Only even rolls before the six
A fair die is rolled until the first appears. Let be the number of rolls, the included.
Given that every roll was even (that is, no , or appeared before the ), find the expected value of .
Fractions, decimals with at least 3 significant digits, and expressions such as sqrt(2)/2, e^-1 or pi/4.
The tempting argument is “the die now only shows , or , so is geometric with parameter and the mean is ”. Simulate a few runs in your head: does a long run of even numbers survive the condition easily?
Stop the rolls at the first number in . The condition says that this number is a . Is the stopping time related to which of the four numbers ended it?
h
s Shown after a correct answer. Revealing it first marks the problem as seen, not solved.
Idea. The condition removes every run that shows an odd number before the , and long runs are much more likely to show one. So conditioning favours short runs. The clean way to see it is to look at the first roll that is either odd or a six.
Step 1: a different stopping time. Let be the first roll that shows , , or . Each roll is one of these with probability , so is geometric with mean .
Step 2: the condition. “All rolls before the first are even” is the same event as “the roll at time is a ”. On that event, .
Step 3: independence. Given that a roll lands in , it is each of the four values with probability , whatever happened before. So the value shown at time is independent of itself. Conditioning on it being a does not change the law of :
Where the wrong answer goes wrong. The answer assumes we only condition each roll to be even. In fact we condition the whole run to avoid odd numbers, which is a far stronger requirement for long runs: a run of length survives with probability relative to the unconditioned law. Short runs dominate.