# Dice probability where sum of seven doesn't precede sum of eight.

Given the question: A pair of fair dice is rolled until the first sum of 8 appears. What is the probability that a sum of seven doesn't precede sum of 8?

My solution is: P(sum of 7 doesn't precede 1st sum of 8) = 1 - P(sum of 7 precedes 1st sum of 8) $= 1 - \left(\frac{6}{36}\right)\left(\frac{5}{36}\right) = \frac{211}{216}$

This solution seems right to me but I thought why can't it be solved using conditional probability: 1 - P(first sum of 8 comes given sum of seven is rolled) = 1 - P(8|7) = 1 - P(8 and 7)/P(7) = 1 - (5/36)(6/36)/6(/36) = 1 - (5/36) = 31/36 or is that just for 8 after 7 on any roll following 7, that is not necessarily 8 right after 7?

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You have found the probability that you don't get $7$ on the first throw and $8$ on the second throw. I see two ways to read the question. Do you want the chance that there is not a seven on the roll immediately preceding the first eight, or that there is not a seven in any of the rolls before the first eight? These are very different questions. – Ross Millikan Jan 21 '14 at 16:06
You can condition on whether the first roll is $7$, $8$, or neither. Call these events $A$, $B$, and $C$, respectively. Let $E$ be your desired event (no $7$'s before the first $8$, presumeably). Note $P(E|B)=1$, $P(E|A)=0$, and $P(E|C)=P(E)$. You thus have $P(E)=P(B)+P(E)P(C)$. Solving for $P(E)$ gives $P(E)={P(B)\over 1-P(C)}=5/11$. – David Mitra Jan 21 '14 at 16:30

Sums other than $7$ or $8$ are irrelevant: if we do not record anything that is not a $7$ or $8$, the game ends in one "step." The ordinary probability of a $7$ is $\frac{6}{38}$, and the probability of an $8$ is $\frac{5}{36}$. Thus if we record only a $7$ or $8$, the probability of $8$ is $\frac{5}{11}$.

More formally, the probability that a toss is an $8$ given that it is a $7$ or an $8$ is, by the usual formula for conditional probabilities, equal to $\frac{\frac{5}{36}}{\frac{6}{36}+\frac{5}{36}}$, that is, $\frac{5}{11}$.

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Let the chance getting a 7 be q, the chance getting an 8 be p

The probability is $p+(1-p-q)p+(1-p-q)^2p+...=\frac{p}{p+q}$

What are $p$ and $q$? Do you see why this is the answer?

Final answer: $5/11$

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Sorry, could you explain how lhs equals p/(p+q) and how does that equal 5/11? – Vector_13 Jan 21 '14 at 18:02
@YuriSwarovski it is the sum of a geometric series with first time p, ratio $(1-p-q)$ – Lost1 Jan 21 '14 at 18:04
I see, and p and q are just rations of their prob. thanks for your answer. – Vector_13 Jan 21 '14 at 18:09
@YuriSwarovski indeed, np – Lost1 Jan 21 '14 at 18:12

You'll need to account for all possible numbers of rolls.

The probability that you roll anything but $7$ or $8$, $n$ times in a row, is $(\frac{25}{36})^n.$

The probability that you roll an $8$ is $\frac{5}{36}$.

So the total probability is

$$\sum_{k=0}^{\infty}\left(\frac{25}{36}\right)^{k}\left(\frac{5}{36}\right)=\frac{5}{36}\sum_{k=0}^{\infty}\left(\frac{25}{36}\right)^k = \frac{5}{36}\frac{36}{11} = \frac{5}{11}.$$

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Errrrr... not quite. You need to use the probability that "you roll anything but a 7 or a 8, $n$ times in a row" and not "you roll anything but 7", as in Lost1's answer. – Dilip Sarwate Jan 21 '14 at 16:21
Yeah, I caught that a few minutes later. Already edited. :) My previous answer did seem a bit high. – John Jan 21 '14 at 16:24
Why does the sum of (25/36)^k equal 36/11? – Vector_13 Jan 21 '14 at 18:01
It's a geometric series: $\sum_{k=0}^{\infty}r^k = (1-r)^{-1}.$ – John Jan 21 '14 at 18:04
I see thank you. – Vector_13 Jan 21 '14 at 18:12