E(X): Find Probability of Rolling 4 Consecutive 6's with a Fair Dice

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The discussion focuses on calculating the expected number of rolls needed to achieve four consecutive sixes with a fair six-sided die, denoted as E(X), which is determined to be 1554. Participants express confusion about the probability calculations, particularly regarding the condition that the last four rolls must be sixes while the roll before them must not be. The probability of rolling four consecutive sixes is calculated as q = 1/6^4, leading to the conclusion that X follows a geometric distribution with this success probability. An alternative method suggested involves using the binomial formula to find the probability of achieving a four-run of sixes by subtracting the probability of not achieving it. Overall, the discussion emphasizes the complexity of calculating expected rolls for consecutive outcomes in probability theory.
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Let X be a random variable representing the number of times you need to roll (including the last roll) a fair six-sided dice until you get 4 consecutive 6's. Find E(X)?
answer is 1554.

I get confused with this, probability { X > n-5 }. I know that the last for throws must be 6's and the one before 'n-4 throws' must not be a 6. Any input please?
 
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anyone?
 
The probability that 4 of 4 throws are all sixes is q = 1/6^4. X is distributed Geometric with success probability q, and each "trial" represents four consecutive throws.

An alternative approach may be to calculate Prob{a 4-run of sixes} = 1 - Prob{not having a 4-run of sixes} using the binomial formula.
 
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