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 Advanced Statistics Advanced Probability and Statistics Math Forum

 November 6th, 2017, 05:36 PM #1 Newbie   Joined: Nov 2017 From: London Posts: 3 Thanks: 0 Markov chain or another solution? 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! November 6th, 2017, 07:27 PM #2 Senior Member   Joined: Sep 2015 From: USA Posts: 2,553 Thanks: 1403 This is how I did it. Consider sequences of rolls that lead to a win in $n$ rolls. Note $n \geq 2$ Roll $n$ will either be a $5$ or a $9$ Rolls $1 \text{ through }(n-1)$ will not be roll $n$ and will contain no $7$ Now consider roll sequences of length $n-1$ that satisfy this. There will be at least $1$, and up to $n-1$ rolls that are either $5$ or $9$ whichever isn't roll $n$, and none of these will be $7$. Otherwise they can be any roll. so we end up with $P[\text{win}] = 2 \sum \limits_{n=2}^\infty \dfrac 1 9\left(\sum \limits_{k=1}^{n-1}\dbinom{n-1}{k}\left(\dfrac 1 9\right)^k\left(\dfrac{11}{18}\right)^{n-1-k}\right)$ Digesting this a bit. The $2$ in front is because roll $n$ can be either $5$ or $9$, and the probability is identical for both. The $\dfrac 1 9$ is the probability of roll $n$ being the $5$ (or the $9$) What's inside the parens is the sum of probabilities that there are $k$ occurrences of $5$ or $9$ whichever isn't roll $n$ This can be simplified to $P[\text{win}] = 2 \sum\limits_{n=2}^\infty \left(\left(\dfrac{13}{18}\right)^{n-1} - \left(\dfrac{11}{18}\right)^{n-1}\right)$ and then finally to $P[\text{win}] =\dfrac{8}{35}$ I leave it to you to digest all of this and reproduce the simplifications in more detail. ps. $p=\dfrac {8}{35}$ matches my sim of this very well so I'm pretty confident in this answer. Thanks from Falsetto Last edited by romsek; November 6th, 2017 at 08:06 PM. November 6th, 2017, 07:43 PM #3 Newbie   Joined: Nov 2017 From: London Posts: 3 Thanks: 0 THANKYOU!! Exactly what i was hoping to find. And also the answer i had was correct as well But a formula like yours is so much better. I will study it in detail tomorrow when i wake up. Thanks for your time and help. Really appreciate it! November 15th, 2017, 04:58 PM #4 Newbie   Joined: Nov 2017 From: London Posts: 3 Thanks: 0 Trying to digest what you did I figured out that there is a lot I still don't know. I have had a look into geometric sequences and that is helping me get started. Are there any other topics you would recommend me learning more about to gain a better grasp of what's going on here? Haha Also, in playing with the problem I found another way to work it out I think. Since we only care about 5,9 and 7. Can I say that P(5 or 9 | 5,9,7) = 8/14 and then the probability of the next one is p(5 | 5,7) = 4/10 8/14 * 4/10 = 32/140 = 8/35 again. Think this only works is a case like this though where the P(n) is the same for both. How would your formula above change for a problem like rolling 4 and 5 before 7? November 15th, 2017, 08:47 PM   #5
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Quote:
 Originally Posted by Falsetto Trying to digest what you did I figured out that there is a lot I still don't know. I have had a look into geometric sequences and that is helping me get started. Are there any other topics you would recommend me learning more about to gain a better grasp of what's going on here? Haha Also, in playing with the problem I found another way to work it out I think. Since we only care about 5,9 and 7. Can I say that P(5 or 9 | 5,9,7) = 8/14 and then the probability of the next one is p(5 | 5,7) = 4/10 8/14 * 4/10 = 32/140 = 8/35 again. Think this only works is a case like this though where the P(n) is the same for both. How would your formula above change for a problem like rolling 4 and 5 before 7?
Your solution is the way to go. Very nicely thought out. Tags chain, dice, markov, solution Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post kfp22 Advanced Statistics 0 October 8th, 2014 06:43 AM Hemantha Probability and Statistics 1 July 7th, 2014 12:57 PM legendoulis Advanced Statistics 4 April 4th, 2012 11:28 AM butabi Advanced Statistics 1 February 12th, 2012 03:20 PM Turloughmack Linear Algebra 0 February 7th, 2011 05:15 AM

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