Timeline for How to predict the likelihood of multiple future occurrences based on probability
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Dec 27, 2021 at 17:06 | comment | added | user801306 | If you compute $P^2$ and sum the third and fourth entries in the first row you'll get the same exact value I obtain in my answer | |
| Dec 27, 2021 at 13:14 | comment | added | user801306 | The fourth entry in the first row of $P^j$ indicates the probability you observe five $4$'s after the $j^\text{th}$ roll. You don't have to obtain these five $4$'s in one shot for this to take place. | |
| Dec 27, 2021 at 6:53 | comment | added | CyanCoding | How do you know the probability of having at least four 4's is the sum of the third and fourth entries? The fourth entry assumes you get all four matching in one roll. If I understand it correctly, the third entry assumes you get three matching on your first roll. So why is it the sum? | |
| Dec 27, 2021 at 2:58 | comment | added | user801306 | FYI learning about Markov chain's is an exciting and surprisingly painless experience. I highly recommend it! | |
| Dec 27, 2021 at 2:54 | comment | added | user801306 | So, since $$P^3=\begin{pmatrix}\frac{1953125}{10077696}&\frac{1421875}{3359232}&\frac{1035125}{3359232}&\frac{753571}{10077696}\\ 0&\frac{15625}{46656}&\frac{11375}{23328}&\frac{8281}{46656}\\ 0&0&\frac{125}{216}&\frac{91}{216}\\ 0&0&0&1\end{pmatrix}$$ The probability of getting at least four $4$'s after $3$ rolls is $$\frac{1035125}{3359232}+ \frac{753571}{10077696}\approx 0.38$$ | |
| Dec 27, 2021 at 2:54 | comment | added | user801306 | Let $X_0=2$ and $X_j$ be the number of fours that appear after $j$ rolls. Then $\{X_j\}_{j\geq 0}$ is a Markov chain with states $\{2,3,4,5\}$ and has state transition matrix $$P=\begin{pmatrix}\frac{125}{216}&\frac{25}{72}&\frac{5}{72}&\frac{1}{216}\\ 0&\frac{25}{36}&\frac{5}{18}&\frac{1}{36}\\ 0&0&\frac{5}{6}&\frac{1}{6}\\ 0&0&0&1\end{pmatrix}$$ The sum of the third and fourth entries in the first row of $P^j$ indicate the probability of having at least four $4$'s after $j$ rolls. | |
| Dec 27, 2021 at 2:49 | comment | added | CyanCoding | A little above my math knowledge haha. They're referenced in the Yahtzee probability article (which achieves a similar goal of mine but doesn't show all the equations) so I'll do more research on that topic. Thanks! | |
| Dec 27, 2021 at 2:35 | comment | added | user801306 | Are you familiar with Markov Chains? | |
| Dec 27, 2021 at 2:16 | comment | added | CyanCoding |
I've figured it out, but how does the formula change if I was calculating the chances with three turns left? Wouldn't I have a X, Y, and Z, with Z being the number of fours in my third round? How can I implement that?
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| Dec 27, 2021 at 0:59 | vote | accept | CyanCoding | ||
| Dec 26, 2021 at 18:44 | history | answered | user801306 | CC BY-SA 4.0 |