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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?

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    $\begingroup$ Are you familiar with the geometric distribution? $\endgroup$ Commented Sep 22, 2022 at 2:23
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    $\begingroup$ For what it's worth, a more interesting question would have been the expected value of the sum of the dice on each turn, for all of the required turns. Here, during stage 1, you could capitalize on the fact that $(7)$ is neutral with respect to the sum of the dice, so you would expect that the sum of the first $(6)$ rolls (for example) would be $(6 \times 7)$. Stage 2 is another matter, because $(3)$ is below the mean die value of $(3.5)$. $\endgroup$ Commented Sep 22, 2022 at 3:07

1 Answer 1

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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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  • $\begingroup$ Thank you for making me understand, now I see. Upvote! $\endgroup$
    – GWA
    Commented Sep 22, 2022 at 10:20
  • $\begingroup$ You are welcome! $\endgroup$
    – bluemaster
    Commented Sep 22, 2022 at 10:25

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