Expected number of rolls until a number appears 2 times discussion

Question

I have the following problem. How many times you have to roll a die until you get two 6 or two 5 in a row.

Problems similar to this one have been discussed in the forum in links like:

[1] Number of rolls until the same number appears 2 consecutive times

[2] Expected number of rolls for fair die to get same number appear twice in a row?

I "solved" this problem as follows:

Let A the event of two consecutives 6 or two consecutive 5 in a row, and B the event of two consecutive equal numbers in a row. Then, it follows that:

\begin{equation} P(A) = P(A \cap B) = P(A/B)P(B) \end{equation}

$P(A/B)=2/6$ because if you already know that there are two identical numbers in a row then A can happen in 2 ways (5,5) and (6,6) over the sample space (i,i), $1\leq i \leq 6.$

But P(B) is a little bit more tricky. I know that the answer is 1/7 (I run a simple program to compute this number) but for me is counter intuitive.

In that case the Expected value would be = 21 because I expect 1 "win" combination out of 21 rolls.

My confusion is that I would have answered 1/6 instead of 1/7 in $P(B)$ because, no matter what number came out from the first dice, the second has 1/6 probability to match that number. Why this intuition fails? I know the answer is in [2] but I still can not handle the intuition on this.

A generalization of this result can be found here: Expected number of rolls until a number appears $k$ times

Moreover, this problem is often to get confused with the 1/36 probability of throwing two dices simultaneously. 

Links where to the discussion of this topic:

[-] If you roll a fair six sided die twice, what's the probability that you get the same number both times?

[-] Why are the probability of rolling the same number twice and the probability of rolling pairs different?

Answer 1

This sort of thing can easily be handled with states, and the symmetry makes it especially easy in this case.

There are three states where care about: $ S_5, S_6,S_{\emptyset}$ according to whether the prior throw was a $5$, a $6$, or neither (we let $S_{\emptyset}$ include the starting state as well). Symmetry tells us that we expect it to take the same number of rolls to finish from $S_5$ as it does from $S_6$, let $B$ denote that expected number. Let $A$ denote the expected number starting from $S_{\emptyset}$, so the answer to the problem is $A$.

Considering the possible outcomes if you are in $S_{\emptyset}$ we see that $$A=\frac 13\times (B+1)+\frac 23\times (A+1)$$

Similarly, thinking about the possible outcomes if you are in $S_5$ or $S_6$ we see that $$B=\frac 16\times 1+\frac 16\times (B+1)+\frac 23\times (A+1)$$

Solving that system of equations yields $$A=21\quad \&\quad B=18$$ so the answer is $\boxed {21}$.

Answer 2

It seems you are trying to solve the problem by finding the probability that two consecutive throws are either both $5$ or both $6$, and then taking one over this probability. Similar to how the expected number of flips it takes to get heads when $P(heads)=p$ is $1/p$. However, this does not work here, since the sequences of pairs of flips are not independent.

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Let $X$ be the number of rolls it takes to get two fives or two sixes in a row. Starting with zero rolls, three things can happen:

• Your first roll is between $1$ and $4$. In this case, the number of subsequent roll has the same distribution as $X$.

• You first roll is $5$ or $6$, and your second roll is different from the first roll. Again, the number of subsequent rolls is $X$.

• You first two rolls are both fives or both sixes. Here, $X=2$.

We get that $$ X\stackrel{d}= \begin{cases} X+1 & \text{with probability }\frac{2}{3} \\ X+2 & \text{with probability }\frac{1}{3}\cdot \frac56 \\ 2 & \text{with probability }\frac{1}{3}\cdot \frac16 \\ \end{cases}\tag{*} $$ Here is the interpretation of the above equality; $X$ has the same distribution as a random variable which is equal to one of the three options above, selected with the given probabilities independently of $X$.

Now, taking expectations of both sides, you get $$ E[X] = \frac23 \cdot E[X+1] + \frac13\cdot\frac56 \cdot E[X+2] + \frac13\cdot\frac16 \cdot 2 $$ This lets you solve for $E[X]$.

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As a side note, $(*)$ also implies that $$ f(X)\stackrel{d}= \begin{cases} f(X+1) & \text{with probability }\frac{2}{3} \\ f(X+2) & \text{with probability }\frac{1}{3}\cdot \frac56 \\ f(2) & \text{with probability }\frac{1}{3}\cdot \frac16 \\ \end{cases} $$ for any function $f$. Using $f(x)=x^2$, then taking expectations, lets you find $E[X^2]$, and therefore $\text{Var}(X)$. Using $f(x)=e^{tx}$ lets you find the full moment generating function of $X$.