Question 11.5.18: A perpetual craps game goes on at Charley’s. Jones comes int......
A perpetual craps game goes on at Charley’s. Jones comes into Charley’s on an evening when there have already been 100 plays. He plans to play until the next time that snake eyes (a pair of ones) are rolled. Jones wonders how many times he will play. On the one hand he realizes that the average time between snake eyes is 36 so he should play about 18 times as he is equally likely to have come in on either side of the halfway point between occurrences of snake eyes. On the other hand, the dice have no memory, and so it would seem that he would have to play for 36 more times no matter what the previous outcomes have been. Which, if either, of Jones’s arguments do you believe? Using the result of Exercise 17, calculate the expected to reach snake eyes, in equilibrium, and see if this resolves the apparent paradox. If you are still in doubt, simulate the experiment to decide which argument is correct. Can you give an intuitive argument which explains this result?
Form a Markov chain whose states are the possible outcomes of a roll. After 100 rolls we may assume that the chain is in equilibrium. We want to find the mean time in equilibrium to obtain snake eyes for the first time. For this chain mki is the same as ri, since the starting state does not effect the time to reach another state for the first time. The fixed vector has all entries equal to 1/36, so ri = 36. Using this fact, we obtain
\bar{m}_i=\sum\limits_{k}{w_km_{ki}+w_ir_i} = 35+1=36.
We see that the expected time to obtain snake eyes is 36, so the second argument is correct.