Alberto throws a die and a coin at the same time.
On the cube there are digits from 1 to 6, and on one side of the coin is written - and on the other side is written +.
First Alberto throws the dice and the coin and writes down the value obtained (e.g., if he rolls the dice 3 and the coin comes up -, he writes down -3). Alberto then rolls again and distributes the value received. Alberto repeats the action one more time.
What is the smallest value Alberto can get in the end?
To solve this problem, let's follow these steps:
- Step 1: Analyze each throw separately. The lowest value from one roll of the die is 1, and with the negative side of the coin, this gives -1.
- Step 2: Since the lowest die number is 1, and when coupled with the "-" side of the coin, it yields -1. If both throws yield -1, the combined result will be -1 + (-1) = -2.
- Step 3: The smallest die result is 6 (largest face), which maximizes negative impact when the coin is "-", giving -6.
- Step 4: Considering two rolls both yielding -6 (maximum negative outcome), we have the sum -6 + (-6) = -12. However, only one roll is considered directly for the smallest one-time roll, resulting in -6.
- Step 5: This indicates one throw sequence would achieve -6 effectively for a minimum value in an independent roll comparison, focusing on one result, not overall combined outcomes.
Upon reviewing, to achieve the smallest value in a single session roll (not progressive sum): the result after one complete throw session (one die, one coin) is potentially a single value here.
Therefore, the smallest result possible from the entire process is indeed −6, considering drags of progressive rolling assessments.