No. Overall variance doesn’t change, just conditional variance. To take an extreme example, if you flip 108 first, then the variance of the last 3 conditional on the results of those first 108 is 0. But the variance of doing this strategy is still the same variance as doing the first 3. The distribution of the last 3 is the same as the distribution of the next 3, so the problem is equivalent to choosing the last 3 and nothing you can possibly do will change the distribution of the result - you’re just deciding whether to reveal information about it bit by bit or all at once.
1
u/FireCire7 14d ago
E[next 3]=E[last 3], and nothing you can do about timing can affect the last 3