Timeline for How do I run a game of Dread over the Internet?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Jul 30, 2017 at 5:55 | comment | added | Nik Geier | A good little way to trim down on numbers to remember is to simply start the counter at 1 out of 20/30/500/whatever. Each time they roll, advance the counter by 1. If they roll equal to or below the counter, the Tower of Dread falls. This cuts down on bookkeeping, and on going through your list to see if you've rolled that exact number yet. | |
| Jul 30, 2017 at 4:30 | comment | added | user17995 | Have you tried this in past games of Dread? | |
| Jul 30, 2017 at 4:29 | history | edited | user17995 | CC BY-SA 3.0 |
Blew away fluff about necromancy
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| Oct 14, 2016 at 14:20 | comment | added | Longspeak | @Scott - The basic idea might need a bit of a shift then. Perhaps a d30 with a 'threshold' which changes, either at predictable points, or perhaps at points driven by the story. First couple of pulls, rolling a 1 will topple. Next couple, 2 or lower... Dunno. You want to keep the mechanic simple and fast moving, and you want the risk to increase until collapse is inevitable and players are really thinking twice about making a pull... | |
| Oct 14, 2016 at 3:12 | comment | added | Scott | So for a 500 side dice, with 27 rolls, you get 1.2E1134, divided by 3.3 E1061. This gives (1.2/3.3) E (1134-1061), which is 0.3636 E 73. We then divide that number by 500&27 (7.5 E 72). This gives us (0.3636/7.5) E (73-72), or 0.04848E1, which is the same as 0.4848. Multiply by 100 to get it in percentage is 48.48 We need the website to calculate factorials and we divide them that manual way because Excel doesn't like any number bigger than 9.99E99. Alternatively, pretty much any programming language should be able to calculate it easily. | |
| Oct 14, 2016 at 3:11 | comment | added | Scott | Ahh, makes sense. I decided to use the formula on the wikipedia page above instead of the simulations. Results: For a d500, you have a 48.4848% chance of losing after 27 pulls from the deck. To calculate this for any number (n=number of die rolls, f = number of faces), you take f factorial, and divide it by (f-n) factorial, and then divide that answer by (f^n). In order to divide numbers this big, which are expressed as 1.2E1134 (1.2 with 1134 0s after it), you divide the number before the E, and you subtract the numebr after the E. | |
| Oct 14, 2016 at 1:26 | comment | added | Longspeak | I was thinking in terms of what random die to use. If it's online you can use nearly any range of numbers, like a d365 or d400. | |
| Oct 14, 2016 at 1:09 | comment | added | Scott | Yeah, it definitely meets the core requirement of dread and could work well. I think some sort of '3 strikes' system could work. You have 3 'hit points' each time you roll a number that has been rolled before, you lose a HP. The 3rd time you roll a duplicate, you lose. With a d100, this gives you somewhere between 23-25 pulls before losing. 4 HP instead of 3 gives 27-29. For people who only own d20s, you need 9 hp to have 27-28 pulls on average, which I think is unlikely to lead to a fun game and feel similar enough to the 1 HP feel of vanilla. | |
| Oct 13, 2016 at 23:24 | comment | added | Longspeak | That's a good point. I still think the basic idea is a good one, providing increased risk as you proceed. What would you suggest for a decent run, with an average number of pulls between 25 and 30? | |
| Oct 13, 2016 at 22:21 | comment | added | Scott | Just an FYI, this will lead to quite short games. using a D20 leads to an average of roughly 6 pulls from the tower. Using a D100 results in roughly 12. (These numbers would produced by simulation using Excel's RANDBETWEEN, not actually calculated. Using a d365.25 results in a 50% probability of the tower having fallen after 23 pulls. en.wikipedia.org/wiki/… | |
| Oct 12, 2016 at 12:19 | history | answered | Longspeak | CC BY-SA 3.0 |