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I'm aware that the $E[X_n]$ where $X$ is the average number rolls needed to roll $n$ consecutive sixes solves to $E[X_2]=42$ when $n=2$... but why does this differ from rolling other numbers (which are distinct from eachother) such as $5$ and then $6$? For example, if I wanted $X_n$ to be the average number of rolls for getting two consecutive numbers in general, which need not be the same number.

If I begin at $0$ rolls and roll a die $2$ times, the probability of getting $2$ sixes on the first two rolls is $\frac{1}{6}\cdot \frac{1}{6}=\frac{1}{36}$ and the probability of rolling a $5$ and then a $6$ is the same probability.

Intuitively, this led me to believe I could extrapolate the result mentioned above regarding $n$ consecutive sixes and apply it to rolling $n$ consecutive numbers (which need not be the same). But the problem I solved did not have the answer as $42$, the answer for the expected value to roll a $5$ followed immediately by a $6$ was $36$.

ClownInTheMoon
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  • I don't understand how you arrived at $ E[X_2] = 42 $. Could you explain that step? If two rolls (say $ X_1$ and $X_2$) are independent, then $ P(X_1=x_1,X_2=x_2) = \frac{1}{6} \times \frac{1}{6} $ for any number $ x_1 $ and $ x_2 $. – Michael R Dec 16 '16 at 04:07
  • @MichaelR http://math.stackexchange.com/questions/192177/how-many-times-to-roll-a-die-before-getting-two-consecutive-sixes – ClownInTheMoon Dec 16 '16 at 04:08

1 Answers1

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When you roll a six and then a non-six, there is no way you can "complete" your two sixes with one more roll.

But if you roll a five, and then a non-six, occasionally, that non-six is a five, in which case, you can complete your pair in the next roll.

This makes it easier to get a $56$ earlier than an $66$.

For example, the ways to get your first $66$ in exactly three rolls is the five possible sequences $[1-5]66$, while the ways to get your first $56$ in exactly three rolls is the six sequences $[1-6]56$.

One reason this is counter-intuitive: We play the following game with two players. A die is tossed repeatedly, until we get a six immediately followed by a 5 or 6. Player 1 wins if the last two rolls are 66, player 2 if they were 65.

This game is "even" - each player has a probability of $\frac{1}{2}$ of winning.

So this even-ness makes us intuit that the expectation of the first occurrence of each is also the same.

But that is a fallacy. When the first is 66 at roll $n$, it is possible for the first $65$ to be at roll $n+1$. On the other hand, if the first is $65$ at roll $n$, it takes minimum of $n+2$ rolls to get $66$. So when $66$ wins, the first $65$ is expected to follow a little faster than the reverse case.

Thomas Andrews
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