suppose you throw a fair die until $6$ appears for the first time. Compute the conditional expectation of the number of throws, given that until throwing only even numbers. This was an exercise in an old exam and I'm interested in a solution.
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If I understand right, only the even results count. In this case, you need $3$ throws in the average to get a $6$. – Peter May 24 '18 at 09:49
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but how do write it mathematically – wayne May 24 '18 at 09:51
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Just calculate the probability of throwing a $6$ given that the throw gives an even number. The reciprocal is the expected number of throws. You can use Bayes's theorem. – Peter May 24 '18 at 09:54
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@Peter no it isn't. Basically this is because you need to factor into your conditional the fact that if you fail to throw a six on the current throw, the next throw also has to be an even number, but if you do throw a six there is no longer any restriction on the next throw. – Especially Lime May 24 '18 at 09:56
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@EspeciallyLime Well, then I seem to have misinterpreted the exercise. But the formulation of the duplicate indicates that I didn't. Isn't the correct result $3$ ? – Peter May 24 '18 at 09:57
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@Peter No. The question is asking "what is the expected number of rolls to get the first six, given that you don't get any odd numbers before the first six?" What you are doing is conditioning on not getting any odd numbers before or after the first six, which is different. See the answers to the question this is a duplicate of. – Especially Lime May 24 '18 at 10:06
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@EspeciallyLime Yes, I interpreted it this way. I think the obvious interpretation of the formulation "all throws are even" – Peter May 24 '18 at 10:10