Abstract:
A die is thrown until a six turns up on four consecutive throws. What is the expected number of times the die must be thrown to achieve this? Krzys Ostaszewski knew how to compute the answer by a probability distribution technique and by a Markov process technique, but neither method clarified why the answer turns out to be a sum of powers of 6. He asked me whether I could find another argument that might show this. In this talk I will explain how an infinite rooted tree can be used not only to answer the problem but to yield the answer in a form that clarifies why it is a sum of powers of 6. As a second application of the method, I will discuss the following problem, related to the games known as Yahtzee, Yam and Balut. Five dice are thrown simultaneously. Those showing a six are left on the table, and the others are picked up and thrown a second time. Any new sixes are also left on the table; those not yet showing a six are picked up and thrown again. This procedure continues until all five dice show a six. What is the expected number of throws to achieve five sixes?
Acknowledgements: I may include some references to Krzys Ostaszewski's methods of solving the 6666 problem. I will also report on some computer simulations which Shailesh Tipnis has done for both problems.