Dice Analyzer

[Aradhus]

I have some queries about the dice analyzer.

I will post one question at a time, and hopefully I get some rational explanations.

First question:
Ideal stats for 1v1 are 41.67% to win and 58.33% to lose.

I calculated different percentages, using very basic math so I assume I must be wrong.

Can someone explain how these ideal stats are worked out and if they are correct?

(I assume ideal stats are the same for everbody, if that's not the case then its buggy?)

[daydream]

to put this really simply, heres all possible rolls in a 1vs1:

Attacker - Defender
1-1
1-2
1-3
1-4
1-5
1-6
2-1
2-2
2-3
2-4
2-5
2-6
3-1
3-2
3-3
3-4
3-5
3-6
4-1
4-2
4-3
4-4
4-5
4-6
5-1
5-2
5-3
5-4
5-5
5-6
6-1
6-2
6-3
6-4
6-5
6-6

green: attacker wins -> 15 possibilities
red: defender wins -> 21 possibilities

15/36= 41.67%
21/36= 58.33%

[Aradhus]

I know that, I wanted the math, cos its been a long time since I've did anything math related, but I think I figured it out.

[daydream]

I know that, I wanted the math, cos its been a long time since I've did anything math related, but I think I figured it out.

i gave you the math

[Aradhus]

No you didn't. I knew the 15 and 21 out of 36. I wanted the math explaining how to turn that into percentages. Which I figured out correctly, once I was no longer stoned.

[Fruitcake]

No you didn't. I knew the 15 and 21 out of 36. I wanted the math explaining how to turn that into percentages. Which I figured out correctly, once I was no longer stoned.

You surprise me Aradhus. Did you not have the old equation.... 'first number times 100 divided by the total' repeated constantly to you at school?

[Aradhus]

I'm not sure I remember a single lesson from school. What you've bolded does not ring a bell, at all.

Edit: Ironically, i was a child math whizz, perfect scores in every test, head of my class, in which I actually did the work that the kids in the year above were doing. Then I turned 14, my hormones kicked into gear and it all went to shit, lol.

Sometimes I think about finding a course to relearn high school math(and beyond - mibbes), and reacquaint myself with my love of numbers.

[Fruitcake]

I'm not sure I remember a single lesson from school. What you've bolded does not ring a bell, at all.

Well now you have it. 2009 has only just started and already you have gained knowledge!

[Alibeg]

if you need a calculation i can say something...but you said that you have figured it out, anyway ill say it:
for every dice your opponent has, there is appropriate percentage you have to get to beat him, then you sum up all those percentages and there you go...
so you have opponent who has:
1 - you need any dice higher than 1 and there are 5 out of 6 dices that can suit you so chances are 5/6, but this is only percentage of winning when opponent has 1, chances that he gets one are 1/6, so you multiply those and get 5/36
2 - you must get 3, 4, 5 or 6 so chances are 4/6 multiplied with 1/6 (chances of getting 2 in his case) and the result is 4/36
3 - same thing and you get 3/6 * 1/6 = 3/36
4 - 2/6 * 1/6 = 2/36
5 - 1/6 * 1/6 = 1/36
6 - 0/36 (you have no chance of winning if he has 6)
and now goes summing up
5/36 + 4/36 + 3/36 + 2/36 + 1/36 + 0/36 = 15/36 = 0.41666666666 or if you like 41.6666666666% or if you like it this way 41.67%
so chances to win are 41.67% and the rest are chances of losing 58.33%.
This seems to be longer way to compute these chances but if you use this method to calculate 2 vs 2 you'll see that it is easier than to write all possible combinations...
cheers everybody