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In successive rolls of a pair of fair dice, what is the probability of getting 2 sevens before 6 even numbers?

Assume that on the nth roll, the game ends. So this means we roll a seven on the nth round and in the $n-1$ rounds, we have only 1 seven and less than 6 rolls of an even.

So P(1 seven in $n-1$ rounds) = $ { n-1 \choose 1}\frac{1}{6} (1-\frac{1}{6})^{n-1-1}$

P(<6 evens and 1 seven) = P(0 even and 1 seven) + P(1 even and 1 seven) + ...+ P(5 evens and 1 seven) = $ \sum_{n=1,\,0\leq i \leq 5}^{\infty} {n-1 \choose 1}\frac{1}{6} (1-\frac{1}{6})^{n-1-1} {n-2 \choose i} (\frac{1}{2})^i (1-\frac{1}{2})^{n-2-i}$ Can anyone give me some pointers - I know this is incorrect as it is since if the summation starts at n=1, then the first term is undefined

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Hint: You can actually ignore the cases where you get neither an $7$ nor an even number, and use relative probabilities. – Thomas Andrews Dec 3 '12 at 20:58
The probability is zero: most dice only have the numbers 1 to 6 on them. I suspect you mean to say that the scores are added when you throw the dice, i.e. a one and a six scores a total of seven. – Fly by Night Dec 3 '12 at 20:59
Yes, I thought this too, but as I had written it is how it appeared in the question. I took it to mean that the sum is seven or even. – CAF Dec 3 '12 at 21:02
You also stop if you throw a sixth even on the nth round and have only 0 or 1 seven so far. – Ross Millikan Dec 3 '12 at 21:07
Note that the sets are disjoint. P(Even)=P(Odd)=1/2. $E(\text{throws for 6 Even}) = 12$. P(7)=1/6. $E(\text{throws for two 7s})=12$. So, the odds are 1:1 for getting two 7s and 6 even. – karakfa Dec 3 '12 at 21:52

Hint: You can ignore all odd throws except $7$ and consider that $P(7)=\frac 14, P(even)=\frac 34$. Then think of having seven throws-you will have either 6 evens or 2 sevens, but not both. So you win if there are 2 or more sevens among the first seven throws. It is probably easier to calculate the chance you lose-you need 6 or 7 evens among 7 throws.

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Distinguish $12$ states $S_{i,k}$ $\ (0\leq i\leq 1, \ 0\leq k\leq 5)$, plus two terminal states $S_{2\, \cdot}\ $, $S_{\cdot\, 6}\ $.

$S_{i,k}$ denotes the state where so far $i$ sevens and $k$ evens have been thrown. Denote by $P_{i,k}$ the probability that you win when the game is in state $S_{i,k}$. Then you have the $12$ equations $$P_{ik}={1\over6} P_{i+1,k}+{1\over2} P_{i,k+1}+{1\over3} P_{ik}\ ,$$ where $P_{2,k}=1$ and $P_{i,6}=0$. Solving this linear system gives you in particular $P_{0,0}$, the a priori probability that you win the game.

In the end this probability comes out to ${\displaystyle{4547\over8192}\doteq{5\over9}}$.

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Thanks for your reply. Can you tell me what is wrong with what I have done? – CAF Dec 4 '12 at 9:14
@CAF: I think it's correct, but you are left with a nasty infinite sum. - A binomial coefficient ${n\choose k}$ is defined to be $0$ unless $0\leq k\leq n$. – Christian Blatter Dec 4 '12 at 9:47
Should I perhaps take this one step further and try to get it into a form that makes the binomial coefficient defined. I.e split it into one infinite sum with n starting at 2 multiplied by the sum with i starting at 0, I.e something like$$ \sum_{n=2}^{\infty} {n-1 \choose 1} (1/6) (1-1/6)^{n-1-1} \sum_{i=0, n =3}^{n = \infty, i = 5} {n-2 \choose i} (1/2)^i (1/2)^{n-2-i}?$$ I have not used latex extensively, so please excuse the sloppy notation. Thanks! – CAF Dec 4 '12 at 10:19
Is the above ok? – CAF Dec 5 '12 at 13:52
@CAF: It was correct in your question, now not any more. The inner sum is over $i$ only. - Don't bother about binomial coefficients with negative entries. It comes out right automatically. – Christian Blatter Dec 6 '12 at 11:03

There are three possibilities: a 7 is rolled with probability 6/36, an even is rolled with probability 18/36, and a non-7 odd is rolled with probability 12/36. The third can be ignored, giving rescaled probabilities of 6/24 = 1/4 for a 7 and 18/24 = 3/4 for an even.

You 'win' if you roll a 7 on your n-th roll and your previous n-1 rolls include a seven and at most 5 even numbers.

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