Expected value of times we roll a pair of dice

Question

I am trying to solve this using a geometric series. Pretty much the question goes like this: We roll a pair of fair dice repeatedly and independently and stop as soon as the sum of the dice is equal to 7.

My random variable is: $X =$ the number of times we roll.

To calculate the expected value of time we will roll I do somehting like this:

$$E(X) = X_1P_1+X_2P_2+X_3P_3+...+X_nP_n$$

The probability or rolling a 7 using 2 dice is $\frac 6 {36}=\frac1 6$ so then,

$$E(X)=1*\frac 1 6+2*\frac 1 6+3*\frac 1 6+...+n*\frac 1 6$$

Not to sure about this:

$$E(X)=\frac 1 6 \sum_{n=1}^{\infty} n*\frac 1 6$$

And if this is the correct sequence what can I do with it next?

Answer 1

No, the probability that it takes you one turn (meaning you roll the pair of dice) is $\frac{6}{36}$. The probability that you take two turns (roll the pair once and fail, roll the pair again and succeed), that is $\frac{30}{36}\cdot\frac{6}{36}$. Thus, you succeed on $k$th roll is $$P(X = k) = \left(\frac{30}{36}\right)^{k-1}\left(\frac{6}{36}\right).\tag{$\star$}$$ Then, calculating the expectation would be $$E[X] = \sum_{k=1}^\infty kP(X = k).$$ You can proceed like this, or you can notice (as $(\star)$ suggests) that $$X\sim\text{Geom}(p = 6/36)$$ over $\{1,2,\dotsc\}$. Therefore $$E[X] = \frac{1}{p} = \frac{36}{6} = 6.$$ You can check your method against this.

Answer 2

The probability of rolling a 7 at any point is $\frac { 6}{36}$.

Since we roll the dice until the first sum appears that is 7 we end up with a geometric series: $$E(X)=\frac {6}{36}*\sum_{k=1}^{\infty} k*(\frac {30}{36})^{k-1}$$

Taking the derivative of the sum we get $$\frac {6}{36}* \frac 1 {(1- \frac {30} {36})^2}$$ which equals to $6$.