Markov chain or a different method? I am rolling 2 fair dice repeatedly...

Falsetto

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Nov 6, 2017
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Ok, here's the problem.

I am rolling 2 fair dice repeatedly and I want to know what the chance is of rolling both a total of 5 and a total of 9 before rolling a total of 7.

Let y equal the probability of a 5 or 9 or 7 being rolled. There are 4 ways each to roll a 5 and 9 and 6 ways to roll a seven for a total of 14/36 combinations.

I calculate P(5|y) = 1/9 / 7/18 or 1/9 * 18/7 = 18/63 = 2/7

P(9|y) must equal P(5|y) so also 2/7

P(7|y) = 1 - (P(5|y) + P(9|y)) = 3/7

So far so good?


Now I'm not sure if I'm on the right track next or not but . . .

If I imagine this problem as having 5 states:-

a) Having made 0 of the numbers yet.
b) Having rolled a 5 but neither a 9 nor 7.
c) Having rolled a 9 but neither a 5 nor 7.
d) Having succeeded in rolling both a 5 and a 9 without the 7 (won).
e) Having rolled a 7 and failed (lost).

I then made a transition matrix from that data.

a b c d e
a 0 2/7 2/7 0 3/7

b 0 2/7 0 2/7 3/7

c 0 0 2/7 2/7 3/7

d 0 0 0 1 0

e 0 0 0 0 1

(EXCUSE THE FORMATTING)


Since I always start in state 'a' I multiplied an initial distribution vector of
[1 0 0 0 0] by the above matrix raised to the power of 50 to try and find an approximation of the steady state probabilities.

This told me that P(5 and 9 before 7) = 0.2286 (ish)

Anyone make any sense of that and tell me if I got the answer right? If yes, could I have done it in an easier way? If not, what should I have done?

Thanks in advance for your help folks!
 
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