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The geometric distribution tells us that when you have independent events and you want to obtain the expected value till you have a success, you know that E[x] = 1/p.

P(2 sixes in a row) = (1/6)(1/6) = 1/36

Since rolling dice is independent trials, why is the expected number of rolls till you get 2 6s not equal to 1/(1/36), or 36? The answer is actually 42.

Mario Diw
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  • https://math.stackexchange.com/questions/192177/what-is-the-expected-number-of-times-a-dice-has-to-be-rolled-to-get-two-consecut – Mario Diw Apr 14 '23 at 17:23
  • It would have been good to reference that link earlier - it would have saved me typing up a whole wasted answer... I suppose I'll add my text to that link – FShrike Apr 14 '23 at 17:25
  • $1/36$ is the answer to the question, if I have two dice and roll them both, counting this as "one roll", how may times do I expect to roll before the dice both come up $6$ on the same roll? So if I rolled $5,6$ the first time and $6,2$ the second time, I would still have to continue rolling because the two $6$s occurred on different rolls of the two dice. In the question you got the answer $42$ from, after rolling $5$, $6,$ and $6$ you would be finished already. – David K Apr 14 '23 at 17:33
  • If the dice rolls are represented by , say , $X_1,X_2,X_3,\ldots$ then it is NOT true that the event ${X_1=6,X_2=6}$ is independent of the event ${X_1 \neq 6,X_2=6,X_3=6}$. The "middle" roll $X_2$ forces dependence of the events. Hence, the independence argument falls apart and a different argument is necessary, one that might use Markov chains, or example. – Sarvesh Ravichandran Iyer Apr 14 '23 at 17:51

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