Percentile probability question...

[Daniel Solis]

Say the task resolution is a percentile roll vs. another person's percentile roll, whoever is higher wins. Now consider that it is possible to roll several d10s and only take the highest two results as the "tens" and "ones," respectively.

For example, I roll three dice. The results are 4, 6, 2. "6" is the highest result, "4" is the second highest. Together, they become a result of "64."

I'm terrible with numbers, but I'm very curious. Can someone with a mathematical inclination figure out how much each extra die increases the possible result?

[DevP]

First off I'd say forget the ones-digit, at least because 20% is close to 25%, and because I have no idea how to do that...

Otherwise, I think the equation you want is something like:
.9*(chance of rolling any nines) + .8 (chance of rolling any eights and no nines) + .7 (chance of rolling any sevens, given no eights and nines) and so on, and I think that might give you some semblance for an average. This is where the binomial probabilty equation would kick in, but it's a bit messy.

Hope this is a good hint.

 - dev in a hurry

[Brian Leybourne]

Even with only 2 dice, just the simple fact that you get to choose which one is the 10's and which one is the 1's changes the curve hugely (and many results become impossible - no number less than 10 is possible, for example, and after 11 the next lowest number is 20, then 21, 22, 30...)

If I get time tomorrow I'll sit down and work out the actual figures for you.

Brian.

[John Kim]

Well, here's the results.  I am assuming that 00 is low (zero rather than one hundred).  It produces a somewhat odd result otherwise -- i.e. one zero is bad, but two zeroes are the best.  

                   Number of Dice
              2       3       4       5
           ------  ------  ------  ------
00 - 09 :   1.00%   0.10%   0.01%   0.00% 
10 - 19 :   3.00%   0.70%   0.15%   0.03% 
20 - 29 :   5.00%   1.90%   0.65%   0.21% 
30 - 39 :   7.00%   3.70%   1.75%   0.78% 
40 - 49 :   9.00%   6.10%   3.69%   2.10% 
50 - 59 :  11.00%   9.10%   6.71%   4.65% 
60 - 69 :  13.00%  12.70%  11.05%   9.03% 
70 - 79 :  15.00%  16.90%  16.95%  15.96% 
80 - 89 :  17.00%  21.70%  24.65%  26.28% 
90 - 99 :  19.00%  27.10%  34.39%  40.95% 

Average =  64.35   74.25   80.17   84.08

[MachMoth]

Well, a similar (though not exactly the same) way of looking at is saying, roll Xd10.  If one player's highest is higher than the other's highest, he wins.  If there's a tie, go to the next highest.

[Ron Edwards]

Hello,

And that would be the Sorcerer system, in that case.

Especially when a person recognizes that any single die is actually a percentile die, differing only in the size of its increments. A 10-sider is a d100 divided into 10% units. A d4 is a d100 divided into 25% units. And so on.

Best,
Ron

[timfire]

Given that its one person's roll vs another person's roll (not against a target number), also given that there are no modifers involved and that both players are using the same number of dice, then it shouldn't matter if the players can pick their dice or not, each player should have a 50% chance of beating he other.

Now, if there are modifers involved, or if the players are using a different number of dice...sorry, its a bit too early in the morning for me to do that math.

[Ron Edwards]

Hi Tim,

Right! Equal numbers of dice (of the same type) against one another means 50%.

No target numbers at all. Degree of success determined by number of dice which are higher than the loser's highest value (after ties).

Therefore the only modulation of better/worse odds come through changing the number of dice in the rolled pools.

All modifiers are expressed only through adding and subtracting dice prior to the roll.

Sorcerer in a nutshell.

Best,
Ron