Can’t strategy: some hints
Can’t Stop is basically a “risk management” game. Once you understand the game mechanic, all that you need
to do is learn when to go and when to stop. I know different players have different strategies and you’re strategy
must be flexible enough to take into account the board position and the opponents you are up against.
Hopefully we can get a nice discussion of strategies.
I’ll start off by giving a few general hints and tips.
1) It is usually best to hold off on placing your 2nd and 3rd markers for as a long as possible.
There are times your roll gives you the best possible location for your 3rd marker, and in that case I
would place it.
2) Try and avoid having all 3 markers higher than the 6 or lower than the 8. If you have only high or only low
markers, you have much less flexibility.
3) It is better to have all even markers such as 4,6,8 than all odd markers such as 5,7,9.
The reason for this is because to make an odd number , you must have both an even and an odd die roll.
Whereas, to make an even number, you could have 2 evens or 2 odds..
4) This isn’t so much a hint and tip, but a source of information. I created a puppet named Scarne
Scarne lives in room C10-13 and you can get odds information from him. If you tell Scarne what your 3
numbers are, he will tell you the odds of a sucessful roll. Just use the command.
If your 3 numbers were 2 7 8 then if you are in C10-13, just type
@rolls 2 7 8
if you aren’t in that room, then use a /tell command
/tell Scarne @rolls 2 7 8
Boomer
April 4th, 2005 at 5:06 am
Tip 3 is wrong. There are 4 possibilities with two dice. odd-odd, odd-even, even-odd, and even-even. It is just as easy to make an odd number as an even number.
The probabilities of each roll with two dice are:
p(2)=1/36, p(3)=2/36, p(4)=3/36, p(5)=4/36, p(6)=5/36, p(7)=6/36, p(8)=5/36. p(9)=4/36, p(10)=3/36, p(11)=2/36, p(12)=1/36
The probability of getting an odd number with two dice is then p(3)+p(5)+p(7)+p(9)+p(11),
which is 2/36+4/36+6/36+4/36+2/36 = 18/36 = 1/2
The situation with four dice is more complex, but the result is the same.
April 4th, 2005 at 5:16 am
Tip three is wrong. Even and odd are equally likely for the sum of any number of six-sided dice. Imagine you have rolled all the dice but one. The even-odd outcome depends entirely on the last die, matching or not matching the evenness of the sum so far. Since there are three even numbers and three odd numbers on the last die, the probibility of the sum being even or odd is the same, 1/2 each.
January 2nd, 2006 at 12:11 am
when one plays “can’t stop” on BSW one doesn’t actually roll dice. One uses a random number generator (whose period is even). Therefore even and odd numbers cannot be equally likely. The boomerknowing the exact algorithm, knows which is more likely and therefore I would assume his tip is correct
April 4th, 2006 at 6:01 am
No, tip three is correct. The point is that you get to divide the four dice into pairs any way you want. You WILL get two odds or two evens. You may or may not get an odd and an even.
4,7,10 is better than 5,7,9.