From Michael Montuori comes an extension of a puzzle that first ran seven years ago!
Way back in 2017, The Riddler featured a puzzle about Showcase Showdown from the game show, “The Price Is Right.” Players spin a giant wheel, which has 20 segments labeled in 5-cent intervals from 5 cents up to 100 cents. This week, Michael is posing the continuous version of this problem.
Suppose you have two players: A and B. Player A is the first to spin a giant wheel, which spits out a real number chosen randomly and uniformly between 0 and 1. All spins are independent of each other. After spinning, A can either stick with the number they just got or spin the wheel one more time. If they spin again, their assigned number is the sum of the two spins, as long as that sum is less than or equal to 1. If the sum exceeds 1, A is immediately declared the loser.
After A is done spinning (whether once or twice), B steps up to the wheel. Like A, they can choose to spin once or twice. If they spin twice and the sum exceeds 1, they are similarly declared the loser.
After both players are done, whoever has the greater value (that does not exceed 1) is declared the winner.
Assuming both players play the game optimally, what are Player A’s chances of winning?
This game is sequential and with perfect information. If Player 2's first spin is not enough to win, Player 2 should always spin again. Even if they go over 1, they were going to lose anyway.
Player 1 knows that and should try to increase their sum. I ran simulations and set Player 1 to spin again if their first spin is less than 0.5.
The following tree diagram shows all possibilites. Ties are ignored for continuous because the probability is ~0.
Out of $100,000,000$ games, $45,313,708$ were won by Player 1 and $54,686,292$ by Player 2.
Player 1 wins $~0.453$ of the times. I could not find an more optimal strategy for either.