slightly less than $\frac{1}{3}$. Note that Player 2 has no interest in blocking you from the left (slightly less), because then Player 3 would also block Player 2 in the same manner and effectively have ~$\frac{2}{3}$, Leaving Player 2 with ~$0$ chance of winning. Note that Player 2 has no interest in playing $\$0$, because again, Player 3 would block from the right and again effectively have ~$\frac{2}{3}$ chance of winning.
This can be extended to $N$ players, each playing optimally, have $\frac{1}{N}$ chance of winning.