Forums → Forum Games → Roll A Dozen: Revived
Welcome back to one of my old games, Roll A Dozen! Due to some personal reasons, i was unable to keep the original Ra12 game alive. Thus, I have created a new thread to continue the game!
Once again, here are the objectives:
- Players must reach exactly 12 to win.
How to Play:
- A six-sided die is used.
- The player must roll the die. If an even number is used (e.g, 2, 4, 6),
you add that number to the previous rolled number.
- If an odd number is rolled (1, 3, 5) you subtract that amount from the
- A new count always begins with the number 1.
- The player who restarts the count must roll the dice and add/subtract the
resulting number from 1.
- The count must be restarted if:
- The count reaches zero or goes into negative numbers.
- A player rolls the same number as the previous poster.
P1: 1 (dice rolled 4) 1+4=5
P2: 5 (dice rolled 2) 5+2=7
P1: 7 (dice rolled 3) 7-3=4
P3: 4 (dice rolled 3) restart
Things to Remember:
- You can back-to-back post! Just don't post three times in a row.
Once a successful count has been reached, please notify me in my profile comments.
Now, let's begin!
- 714 Replies
Well, might as well keep the ball rolling.
15. Previous number rolled - 1
15 + 6 = 21
Lets see what we get... 1d6=3
21 + 3 = 24
Correction! (finally, a post that i can correctly correct)
The number should be 18, because you subtract odd numbers. 21 - 3 = 18
18 - 5 = 13
13 - 3 = 10
Current number: 10
okie i'll do a roll.
if i understand correctly, 16-5 = 11 ?
Let's see if I remember how to do this...
Roll = 5
13-5 = 8
(8) let's join this game... also hi Sappy, it's been some time I don't see you in the AG forums 1d6=5
edit: reset :/
My bad, sorry... next try 1d6=5
7-5 = 2
2. So, how's everyone's weekend so far?
2 - 3 = -1
1. My weekend is good... played Minecraft, played AG games, had a German class... YAY! 1d6=5
2 - 5 = -3
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