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We make successive throws of two balanced dice of six faces and we are interested in finding the probability of the event that the sum of 5 (the faces of the two dice) will appear before the sum 7. We assume that the throws are independent.

  1. First calculate the probability of the event E(n): In the first n - 1 throws, neither the sum 5 nor the amount 7 occurred, and in the n throw the amount was 5.

  2. The same question, but replacing 5 with 2.

I'm stuck . I have found that the sum 5 can appear in 4 cases : (1,4);(4,1);(2,3);(3,2).

And 7 in 6 cases : (1,6);(6,1);(2,5);(5,2);(3,4);(4,3).

All possible cases are 6*6=36 .

I found the probability of sum 7 : 6/36 = 1/6 = 0,16 And the probability of sum 5 : 4/36 = 1/9 = 0,11

Now I'm stuck and I don't even know if all I did here it's correct. If I understand how to calculate solution for 1, I can do alone for 2. Please help me !

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  • $\begingroup$ Well, compute $q$, the probability that a random toss is neither $5$ nor $7$. Then the answer to $1$ is $q^{n-1}\times \frac 4{36}$. Note: in $\# 2$ you meant to type $7$ instead of $2$. $\endgroup$ – lulu Oct 14 '17 at 13:17
  • $\begingroup$ Thanks. For #2, i meant what i said, replacing " 5 " with " 2 " , the probability to have 2 before 7. $\endgroup$ – Maria Bluee Oct 16 '17 at 11:06
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Dividing by 36 gives you the probability that a single roll of the dice will give you 7 or 5, respectively. But for your purpose here it is a detour.

It is more direct to say that you keep rolling until one of the ten game-ending rolls show up. Once that happens, you look at whether that roll is a 5 or a 7.

Hopefully it is clear that each of the $10$ rolls of interest is equally likely to be the one that ends the game. Of these $10$ rolls, $4$ win, so the probability of winning is $\frac{4}{10}=\frac25$.

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