The even number dice problem
You roll a fair dice until you get a 6. What is the expected number of rolls, including the roll of 6, conditioned on the event that all previous rolls (if any) were even numbers?
By the way, the "dice" I am talking about is a singular form. Search up on Oxford Dictionary.
The answer is 1.5.
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1 solution
Lots of you guys should have answered 3, but that is wrong. In the event that we roll only even numbers until we roll a 6, we must only roll 2s and 4s until we eventually roll a 6. In other words, the outcome of the rolls is a sequence of 2s and 4s that eventually terminate in a 6. We are interested in the average length of the sequence, including the final 6. But there’s nothing special about 6. Since each roll shows with equal chance, the sequence could also terminate in a 1, or a 3, or a 5, and it would also have the same average length. So we ask: what is the expected number of rolls until we get a 1, 3, 5, or 6, given that all previous rolls involved 2s or 4s? We can solve this by capitalizing on the memoryless property of the geometric distribution. Consider the very first roll. There is a 4/6 chance we roll a 1, 3, 5, or 6, and the experiment ends with 1 roll. But there is also a 2/6 chance we roll a 2 or 4. In that case, we have used 1 roll. As we start out second roll, the dice has no memory of the first roll–it is like we start anew. So the expected rolls to end is now 1 more than when we started. So we have the equations:
L = E(total rolls until 1, 3, 5, 6 | previous rolls are 2, 4)
L = E(L | first roll 1, 3, 5, 6) + E(L | first roll not 1, 3, 5, 6)
L = (1)Pr(first roll 1, 3, 5, 6) + (L + 1)Pr(first roll not 1, 3, 5, 6)
L = (1)(4/6) + (L + 1)(2/6)
(4/6)L = 1
L = 6/4 = 3/2 = 1.5
Conditional on the event you only roll an even number, it only takes an expected 1.5 rolls to see a 6. This is because in most of the cases you will roll a 6 on the very first roll which brings the average down.