<s>[INST] Probability Q on rolling two 8-sided dice Let's say I roll two 8-sided dice. I win if the sums '7' and '11' show up before we see the sum '9' TWICE. What is my probability of winning?
So this is my answer and please correct me if I am wrong:
Prob of getting either 7 or 9 —> 6/64 + 6/64 = 12/64 —> 18.75%
Prob of getting 9 twice —> 8/64 x 8/64 = 1/64 —> 1.6%
But now to find the probability of winning, would it be 18.75 - 1.6 = 17.15% ? Am I right or wrong?
Thanks!
EDIT
From the answers provided below, we have three different results:
84%
49%
60%
Which one would be the correct answer?
 [/INST] This answer assumes the person wins if there is a sum of $\, 7\,$ or $\,11\,$ in any toss before a sum of $\,9\,$ appears twice.
Probability of winning in first toss $ = \frac{12}{64}$
a) If the person gets a sum of $9$ once, the chance of winning from there
$\displaystyle P(W1) = \frac{12}{64} + \frac{44}{64} \times (\frac{12}{64} + \frac{44}{64} \times ...)... = \frac{12}{64} (1 + \frac{44}{64} + (\frac{44}{64})^2 + ...)$
$\displaystyle = \frac{3}{5}$
[Using sum of infinite geometric series $a + ar + ar^2 + ... = \frac{a}{1-r}$]
b) If the first toss is any sum other than $7, 9, 11, \,$starting 2nd toss
$P(W2) = \frac{12}{64} + \frac{8}{64} \times P(W1) + \frac{44}{64} \times (\frac{12}{64} + \frac{8}{64} \times P(W1) + ...)... $
$$= (\frac{12}{64} + \frac{8}{64} \times P(W1))(1 + \frac{44}{64} + (\frac{44}{64})^2 + ...) = \frac{21}{25}$$
$\,$
So person's chance of winning $ \displaystyle = \frac{12}{64} + \frac{8}{64} \times P(W1) + \frac{44}{64} \times P(W2) = \frac{21}{25}$
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