Final answer:
The expected payout for rolling a die up to three times in a game is calculated by considering the average outcomes and the probabilities of each roll. Strategic decisions can be made after each roll to potentially increase the expected payout, which is higher than 3.5 when playing optimally.
Step-by-step explanation:
The question at hand entails calculating the expected payout for a game involving rolling a die thrice with the option to stop after any number of rolls. The expected payout is a concept from probability theory, which is used to determine the average outcome of an event if it were to be repeated an infinite number of times. For each roll of the die, there's an equal chance to land on any of the numbers 1 through 6, each occurring with probability 1/6. After the first roll, you can choose to take the number shown on the die or try again, up to two more times, for potentially higher numbers. The best strategy, maximizing expected payout, involves rolling again unless you roll a particularly high number, typically 4, 5, or 6, on your first or second try.
To calculate the expected value after the first roll, you'd take the sum of all the possible outcomes (1 through 6) each multiplied by their probability of occurring (1/6), unless you decide to stop after a high roll. After the second roll, if you haven't stopped, you again face the decision to take your number or roll a third and final time. Again you'd calculate the expected value based on potential outcomes. On the third roll, you are forced to take whatever value you roll, thereby the expected value is once more the average of possible outcomes from 1 to 6, equal to (1+2+3+4+5+6)/6, which is 3.5. Overall, if deciding strategically, the game would provide an expected payout higher than 3.5 since you are unlikely to keep a low initial or second roll. The precise expected payout would require a detailed calculation factoring in the probabilities and decisions made after each roll, optimizing for the highest possible outcome.