Dice Game for the Horny Girls.....and Boys...

[ballsfull]

Why don't you fill people in the upper slots? Say just make him slot 8 instead of yourself and that's it. If someone else turns out, make him 9th. After all, it's their fault for not being fast enough, and it's not like upper slots are so much worse.

Or do you treasure your advantage that much Greeeedy, gtreeeedy!!!

I thought about it, but as is well pointed out, the higher numbers are going to suck pretty quickly....
If you wind-up working your way there, then yes, that's just your bad luck....
And I'd actually feel bad putting someone up there from the get-go.... They'd have bad odds initially, and it'll only get worse....

At least the players now have odds that get "better" over time, before they start getting worse....

[les]

Last Cum Feb17th
Les

[Nezhul]

Take a look at this table. It's the number of combinations of 2 dies for each number, total 21 combinations are avaliable.

1 - 0 (can not be achieved)
2 - 1
3 - 1
4 - 2
5 - 2
6 - 3
7 - 3
8 - 3
9 - 2
10 - 2
11 - 1
12 - 1

As you can probably see, 6-7-8 are the best numbers with the most chances to pop up, that is ~14.285% each.
4-5-9-10 has equal chances of popping up too, tho they are a bit lower ~9.5%
2-3-11-12 has yet again equal chances, about 4.7%

So as you can see, the chances above 7 are not worse than chances BELOW 7. The only difference is the number of edges, and that lesser numbers will be shifted upwards more often.

As you can see tho the difference between chances are pretty severe (3 times), so it's really not that cool. There's a d10 die, if you are interested in my suggestion, wich provide equal chances for all numbers.

[les]

Take a look at this table. It's the number of combinations of 2 dies for each number, total 21 combinations are available.

1 - 0 (can not be achieved)
2 - 1
3 - 1
4 - 2
5 - 2
6 - 3
7 - 3
8 - 3
9 - 2
10 - 2
11 - 1
12 - 1

As you can probably see, 6-7-8 are the best numbers with the most chances to pop up, that is ~14.285% each.
4-5-9-10 has equal chances of popping up too, tho they are a bit lower ~9.5%
2-3-11-12 has yet again equal chances, about 4.7%

So as you can see, the chances above 7 are not worse than chances BELOW 7. The only difference is the number of edges, and that lesser numbers will be shifted upwards more often.

As you can see tho the difference between chances are pretty severe (3 times), so it's really not that cool. There's a d10 die, if you are interested in my suggestion, wich provide equal chances for all numbers.

there are in fact 36 combinations as shown in one of my previous posts,the reasoning is as follows dice a may show any one of six numbers dice b may also show any one of six numbers.
therefore 6 times 6 is 36.
a simpler proof is to use two die of different colours.

repeated here.



Les

[RoniDev]

just use "one" 12 sided dice. It isnt hard....

[ballsfull]

Ummmm....you're right.... It's not hard at all....
And that would be ~another~ perfectly valid way to do this also....

I'm just not doing it that way in this particular game.....

[les]

Ummmm....you're right.... It's not hard at all....
And that would be ~another~ perfectly valid way to do this also....

I'm just not doing it that way in this particular game.....

ballsful has taken up the challenge let him do it his way chaps and i am sure if it doesn't work as he wants then he can adapt it.

Les

[Nezhul]

there are in fact 36 combinations as shown in one of my previous posts

36, is when the order of the dies matter in any way. i.e. when 5&2 and 2&5 combinations are considered different. In out case that doesn't matter, because only that matters is the sum.

All possible combinations are:
1-1,1-2,1-3,1-4,1-5,1-6,2-2,2-3,2-4,2-5,2-6,3-3,3-4,3-5,3-6,4-4,4-5,4-6,5-5,5-6,6-6
Total 21.

[les]

obviously English University education is flawed.

Les

[Nezhul]

It's just that I'm a programmer (a lot of math) and we were told "Probability and statistics" among all others.

And yes, I think there is belief that Russian education is more... full. That's not that good though, if you ask me, because sometimes you get the feeling you spend just too much time on stuff that you would never ever need again, and go too much in depth of some matters. Math is not among those, tho. Math is good and needed for me.

[Telly]

obviously English University education is flawed.

Les

Obviously, so is German education, as I would have said exactly as you did.

By the way, if you use Nezhul's formula for 2 out of 6, you get 15 (6! = 720; 2! = 2; 4! = 24
720/2/24=15). The reason for this is that this formula applies for any combination of two different numbers, excluding 1-1, 2-2, 3-3, 4-4, 5-5 and 6-6

[Nezhul]

oh yes, I totally forgot about that.

btw wasnt it easyer not to calculate big factorials, but just do like that:
6! / (2!*4!) = 5*6 / 2! = 15?

[Telly]

You're right, that would have been easier, as the fraction can be reduced by 4!

[les]

So with the corrected details

Take another look at Nezhul's table. It's the number of combinations of 2 dies for each number, total 15 combinations are available.

1 - 0 (can not be achieved)
2 - 0 (can not be achieved)??
3 - 1
4 - 1
5 - 2
6 - 2
7 - 3
8 - 2
9 - 2
10 - 1
11 - 1
12 - 0 (can not be achieved)??


Silly
Les

[Nezhul]

no, why? 2 and 12 can be achieved. You can't achieve 1 because each die will roll at least 1, the sum will be at least two (unless, off course, under some misterious circumstances one of the dies will stay floating in the air)

It's just that I forgor that on this formula the combinations of 2 similar numbers can not be achieved. In reality they can, and we just make it 15+6=21

my table above showed how may combinations can be there to get that sum. thus, if you want

1 - can not be achieved
2 - is only 1+1
3 - is only 1+2
4 - is eather 3+1 or 2+2
5 - is eather 4+1 or 3+2
6 - is eather 5+1 or 4+2 or 3+3
7 - is eather 6+1 or 5+2 or 4+3
and so on, and it degrades because
10 is 5+5 or 6+4 - there's no 3rd combination for 10.
11 is ONLY 5+6.