'Lottery' odds resolution

[Anthony]

One theory of balance I've played around with is that your odds of success at a task should be equivalent to buying lottery tickets: if you have 10 points in a skill, and someone else has 4 points, your odds of winning are 10/14 (or scaling success the same way, success chance vs difficulty 4 is 10/14).

It would be technically possible to do this in a game (just toss colored beads in a bowl and pull one out), but seems like a hassle. Unfortunately, the alternatives I can think of are either messy or imprecise.

On a computer, I could check for random()%14 >= 4, but on a tabletop I don't have d14s. I could roll a d20 and reroll everything over 14 (to be a bit more generic, if skill and difficulty are both on a 1-10 scale, 1-(skill) is success, 11-(difficulty+10) is fail, reroll anything else) but that can be a bunch of rerolling.

I could roll a dice pool on each side; highest die wins. On a tie, reroll all tied dice, repeat until no ties.

You can approximate it by assigning exponential costs to skills; a 3d6 vs 3d6 quick contest works out to about x1.5 per point of skill difference. However, this winds up messy for adding abilities.

Any clever ideas out there?

[whswhs]

I don't think that quite works.

Suppose that I have 8 points in Games (Chess) and you have 2 point. By your approach, that would give me an 80% change of winning.

But suppose that I have IQ 10, and you have IQ 13. Now your skill is Games (Chess)-14, and mine is Games (Chess)-13. Your skill is higher than mine, and I ought to have less than a 50% chance of winning.

Now, here's an alternate approach. For a player with IQ 10, it takes 8 points to get to skill 13, and 12 points to get to skill 14. Use those to figure the odds, and you get a 60% chance of your winning. That seems like a fairer representation of the odds, and less likely to produce counterintuitive results.

[Anthony]

Originally Posted by whswhs

I don't think that quite works.

Suppose that I have 8 points in Games (Chess) and you have 2 point. By your approach, that would give me an 80% change of winning.

But suppose that I have IQ 10, and you have IQ 13. Now your skill is Games (Chess)-14, and mine is Games (Chess)-13. Your skill is higher than mine, and I ought to have less than a 50% chance of winning.

This is not in the GURPS section for a reason. I was assuming stats don't exist.

[khorboth]

It suffers from a bit of a problem of scale.

If I have a skill 5 and someone else a skill of 3, odds are significantly (5/8 0.625) in favor of me winning. But, if we are both more skilled but with the same difference at 15 and 13... then the odds are much more even (15/28 = 0.536) and it gets more serious as the numbers get larger with luck being MORE important with more skilled folk. This seems like the opposite of what one would want.

On the other hand, modifiers matter more for low-skilled folk, which does seem like what one would want.

As a practical matter, what would you use to resolve conflicts? Die rolls are fast and accepted. What would you use?

[Bengt]

Originally Posted by Anthony

I could roll a dice pool on each side; highest die wins. On a tie, reroll all tied dice, repeat until no ties.

Dice pool systems are usually counting the number of dice that hits a target number, so there will be less ties than if you just look at the highest die, especially if you roll a relatively large number compared to the number of sides.

[Anthony]

Originally Posted by khorboth

It suffers from a bit of a problem of scale.

If I have a skill 5 and someone else a skill of 3, odds are significantly (5/8 0.625) in favor of me winning. But, if we are both more skilled but with the same difference at 15 and 13... then the odds are much more even (15/28 = 0.536) and it gets more serious as the numbers get larger with luck being MORE important with more skilled folk. This seems like the opposite of what one would want.

The intent is based on points (of whatever currency the game uses), not skill level. I think it's perfect reasonable to think that there is a bigger relative difference between a person with five months experience and a person with three months experience than there is between a person with fifteen months and thirteen months.

Originally Posted by khorboth

As a practical matter, what would you use to resolve conflicts? Die rolls are fast and accepted. What would you use?

The point is to come up with a fast way to resolve that works for this model.

Originally Posted by Bengt

Dice pool systems are usually counting the number of dice that hits a target number, so there will be less ties than if you just look at the highest die, especially if you roll a relatively large number compared to the number of sides.

I'm aware of how dice pool systems normally work, but that doesn't mean I need to follow that.

[Anaraxes]

Dice go up by multiples of two for reasons of physical symmetry on the object. Coming up with a d7 or a d11 is possible (cylinders with N sides, for instance), but uncommon. But you can divide the common polyhedral dice by two and get a pretty solid range from 1-10. (The odd ones out are 7 and 9 -- for now, assume you do that with a d8, re-roll 8s, and d10, re-roll 10s.)

Which observation leads to the notion that the system will be better off with small numbers. If you measure skills on a single-digit point scale, you're better off than if you have double or triple digits worth of currency. Small numbers are easier in play, and a much easier match with this system than, say, a 1-100 system. (The tradeoff is of course granularity -- but how fine-grained does the question "am I better than you" really need to be?) So it's worth asking "how wide of a range do I actually need to cover", rather than "how do I solve this mathematical problem for any two integers".

Or since it's the 21st century, you could just go with dice-roller apps. New tools for a new system.

[ak_aramis]

Originally Posted by Bengt

Dice pool systems are usually counting the number of dice that hits a target number, so there will be less ties than if you just look at the highest die, especially if you roll a relatively large number compared to the number of sides.

Showing some of the supporting math...

with 3d6 getting 1S on 5 or 6, we can reduce this to a d3 marked 0,0,1...
this gives the following range space:
0 x y N=2
0 0 1
0 0 1
1 1 2
and
1 x y N=1
1 1 2
1 1 2
2 2 3

Collapsing this
4@0, 8@1, 5@2, 1@3. N=27. highest odds are 8/27; odds of a tie are
sum(n(x)²/N²) so (4²+8²+5²+1²)/27² = 16+64+25+1/ = 106/729 = 14.54

adding another die, for 4d of 5-6...
2@(4@0, 8@1, 5@2, 1@3)+1@(4@1, 8@2, 5@3, 1@4) & N=81
= 8@0, 20@1, 18@2, 7@3, 1@4 N=81
so (64+400+324+49+1)/6561= 838/6561=12.72%,

And 5d
2@(8@0, 20@1, 18@2, 7@3, 1@4) + 1@(8@1, 20@2, 18@3, 7@4, 1@5) N=243
=16@0, 48@1, 56@2, 32@3, 9@4, 1@5
Tie (16²+48²+56²+32²+9²+1²)/243² = 11.52%

Now, for XdYkh1...
iterations of A = (A^Y)-({A-1}^Y)
So, on 3d6kh1...
(1³)-({1-1}³)@1 (2³)-({2-1}³)@2 (3³-{3-1}³)@3 and so on...

doing the calcs via spreadsheet, I get the following ...

Dice:	1	2	3	4	5	6
A@1	1	1	1	1	1	1
A@2	1	3	7	15	31	63
A@3	1	5	19	65	211	665
A@4	1	7	37	175	781	3367
A@5	1	9	61	369	2101	11529
A@6	1	11	91	671	4651	31031
n	6	36	216	1296	7776	46656	279936	1679616
ties	16.67%	22.07%	29.54%	37.00%	44.16%	50.88%	57.10%	62.77%

so 5d6s5+ is 11.52% tie, while 5d6kh1 is 44.16% of a tie...

[Irish Wolf]

Originally Posted by ak_aramis

Showing some of the supporting math...

with 3d6 getting 1S on 5 or 6, we can reduce this to a d3 marked 0,0,1...
this gives the following range space:
0 x y N=2
0 0 1
0 0 1
1 1 2
and
1 x y N=1
1 1 2
1 1 2
2 2 3

Collapsing this
4@0, 8@1, 5@2, 1@3. N=27. highest odds are 8/27; odds of a tie are
sum(n(x)²/N²) so (4²+8²+5²+1²)/27² = 16+64+25+1/ = 106/729 = 14.54

adding another die, for 4d of 5-6...
2@(4@0, 8@1, 5@2, 1@3)+1@(4@1, 8@2, 5@3, 1@4) & N=81
= 8@0, 20@1, 18@2, 7@3, 1@4 N=81
so (64+400+324+49+1)/6561= 838/6561=12.72%,

And 5d
2@(8@0, 20@1, 18@2, 7@3, 1@4) + 1@(8@1, 20@2, 18@3, 7@4, 1@5) N=243
=16@0, 48@1, 56@2, 32@3, 9@4, 1@5
Tie (16²+48²+56²+32²+9²+1²)/243² = 11.52%

Now, for XdYkh1...
iterations of A = (A^Y)-({A-1}^Y)
So, on 3d6kh1...
(1³)-({1-1}³)@1 (2³)-({2-1}³)@2 (3³-{3-1}³)@3 and so on...

doing the calcs via spreadsheet, I get the following ...

Dice:	1	2	3	4	5	6
A@1	1	1	1	1	1	1
A@2	1	3	7	15	31	63
A@3	1	5	19	65	211	665
A@4	1	7	37	175	781	3367
A@5	1	9	61	369	2101	11529
A@6	1	11	91	671	4651	31031
n	6	36	216	1296	7776	46656	279936	1679616
ties	16.67%	22.07%	29.54%	37.00%	44.16%	50.88%	57.10%	62.77%

so 5d6s5+ is 11.52% tie, while 5d6kh1 is 44.16% of a tie...

Is this going to become like that episode of Futurama where in order to resolve a major plot point, the writers wound up proving a new mathematical theorem?

[ak_aramis]

Originally Posted by Irish Wolf

Is this going to become like that episode of Futurama where in order to resolve a major plot point, the writers wound up proving a new mathematical theorem?

I wouldn't know. I hated the few episodes of Futurama I've seen...