# What is the probability of rolling 2 7's before 6 even numbers on successive rolls of a pair of fair dice?

Would this equal Pr( rolling 2 7's)/Pr(rolling 2 7's or rolling 6 even numbers)? Or could I approach the problem as follows:

1- (Pr(rolling 6 even numbers)+ Pr( rolling 5 even numbers, 7 and an even number)) ?

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What exactly is "rolling 6 even numbers"? Three times rolling two even numbers? Or six times a sum that is even? –  TMM Sep 28 '11 at 1:26
A sum that is even. –  lord12 Sep 28 '11 at 1:28

On the assumption that your question means that you roll a pair of dice several times and stop when you have rolled either six even numbers or two sevens during the sequence, then neither of your answers are correct.

What you need to do is look at the probability of rolling a seven before an even number: this is $\dfrac{\frac{1}{6}}{\frac{1}{6}+\frac{1}{2}}=\dfrac{1}{4}$.

You can then work out the probabilities of getting to the position of $a$ sevens and $b$ even numbers without having previously stopped. You have

• $\Pr(0,0)=1$,
• $\Pr(0,b)=\frac{3}{4}\Pr(0,b-1)$ for $1\le b \le 6$,
• $\Pr(a,0)=\frac{1}{4}\Pr(a-1,0)$ for $1\le a \le 2$,
• $\Pr(1,b)=\frac{3}{4}\Pr(1,b-1)+\frac{1}{4}\Pr(0,b)$ for $1\le b \le 5$,
• $\Pr(1,6)=\frac{3}{4}\Pr(1,5)$, and
• $\Pr(2,b)=\frac{1}{4}\Pr(1,b)$ for $1\le b \le 5$.

The answer is $$\sum_{b=0}^5 \Pr(2,b)= \dfrac{4547}{8192} \approx 0.555.$$

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On each meaningful roll the probability of getting a seven is $$\frac{\frac16}{\frac16+\frac12} = \frac14,$$ so the probability of getting at most one seven in seven meaningful rolls is $$\binom70\left(\frac34\right)^7 + \binom71\left(\frac14\right)\left(\frac34\right)^6 = \frac{3^7+7\cdot 3^6}{4^7} = \frac{10\cdot 3^6}{4^7} = \frac{3645}{8192}.$$
The probability of getting at least two sevens in seven meaningful rolls is therefore $$1 = \frac{3645}{8192} = \frac{4547}{8192}.$$