Timeline for What's the probability for a 14-card hand to contain at least one full-house?
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 29, 2020 at 15:06 | vote | accept | user11718766 | ||
| Mar 29, 2020 at 15:05 | vote | accept | user11718766 | ||
| Mar 29, 2020 at 15:06 | |||||
| Mar 29, 2020 at 15:04 | vote | accept | user11718766 | ||
| Mar 29, 2020 at 15:05 | |||||
| Mar 29, 2020 at 14:46 | answer | added | Daniel Mathias | timeline score: 1 | |
| Mar 29, 2020 at 14:43 | vote | accept | user11718766 | ||
| Mar 29, 2020 at 15:04 | |||||
| Mar 29, 2020 at 14:28 | comment | added | joriki | @greg.: Those are my own; here they are: Binomials, BallsInBins | |
| Mar 29, 2020 at 14:19 | answer | added | joriki | timeline score: 1 | |
| Mar 29, 2020 at 14:11 | comment | added | user11718766 | @joriki: Where did you get the Binomials and BallsInBins classes used in you code from? Can you make them available? | |
| Mar 29, 2020 at 12:46 | comment | added | Daniel Mathias | That result is from enumeration via code, similar to the Java code joriki used to confirm the result for consecutive pairs. (Speaking of...) | |
| Mar 29, 2020 at 12:45 | comment | added | joriki | @greg.: I get the same count of $1024024781208$ by enumeration (using this Java code). Your doubts about your treatment of the $3$-card rank are right. I'll think about how this can be done properly; I think it's a bit more difficult than the other one. | |
| Mar 29, 2020 at 12:40 | comment | added | user11718766 | How did you come up with that answer? Did you also use inclusion-exclusion? | |
| Mar 29, 2020 at 12:34 | comment | added | Daniel Mathias | The probability is $\frac{1024024781208}{\binom{52}{14}}\approx 57.888313%$. I'll see if I can spot your error. | |
| Mar 29, 2020 at 12:23 | history | asked | user11718766 | CC BY-SA 4.0 |