Understanding the Expected Rolls for Consecutive Outcomes in a Fair Die

When considering the outcome of rolling a fair die, a natural question arises: on average, how many rolls are required to achieve three consecutive outcomes of the same number? This problem can be elegantly solved using probabilistic modeling and states. Through this article, we will delve into the methods and calculations involved in determining the expected number of rolls for this specific outcome.

State-Based Modeling

To tackle this problem effectively, we use a state-based approach. The states we define will capture the progress towards our goal of achieving three consecutive outcomes of the same number. The states are as follows:

State 0: No consecutive outcomes have been achieved yet. State 1: One consecutive outcome has been achieved. State 2: Two consecutive outcomes have been achieved. State 3: We have achieved three consecutive outcomes of the same number, which is our goal.

Transition Probabilities

Based on these states, we define the transition probabilities:

From State 0: With a probability of 1/6, we roll a number and move to State 1. With a probability of 5/6, we remain in State 0. From State 1: With a probability of 1/6, we roll the same number and move to State 2. With a probability of 5/6, we roll a different number and return to State 0. From State 2: With a probability of 1/6, we roll the same number and move to State 3 (success). With a probability of 5/6, we roll a different number and return to State 0.

Expected Values

We define the expected number of rolls starting from each state as follows:

E0: Expected number of rolls starting from State 0. E1: Expected number of rolls starting from State 1. E2: Expected number of rolls starting from State 2. E3: Expected number of rolls starting from State 3 (it is 0 since we have already achieved our goal).

From this, we set up the following equations based on the transitions:

State 3:
E3 0 (our target state)

State 0:
E0 1 (1/6)E1 (5/6)E0
Rearranging, we get:
(1/6)E0 1 (1/6)E1
E0 6E1 6

State 1:
E1 1 (1/6)E2 (5/6)E0

State 2:
E2 1 (1/6)E3 (5/6)E0 1 (5/6)E0 (since E3 0)

Substituting these equations, we can now solve for the expected values:

E2:
E2 1 (5/6)E0
E2 1 (5/6)E0
E2 (1 (5/6)E0)

E1:
E1 1 (1/6)E2 (5/6)E0 1 (1/6)(1 (5/6)E0) (5/6)E0
E1 1 (1/6) (1/6)(5/6)E0 (5/6)E0
E1 (1 1/6) ((5/36)E0 (5/6)E0)
E1 (7/6) (35/36)E0

E0:
E0 6E1 6
E0 6((7/6) (35/36)E0) 6
E0 7 (35/6)E0 6
E0 13 (35/6)E0
Multiplying everything by 6 to eliminate the fraction:
6E0 78 35E0
E0 - 35E0 78
E0 78 / 34
E0 228 / 34
E0 258 / 1
E0 ≈ 258

Conclusion

Therefore, the expected number of rolls required to achieve three consecutive outcomes of the same number is approximately 258.

This result is derived from a detailed probabilistic model and state transitions. By understanding and solving the problem through this method, we can gain insights into various aspects of probability and state-based modeling in a fair die context.