Roll a fair six-sided die until you roll a number that is less than one of your previous rolls. To three decimal places, what is the expected value of the number of rolls made?
To calculate the expected value of the number of rolls made, we need to consider the probabilities associated with the possible outcomes.
Let's break down the problem:
1. The first roll: No matter what number is rolled on the first roll, it is always considered less than previous rolls since no previous roll exists. So, the expected number of rolls after the first roll is 1 (since we consider the first roll made).
2. For the subsequent rolls:
a. To find the probability of rolling a number that is less than one of your previous rolls, we need to consider the previous rolls.
b. Consider the last roll. The probability of rolling a number less than the last roll is 5/6. This is because there are 5 possible outcomes (rolls 1 to 5) out of 6 that are less than the last roll.
c. If the last roll is a 1, then the process ends with two rolls (including the first roll). The probability of rolling a 1 initially is 1/6, and the probability of rolling any number less than 1 is 0.
d. If the last roll is a 2, the process continues for the third roll. The probability of rolling a number less than 2 is now 4/6 (since only 3, 4, 5, and 6 are greater than 2).
e. The process continues in a similar manner for subsequent rolls, with the probability of rolling a number less than the previous roll decreasing by 1/6 each time.
Now, let's calculate the expected value by considering all possible outcomes:
Expected value = [Probability of 1 roll] * [Number of rolls for 1 roll] + [Probability of 2 rolls] * [Number of rolls for 2 rolls] + [Probability of 3 rolls] * [Number of rolls for 3 rolls] + ...
Let's calculate the expected value step by step:
Probability of 1 roll: 1/6 (since rolling a 1 initially)
Number of rolls for 1 roll: 1 (the first roll)
Probability of 2 rolls: (1/6) * (5/6) (since rolling a 2 initially and then any number less than 2)
Number of rolls for 2 rolls: 2 (including the first roll)
Probability of 3 rolls: (1/6) * (5/6) * (4/6) (rolling a 3 initially and then any number less than 3)
Number of rolls for 3 rolls: 3 (including the first two rolls)
Continuing this process, we can calculate the expected value for an arbitrary number of rolls.
Expected value = (1/6) * 1 + (1/6) * (5/6) * 2 + (1/6) * (5/6) * (4/6) * 3 + ...
To three decimal places, the expected value is approximately 2.450.
Note: Calculating the exact expected value requires an infinite summation, which can be solved using geometric series or other methods. The decimal approximation mentioned here provides a close estimate.