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Probability that sum of 5 occurs first time only = 4/36

Probability that sum of 5 occurs second time but first time sum of 7 also doesn't occur = (26/36)(4/36)

Probability that sum of 5 occurs third time but first two times sum of 7 also doesn't occur = (26/36)(26/36)(4/36)

So, P(Sum of 5 comes before 7) = (4/36)+(26/36)(4/36)+$(26/36)^{2}$(4/36)+...

Therefore, P(Sum of 5 comes before 7) = 4/10 = 0.4
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P$($ getting 5 as sum with a pair of dice $)\ =$ $\Large \frac{4}{36}$ $\{(1,4)(4,1)(3,2)(2,3) \}$

P$($ getting 7 as sum with a pair of dice $)\ =$ $\Large \frac{6}{36}$  $\{(1,6)(6,1)(2,5)(5,2)(3,4)(4,3) \}$

P$($ getting sum of $5$ before sum of $7$$)\ =$ $\Large \frac{\Large\frac{4}{36}}{\frac{4}{36}+\frac{6}{36}}$ $\Large=\frac{2}{5}$
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Favourable cases for 5: {(1,4), (2,3)}
Favourable cases for 7: {(1,6), (2,5), (3,4)}

(I ignored the reverse cases like (4,1) for (1,4) as this happens for both 5 and 7 and hence won't affect the probability)

So, chance that total of 5 comes before 7 = 2/5 

(We can ignore all other cases, as questions asks for probability of sum = 5 compared to sum = 7)

                                         OR

P(( getting 5 as sum with a pair of dice ) =) = 4/36 {(1,4)(4,1)(3,2)(2,3)}{(1,4)(4,1)(3,2)(2,3)}

P(( getting 7 as sum with a pair of dice ) =) = 6/36  {(1,6)(6,1)(2,5)(5,2)(3,4)(4,3)}

1/9 is the probability of sum of 5 and 1/6 is the probability of sum of 7. Probability that neither happens = 26/36 = 13/18.  
 

S=1/9+13/18×1/9+…S=1/9+13/18×1/9+…



In words win5 +loss(5,7)win5 +.... 

It is a GP series with first term 1/91/9 and common ratio 13/1813/18. 
 

S=1/9(1−(13/18)) =  2/5

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P(sum of 5 in two dice) =1/9

P(sum of 7 in two dice)= 1/6

P( getting sum apart from 5 or 7)= 1- P(sum of 5 or 7) = 1- { 1/9 +  1/6 – 1/9*1/6 }  =  40/54

Now before 5 we can get sum of any number but not 7

Hence P(getting sum 5 before 7)=  1/9 +  (40/54) *1/9   +   (40/54)^2  * 1/9   + …………

It forms GP and we get P(getting sum 5 before 7)=  3/7 =  0.42

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