<s>[INST] Roll two balanced dice until the sum of the faces equals 7 appears for the first time. Determine the expected value of tosses in this experiment. Roll two balanced dice until the sum of the faces equals 7 appears for the first time. After that, roll the same two dice until some face 3 appears for the first time. Determine the expected value of tosses in this experiment.
All the possibilities are bellow:
(1,1)(2,1)(3,1)(4,1)(5,1)(6,1)
(1,2)(2,2)(3,2)(4,2)(5,2)(6,2)
(1,3)(2,3)(3,3)(4,3)(5,3)(6,3)
(1,4)(2,4)(3,4)(4,4)(5,4)(6,4)
(1,5)(2,5)(3,5)(4,5)(5,5)(6,5)
(1,6)(2,6)(3,6)(4,6)(5,6)(6,6)\
So, there are 6 possibilities for the sum to be 7. So the probability is $6\over36$ $=$ $1 \over 6$.\
And the probability of face three is going to be $11 \over 36$.
But how do I determine the number of expected tosses?
 [/INST] Let us represent

*

*$X$: number of tosses until dice sum is 7, $X\sim$ Geometric$(p=1/6)$, $E(X)=1/p=6$.

*$Y$: number of tosses until some face is 3, $Y\sim$ Geometric$(q=11/36)$, $E(Y)=1/q=36/11$.

*$X+Y$: total number of tosses

It is true that $E(X+Y)=E(X)+E(Y)=6+36/11=102/11.$
Therefore the expected number of tosses is $E(X+Y)=102/11$.
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