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Rolls to a six, given all even

HardConditional Probability & ExpectationExpected Value~9m

You roll a fair die until you get a 66. Conditioned on the event that every roll (including the final 66) shows an even number, what is the expected number of rolls?

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  1. Compute P(N=n,all even)=(1/3)n−1(1/6)\mathbb{P}(N=n, \text{all even}) = (1/3)^{n-1}(1/6) and its normalizer P(all even)\mathbb{P}(\text{all even}).
  2. E[N∣all even]=E[N;all even]/P(all even)\mathbb{E}[N\mid\text{all even}]=\mathbb{E}[N;\text{all even}]/\mathbb{P}(\text{all even}). It's 3/23/2, not 33 - long runs are penalized.

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1.5

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