Probability of obtaining 7 before 8 with pair of fair dice

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SUMMARY

The probability of obtaining a sum of 7 before a sum of 8 when rolling a pair of fair dice is calculated using a geometric series. The derived probability is 5/11, which is obtained by summing the series where 5/36 represents the probability of rolling an 8 and 25/36 represents the probability of rolling neither a 7 nor an 8. The discussion confirms that a geometric series is necessary to solve this problem due to its nature of infinite sums.

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squaremeplz
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Homework Statement



A pair of fair dice is rolled until the first 8 appears. What is the probability that a sum of 7 does not precede a sum of 8.

Homework Equations



Geometric series


The Attempt at a Solution



P(sum of 7 does not appear before sum of 8) =

5/36 + 5/36 * 25/36 + 5/36 *(25/36)^2 + ...

= (5/36)/(1-25/36) = 5/11

do I have to use geometric series for this or just find the prob of 7 before 8?

Thanks!
 
Last edited:
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Yes, I think you need to use a geometric series. Either 7 or 8 appearing ends the game, right? So I think you need to find the probability that neither 7 nor 8 appears in k-1 rolls times the probability that 8 appears on the kth roll. Then sum over all k.
 


squaremeplease said:

Homework Statement



A pair of fair dice is rolled until the first 8 appears. What is the probability that a sum of 7 does not precede a sum of 8.

Homework Equations



Geometric series

The Attempt at a Solution



P(sum of 7 does not appear before sum of 8) =

5/36 + 5/36 * 25/36 + 5/36 *(25/36)^2 + ...

= (5/36)/(1-25/36) = 5/11

do I have to use geometric series for this or just find the prob of 7 before 8?

Thanks!

I think that's correct 5/36 is the probability of 8 and 25/36 is the probability of neither 7 nor 8. You did do a geometric series. How can you compute the probability of 7 before 8 without it?
 
Last edited:


I don't think it is possible without using a geometric series because it's a problem of infinite sum.
 

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