Rolling a die until two rolls sum to seven

Question

Here's the question:

You have a standard six-sided die and you roll it repeatedly, writing down the numbers that come up, and you win when two of your rolled numbers add up to $7$. (You will almost surely win.) Necessarily, one of the winning summands is the number rolled on the winning turn. A typical game could go like this: $1, 1, 4, 5, 3$; you win on the 5^th turn because $3 + 4 = 7$. How many turns do you expect to play?

Here's what I've tried: We seek $E(N)$ where $N$ is a random variable counting the number of turns it takes to win. Then $N \ge 2$, and $$E(N) = \sum_{n=2}^\infty n P(N=n) = \sum_{n=1}^\infty P(N > n).$$ I want to find either $P(N=n)$, the probability that I win on the $n$^th turn, or $P(N > n)$, the probability that after $n$ turns I still haven't won. Note that $P(N = 1) = 0$. Let $X_k$ be the number rolled on the $k$^th turn. Then $$P(N = 2) = P(X_1 + X_2 = 7) = \sum_{x=1}^6 P(X_1 = x)P(X_2 = 7-x) = 6\cdot \frac{1}{6}\cdot\frac{1}{6} = \frac{1}{6}.$$ So far so good. To compute $P(N > 3)$ I let $A_{i, j} = \{\omega \in \{1, \dotsc, 6\}^3 : w_i + w_j = 7\}$ and used the inclusion-exclusion principle and symmetry to find $$|A_{1,2} \cup A_{2,3} \cup A_{1,3}| = 3|A_{1,2}| - 3|A_{1,2}\cap A_{1,3}| = 90$$ so $P(N > 3) = \frac{126}{216} = \frac{7}{12}$. This is the probability that no two of three dice sum to seven. Similarly, I found $P(N > 4)$ to be $\frac{77}{216}$.

I don't see how to generalize the above. I also thought that $$P(N > n) = P(X_i + X_j \ne 7 \text{ for all }1 \le i\ne j \le n) = (1 - P(X_i + X_j = 7))^{\binom{n}{2}} = \left(\frac{5}{6}\right)^{n(n-1)/2}$$ but that's false because the events are not independent.

I also tried $$P(N = n) = P(X_n = 7 - X_k \text{ for some } 1 \le k < n \text{ and }N \ne n - 1)$$ where that last clause is shorthand for "and the previous rolls did not secure your victory". This yields the recursion $p_n = (1-(5/6)^{n-1})(1-p_{n-1})$, $p_1 = 0$, which didn't agree with my previously computed probabilities. (Perhaps I made an error.)

Answer 1

To be clear: What follows assumes that the pair that adds to $7$ need not be consecutive. Thus, for example, I am assuming that the sequence $\{1,3,6\}$ ends the game in three rolls.

Let $E=E_0$ be the answer. If you have rolled a collection $S$ which contains no pair which adds to $7$ then let $E_S$ denote the expected number of rolls it will take from there. As every roll is equally probable, all that really matters is the size of $S$, so let $E_S=E_n$ if $S$ has $n$ elements. Of course the only possible $n$ are $\{0,1,2,3\}$.

We start with $E_3$. We note that rolling again has a $\frac 12$ chance of completing the $7$ and a $\frac 12$ chance of getting a useless duplicate, thus $$E_3=\frac 12\times 1+\frac 12 \times (E_3+1)\implies E_3=2$$

Now for $E_2$. As before we consider the next toss and write $$E_2=\frac 13\times 1+\frac 13 \times (E_2+1)+\frac 13 \times (E_3+1)\implies E_2=\frac 52$$

And $E_1$. As before $$E_1=\frac 16\times 1+\frac 16 \times (E_1+1)+\frac 46\times (E_2+1)\implies E_1=\frac {16}5$$

And then of course $$E=E_1+1=\frac {21}5$$

Answer 2

assuming that rolls need not be consecutive I.e. $1,3,6$ would be a terminating sequence of rolls since $1+6=7$ despite the $1$ and the $6$ not being adjacent.

I will admit I didn't read through all of your work, but this is the method I would approach the problem from. We can describe this using an absorbing markov chain with the following states:

$A_1 = $ Exactly one type of number has been seen so far. (This can be used as the starting state as we are guaranteed to enter this state after the first roll)

$A_2 = $ Exactly two types of numbers have been seen so far and they do not sum to seven. (The two numbers would be both from $\{1,2,3\}$, both from $\{4,5,6\}$, or one from each)

$A_3 = $ Exactly three types of numbers have been seen so far and no pair of which sum to seven. (The three numbers would be all from $\{1,2,3\}, \{4,5,6\}$ or from some combination thereof)

Finally, $B$ will be the endstate where we have some pair of numbers adding to seven. Convince yourself that these are indeed the only possible states.

From each state, you may either travel from $A_i$ to $A_{i+1}$, from $A_i$ to itself, or from $A_i$ to $B$ according to the following stochastic matrix with order of rows and columns as $A_1,A_2,A_3,B$

$$\begin{bmatrix} \frac{1}{6} & 0 & 0 & 0\\ \frac{2}{3} & \frac{1}{3} & 0 & 0\\ 0&\frac{1}{3}&\frac{1}{2}&0\\ \frac{1}{6}&\frac{1}{3}&\frac{1}{2}&1\end{bmatrix}$$

Recognizing this as an absorbing stochastic matrix, we rearrange the rows/columns into standard form: $\left[\begin{array}{c|c} I&S\\\hline 0&R\end{array}\right]$. We will instead use order $B,A_1,A_2,A_3$ for rows and columns

$$\begin{bmatrix} 1&\frac{1}{6}&\frac{1}{3}&\frac{1}{2}\\ 0&\frac{1}{6}&0&0\\ 0&\frac{2}{3}&\frac{1}{3}&0\\ 0&0&\frac{1}{3}&\frac{1}{2}\end{bmatrix}$$

In solving for the long-term state in the case of multiple possible absorbing states, we would calculate it using the limiting matrix: $\left[\begin{array}{c|c}I&S(I-R)^{-1}\\0&0\end{array}\right]$. Of course, in the current problem, there is only one ending state, so we know it will become $1$ in the entire top row and zeroes elsewhere. Still, the matrix $(I-R)^{-1}$ contains incredibly useful information and is referred to as the fundamental matrix for the markov chain. Depending on starting position, the sum of the column will give the expected number of turns until reaching an endstate. Alternatively, if the starting position is unknown, multiplying by an appropriate probability vector will give the desired information.

In our case: $I-R = \begin{bmatrix}\frac{5}{6}&0&0\\-\frac{2}{3}&\frac{2}{3}&0\\0&-\frac{1}{3}&\frac{1}{2}\end{bmatrix}$

$(I-R)^{-1} = \begin{bmatrix}1.2&0&0\\1.2&1.5&0\\0.8&1&2\end{bmatrix}$

Summing along the first column, we expect from having started in state $A_1$ it to take $1.2+1.2+.8 = 3.2$ turns to complete. Accounting for the fact that it takes one turn to enter state $A_1$, this gives a total expected time as $4.2$ turns.

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As an aside, one can extract $Pr(N=n)$ from the stochastic matrix. Letting $A$ be the matrix in standard form and $v$ be the column vector with a $1$ in the second entry and zeroes elsewhere, you have the first entry of $A^{n-1}v - A^{n-2}v$ will give the probability that you arrive at the endstate on turn $n$ but not having already been there on turn $n-1$.

Answer 3

While not exactly stated, it seems that you win only if the sum of consecutive numbers sums to 7.

Obviously you can't win on the first roll. For every roll thereafter, you have a $1/6$ chance of winning on the next roll, (and a $5$ in $6$ chance of the game continuing).

Now, $P(N=2) = \frac{1}{6}$, $P(N=3) = \frac{5}{6}\times\frac{1}{6}$, $P(N=4) = \left(\frac{5}{6}\right)^2\times\frac{1}{6}$, so $P(N=n) = \left(\frac{5}{6}\right)^{n-2}\times\frac{1}{6}$.

Thus $$E(N) = \sum P(N=n)n = 2\times\frac{1}{6} + 3\times\frac{5}{6}\times\frac{1}{6} + 4\times\left(\frac{5}{6}\right)^2\times\frac{1}{6} + \dotsb = \frac{1}{6} \sum_{n=0}^{\infty}(n+2)\left(\frac{5}{6}\right)^n.$$

I don't know if you need to derive this, but, $$\sum_{n=1}^{\infty} n x^{n-1} = \sum_{n=0}^{\infty}(n+1) x^n = \frac{1}{1-x}\sum_{n=0}^\infty x^n = \frac{1}{(1-x)^2}$$ so $$\frac{1}{6}\sum_{n=0}^{\infty}(n+2)\left(\frac{5}{6}\right)^n = \frac{1}{6}\sum_{n=0}^\infty(n+1)\left(\frac{5}{6}\right)^n + \frac{1}{6}\sum_{n=0}^\infty \left(\frac{5}{6}\right)^n$$ which equals $$\frac{1}{6}\times\frac{1}{\left(1-\frac{5}{6}\right)^2} + \frac{1}{6}\times\frac{1}{1-\frac{5}{6}} = 7.$$

Answer 4

The answer is: 4.

Reasoning: A six sided die presumably has numbers in $\{1,...,6\}$. Every one of those numbers can be used to add to $7$. So take your first roll. It doesn't matter what the number is. Now look at your die. There is one number that can add to seven with your first number, and you can roll it with probability $1/6$. Thus, the probability of winning on each successive turn is $1/6$. The probability of winning on a successive turn is given by the binomial distribution where $p = 1/6$.

The interesting part about this problem is how it is phrased: it is asking for how many turns you expect to need until you win. To me, that means the number of rolls required to find a probability of you winning $\ge 1/2$.

The expected value of the binomial distribution is $np$. So, we want $np \ge 1/2$. Since $p = 1/6$, $n$ must be at least $3$. Therefore, you need your first turn ($1$) plus your $3$ "expected" turns to win: $1 + 3 = 4$.