Expectation of two dice game The game plays like this:
You roll two dices at the same time.
If you get same number on both dices, you have to roll again, until you get different numbers.
If you get different numbers, you stop.
your payout is the sum of the number on both dices, at the final roll.
What is the expectation of your payout
Follow up questions Guys. 
I encountered this question in a phone interview.
How can I do it quickly and explain it clearly.
The methods in the answers are clear and correct, but obviously I couldn't list down all the scenarios. That will take too long and make recruiter feel boring.
Please do advice a good way to explain while thinking.Thanks alot !!
 A: EDIT: I understand that the gain is the sum of the dice values on the terminating throw, that is, when you get different values, you only gain the sum.
There are obviously 6 cases of "continuing the game": 1+1,2+2,3+3,4+4,5+5,6+6. All other cases are terminating throws, and then you get up the gain of every case:
$$\begin{array}{ccccccccc}
1+2&=&3&&&&&&&&&&&&&&&&&\\
1+3&=&4&&2+3&=&5&&&&&&&&&&&&\\
1+4&=&5&&2+4&=&6&&3+4&=&7&&&&&&&&\\
1+5&=&6&&2+5&=&7&&3+5&=&8&&4+5&=&9&&&&\\
1+6&=&7&&2+6&=&8&&3+6&=&9&&4+6&=&10&&5+6&=&11\\
\end{array}$$
Note these cases repeat twice: $X=1;Y=2$ and $X=2;Y=1$ are different events.
So, the expected value of the gain  of this "terminating throw" is the average of the possible gains, over all possible $36-6=30$ cases, that is,
$$\mathbb{E}\{\text{gain}\}=2\cdot\frac{3+4+5+6+7+5+6+7+8+7+8+9+9+10+11}{30}=7$$
This is because all sums are equiprobable. If the events of gain were not equiprobable, you should calculate the wheighted average.
Why are they equiprobable? Let's see the probability of having a repeating throw, that is, equal values on both dices:
$$P_e=P(\text{dice 1}=\text{dice 2})=\sum_{k=1}^6P(X=k\cap Y=k)=6\cdot P(X=k)^2=\frac{1}{6}$$
Then, the probability of having different values is:
$$P_n=P(\text{dice 1}\neq\text{dice 2})=1-\frac{1}{6}=\frac{5}{6}$$
And it does not depend on the values!

EDIT 2: this observation comes from comments made during the writing of this answer.
Now, that was the expected gain of one "terminating throw", given that I understand that you only gain the sum of the throw with different values. If you understand that the total gain is the sum of every throw (repeating and terminating throws), we should calculate the probability of having $n$ throws, the associated gain, and then calculate the expected value (using total probability or something like that). 

EDIT 3: this is yet another observation from comments.
If you were to define a "winning strategy", that is, try to have a different outcome given that the present "repeating throw" is (1,1) or (2,2), it just doesn't matter because all throws are independent!
A: The expected value of the game is the conditional expectation of the sum of two dice given that you do not roll doubles.
Let $X$ be the sum of the dice. Let $A$ be the event of rolling doubles. So we are looking for
$$E(X|A^\complement).$$
Now, the unconditional expectation of the sum (roll the dice once and add them regardless of doubles or not) is well-known to be $E(X) = 7$ (twice the expected value of one die).
The conditional expectation if you do roll doubles is obviously $E(X|A) = 7.$
(Once again it's twice the expected value of one die--the same six equally likely
outcomes, only this time each one is simply doubled by the second die.)
Now it must be the case that
$$E(X) = E(X|A) P(A) + E(X|A^\complement) P(A^\complement).$$
Since $P(A) + P(A^\complement) = 1,$
the only way for this to happen is if $E(X|A^\complement) = E(X|A) = 7.$
Informally, if the expected value without
doubles were any higher or lower than the expected value with doubles, when you
averaged them together (producing the expected value over all rolls)
the "without doubles" value would drag the average upward or downward from the
"with doubles" value. But it doesn't, and therefore it's the same value.
A: The expectation of the sum of two (independent) dice is the sum of expectations of each die, which is $$\frac{1+2+3+4+5+6}{6}=\frac{7\times6}{2\times6}=\frac72$$
Hence for two dice the expectation is $7$.
But we have not yet taken into account the fact that at the final roll, the two dice have different value.
The final expectation is thus (with sums indices ranging from $1$ to $6$)
$$\sum_{i\neq j}\frac{i+j}{36}=\sum_{i,j}\frac{i+j}{36}-\sum_{i=j}\frac{i+j}{36}$$
The first term in RHS is the preceding expectation ($7$), the last is
$$\frac{1+2+3+4+5+6}{18}=\frac{7\times6}{2\times3\times6}=\frac{7}{6}$$
Thus your final expectation is $$7-\frac{7}{6}=\frac{35}{6}$$
A: when you get different numbers on your dices, your expectation is no longer 7.So the game divided in to two scenarios. First scenario(prob=1/6), you get same numbers and expectation is 7. Second scenario(prob=5/6), then this expectation should still be 7 right..? (36*7-7*6)/30=7.. so that's why
A: By simple symmetry the answer is 7,i.e., it is the same expected value as you roll 2 dice doesnt matter if you roll again if take doubles.
