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Anyone try Fudge "wild dice"?
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 I constantly seem to be coming up with new ways to tweak the FUdge system. I can't help it. Maybe that's because the system really encourages it, but there are just so many cool and unique things that oen can do with it.
Anyway, after our Psipunk playtest tonight one of the guys in the group was talking up an upcoming Hellfrost (Savage Worlds) game that he's going to start pretty soon. For no particular reason that got me thinking about possible Fudge "wild dice" mechanics.
Here's what I was thinking:
Roll the standard 4dF, but designate one of them as a unique "wild dice". The easiest way to do this would be to make the wild dice a different color (black instead of white, red instead of blue, whatever).
If the wild dice rolls a +, give the player the *option* of rerolling it and adding the result. There would be a 1/3 chance of it hinering him (a minus), a 1/3 chance of doing nothing (blank) and a 1/3 chance of adding +1.
This dice could keep "exploding" as long as it continued rolling plusses, but due to the nature of Fudge dice I don't think the percentage chance of getting absurdly high results would be very high (though I lack the statistical knowhow to really calculate that out).
Has anyone else tried something like this to positive effect?
I do like the idea, but I don't think it would fit into Psipunk at this stage so I'd have to reserve it, I think, for another day. Still, I just can't help but throw these ideas out there and see what sticks!  This is an interesting idea. I wish I could add more except that I might
try this one day.
~Jonathan S.On Jul 3, 2012 2:08 AM, "munkwunk" <munkwunk@...> wrote:
> **
>
>
> I constantly seem to be coming up with new ways to tweak the FUdge system.
> I can't help it. Maybe that's because the system really encourages it, but
> there are just so many cool and unique things that oen can do with it.
>
> Anyway, after our Psipunk playtest tonight one of the guys in the group
> was talking up an upcoming Hellfrost (Savage Worlds) game that he's going
> to start pretty soon. For no particular reason that got me thinking about
> possible Fudge "wild dice" mechanics.
>
> Here's what I was thinking:
>
> Roll the standard 4dF, but designate one of them as a unique "wild dice".
> The easiest way to do this would be to make the wild dice a different color
> (black instead of white, red instead of blue, whatever).
>
> If the wild dice rolls a +, give the player the *option* of rerolling it
> and adding the result. There would be a 1/3 chance of it hinering him (a
> minus), a 1/3 chance of doing nothing (blank) and a 1/3 chance of adding +1.
>
> This dice could keep "exploding" as long as it continued rolling plusses,
> but due to the nature of Fudge dice I don't think the percentage chance of
> getting absurdly high results would be very high (though I lack the
> statistical knowhow to really calculate that out).
>
> Has anyone else tried something like this to positive effect?
>
> I do like the idea, but I don't think it would fit into Psipunk at this
> stage so I'd have to reserve it, I think, for another day. Still, I just
> can't help but throw these ideas out there and see what sticks!
>
>
>
[Nontext portions of this message have been removed]  I've considered doing almost the same thing, munkwumk, but like you, my
games have tended more towards gritty than wild.
The old DC Heroes game, back in the late 80s, had a similar system. You
rolled 2d10 and if you rolled doubles, you could reroll and add the result
to your original roll, to get a better result. It was cumulative, so if
you continued rolling doubles you could keep rolling. If you rolled two
ones, however, it was an automatic failure, regardless of if you rolled it
on your first roll or your tenth.
Neat idea!
Robb
On Tue, Jul 3, 2012 at 9:51 AM, Jonathan Snyder <morg223@...> wrote:
> This is an interesting idea. I wish I could add more except that I might
> try this one day.
>
> ~Jonathan S.
> On Jul 3, 2012 2:08 AM, "munkwunk" <munkwunk@...> wrote:
>
> > **
> >
> >
> > I constantly seem to be coming up with new ways to tweak the FUdge
> system.
> > I can't help it. Maybe that's because the system really encourages it,
> but
> > there are just so many cool and unique things that oen can do with it.
> >
> > Anyway, after our Psipunk playtest tonight one of the guys in the group
> > was talking up an upcoming Hellfrost (Savage Worlds) game that he's going
> > to start pretty soon. For no particular reason that got me thinking about
> > possible Fudge "wild dice" mechanics.
> >
> > Here's what I was thinking:
> >
> > Roll the standard 4dF, but designate one of them as a unique "wild dice".
> > The easiest way to do this would be to make the wild dice a different
> color
> > (black instead of white, red instead of blue, whatever).
> >
> > If the wild dice rolls a +, give the player the *option* of rerolling it
> > and adding the result. There would be a 1/3 chance of it hinering him (a
> > minus), a 1/3 chance of doing nothing (blank) and a 1/3 chance of adding
> +1.
> >
> > This dice could keep "exploding" as long as it continued rolling plusses,
> > but due to the nature of Fudge dice I don't think the percentage chance
> of
> > getting absurdly high results would be very high (though I lack the
> > statistical knowhow to really calculate that out).
> >
> > Has anyone else tried something like this to positive effect?
> >
> > I do like the idea, but I don't think it would fit into Psipunk at this
> > stage so I'd have to reserve it, I think, for another day. Still, I just
> > can't help but throw these ideas out there and see what sticks!
> >
> >
> >
>
>
> [Nontext portions of this message have been removed]
>
>
>
> 
>
> > Please visit the Fudge Community Yahoo Group home page at
> http://games.groups.yahoo.com/group/fudgecommunity to search the
> archives, upload or download files, and more! Note: No attachments are
> allowed on the fudgecommunity@yahoogroups.com email list, please upload
> any attachments to the Files section instead. Thank you! <Yahoo! Groups
> Links
>
>
>
>
[Nontext portions of this message have been removed] >Roll the standard 4dF, but designate one of them as a unique "wild dice".
I tried it when the old d6 from West End Games came out. I had three white die and 1 black die, but instead of exploding up and out, the wild die caused trouble (a bit like the old Ghostbusters game with the 1 ghost die  Fudge Ghostbusters is still out there, I think). This same concept held up when Grey Ghost Games came out with the "flawed" Olympia dice.
>[snip] make the wild dice a different color
>
And all that reminds me of Lord of the Dice from Fudge Factor:
www.fudgefactor.org/2003/04/01/lord_of_the_dice.html
(currently offline???)
And Patrick Benson's Power Dice/Vampire Dice as a different take, for grittier results:
http://www.sinisterforces.com/2008/12/18/randomizingresultshowitinfluencestherestofthegame/
(also missing???)
(Still enjoying your PsiPunk playtest, just haven't had the chance to post in a while.)
J.Tim I tried something similar, but not as risky. The Wild die reroll was only counted if it was positive  otherwise, it was ignored. So you couldn't reduce the earlier result.
It was okay, gave a little extra thrill, but ultimately I dislike multiple rolls so I dropped it.
 In fudgecommunity@yahoogroups.com, "munkwunk" <munkwunk@...> wrote:
>
> I constantly seem to be coming up with new ways to tweak the FUdge system. I can't help it. Maybe that's because the system really encourages it, but there are just so many cool and unique things that oen can do with it.
>
> Anyway, after our Psipunk playtest tonight one of the guys in the group was talking up an upcoming Hellfrost (Savage Worlds) game that he's going to start pretty soon. For no particular reason that got me thinking about possible Fudge "wild dice" mechanics.
>
> Here's what I was thinking:
>
> Roll the standard 4dF, but designate one of them as a unique "wild dice". The easiest way to do this would be to make the wild dice a different color (black instead of white, red instead of blue, whatever).
>
> If the wild dice rolls a +, give the player the *option* of rerolling it and adding the result. There would be a 1/3 chance of it hinering him (a minus), a 1/3 chance of doing nothing (blank) and a 1/3 chance of adding +1.
>
> This dice could keep "exploding" as long as it continued rolling plusses, but due to the nature of Fudge dice I don't think the percentage chance of getting absurdly high results would be very high (though I lack the statistical knowhow to really calculate that out).
>
> Has anyone else tried something like this to positive effect?
>
> I do like the idea, but I don't think it would fit into Psipunk at this stage so I'd have to reserve it, I think, for another day. Still, I just can't help but throw these ideas out there and see what sticks!
>  If you roll 3 normal fudge dice and one wild fudge die that keeps
exploding as long as you roll a plus, then the resulting probability
distribution isn't that much different from an ordinary 4dF roll. For
the purposes of making these calculations, I'm assuming the player
doesn't have a choice about when to stop rolling the wild die, but keeps
rolling until either a 1 or 0 appears. (I can't calculate when a player
would choose to stop rollingchoice isn't chance... [:)] )
A normal 4dF roll has a mean of 0 and a standard deviation of
2*sqrt(2/3) = 1.63299.
The wild dice roll also has a mean of 0, but a standard deviation of
sqrt(3) = 1.73205.
Here are the probabilities for each distribution from 4 to 10.
x 4dF 3dF + Wild dF
4 0.0123457 0.0123457
3 0.0493827 0.0534979
2 0.123457 0.128944
1 0.197531 0.203475
0 0.234568 0.228319
1 0.197531 0.187217
2 0.123457 0.111789
3 0.0493827 0.0496085
4 0.0123457 0.0165362
5 0. 0.00551206
6 0. 0.00183735
7 0. 0.000612451
8 0. 0.00020415
9 0. 0.0000680501
10 0. 0.0000226834
If you're interested, there is an exact formula for the probability mass
function f of this fudge wild dice roll. Let w be the conditional
probability mass function of an exploding dF given that the first roll
is a plus, namely w(x) = 0 if x < 0, 1/3 if x = 0, and 4/(3^(x+1)) if x> 0. The rationale for this is as follows: Given the first roll is 1,
there is no way for x to be less than 0. There is exactly one way for x
to be 0, namely roll a 1 on the second roll. This has probability 1/3.
There are two ways for x to be 1: roll a 0 on the second roll; or roll
another 1 followed by 1; these have probabilities 1/3 and 1/9,
respectively; and 1/3+1/9=4/9. This pattern continues for x larger than
1, namely the probability is 4/(3^(x+1)).
Let v be the probability mass function for one ordinary dF. Let * be the
convolution operator. Let u be the probability mass function for
ordinary 3dF. Then u(x) = (v*v*v)(x).
Then the probability mass function f of the fudge wild dice roll is the
following mixture distribution:
f(x) = (1/3) u(x+1) + (1/3) u(x) + (1/3) (w*u)(x)
The intuition here is that on the first roll, the wild die can be either
1, 0 or 1 with equal probabilities 1/3rd each. This gives rise to the
three terms in the mixture distribution with weights 1/3rd each. The
first term is 3dF shifted left, corresponding to a roll of 1 on the
first (and only) roll of the wild die. The second term corresponds to a
0 on the (nonexploding) wild die. And the third term corresponds to an
exploding roll of 1 on the first roll of the wild die.
I coded this formula in Mathematica to compute the probabilities.
Then I doublechecked this formula by writing a Monte Carlo simulation
in Perl, simulating one million random rolls. This simulation matched
the theoretical values given above:
sub dF {
int(rand(3))  1;
}
$count = 1_000_000;
%prob = ();
sub getWild {
my ($Wild) = @_;
my $w = dF();
if($w == 1){
$Wild = getWild($Wild+1);
} else {
$Wild += $w;
}
}
sub getSum {
my ($x,$y,$z) = (dF(),dF(),dF());
$x+$y+$z+getWild(0);
}
sub printDistrib {
my ($prob) = @_;
my ($x,$p,$sx,$sx2,$v,$cdf);
for $x (sort {$a <=> $b} (keys %$prob)) {
$p=$$prob{$x}/$count;
$cdf+=$p;
$sx+=$p*$x;
$sx2+=$p*$x*$x;
print "P[X=",$x,"]=",$p,", P[X<=",$x,"]=",$cdf,"\n";
}
$v=$sx2($sx*$sx);
print "Sum P[X]=",$cdf,"\n";
print "E[X]=",$sx,"\n";
print "Var[X]=",$v,"\n";
print "sd[X]=",sqrt($v),"\n";
}
for $_ (1..$count) {
$prob{getSum()}++;
}
printDistrib(\%prob);
[Nontext portions of this message have been removed]  To your point, the Wild die adds very little skew to the outcome of the
roll.
On the other hand, it adds to the excitement of the test, and the fear of
the opponent. There is ALWAYS a chance that anyone, can do anything
at any time. That one point orc, could slay the invincible knight if it
was lucky!!
Geek content:
I think your chart is a bit off.
Rolls the yield 3 Rolls that yield +3
1 1 1 [0 ] 1 1 1 [0 ]
1 1 1 [1 1] 1 1 1 [1 1 ]
1 1 0 [1 ] 1 1 0 [1 1 1]
1 0 1 [1 ] 1 0 1 [1 1 1]
0 1 1 [1 ] 0 1 1 [1 1 1]
1 0 0 [1 1 0] *3
1 0 0 [1 1 1 1] *3
0 0 0 [1 1 1 0] *3
0 0 0 [1 1 1 1 1] *3
So the chances of rolling a 3 should be less than the chances of
rolling a +3.
Your perl script consistently shows this, but the chart doesn't. I'm not
that great at statistics, so I don't understand exactly why the math
doesn't work. But as stated above, the differences are less than 1% almost
negligible.
Mike Greene
On Wed, Jul 11, 2012 at 12:49 AM, Aher <elisha_ben_abuyah@...> wrote:
> **
>
>
> If you roll 3 normal fudge dice and one wild fudge die that keeps
> exploding as long as you roll a plus, then the resulting probability
> distribution isn't that much different from an ordinary 4dF roll. For
> the purposes of making these calculations, I'm assuming the player
> doesn't have a choice about when to stop rolling the wild die, but keeps
> rolling until either a 1 or 0 appears. (I can't calculate when a player
> would choose to stop rollingchoice isn't chance... [:)] )
>
> A normal 4dF roll has a mean of 0 and a standard deviation of
> 2*sqrt(2/3) = 1.63299.
>
> The wild dice roll also has a mean of 0, but a standard deviation of
> sqrt(3) = 1.73205.
>
> Here are the probabilities for each distribution from 4 to 10.
>
> x 4dF 3dF + Wild dF
> 4 0.0123457 0.0123457
> 3 0.0493827 0.0534979
> 2 0.123457 0.128944
> 1 0.197531 0.203475
> 0 0.234568 0.228319
> 1 0.197531 0.187217
> 2 0.123457 0.111789
> 3 0.0493827 0.0496085
> 4 0.0123457 0.0165362
> 5 0. 0.00551206
> 6 0. 0.00183735
> 7 0. 0.000612451
> 8 0. 0.00020415
> 9 0. 0.0000680501
> 10 0. 0.0000226834
>
> If you're interested, there is an exact formula for the probability mass
> function f of this fudge wild dice roll. Let w be the conditional
> probability mass function of an exploding dF given that the first roll
> is a plus, namely w(x) = 0 if x < 0, 1/3 if x = 0, and 4/(3^(x+1)) if x
> > 0. The rationale for this is as follows: Given the first roll is 1,
> there is no way for x to be less than 0. There is exactly one way for x
> to be 0, namely roll a 1 on the second roll. This has probability 1/3.
> There are two ways for x to be 1: roll a 0 on the second roll; or roll
> another 1 followed by 1; these have probabilities 1/3 and 1/9,
> respectively; and 1/3+1/9=4/9. This pattern continues for x larger than
> 1, namely the probability is 4/(3^(x+1)).
>
> Let v be the probability mass function for one ordinary dF. Let * be the
> convolution operator. Let u be the probability mass function for
> ordinary 3dF. Then u(x) = (v*v*v)(x).
>
> Then the probability mass function f of the fudge wild dice roll is the
> following mixture distribution:
>
> f(x) = (1/3) u(x+1) + (1/3) u(x) + (1/3) (w*u)(x)
>
> The intuition here is that on the first roll, the wild die can be either
> 1, 0 or 1 with equal probabilities 1/3rd each. This gives rise to the
> three terms in the mixture distribution with weights 1/3rd each. The
> first term is 3dF shifted left, corresponding to a roll of 1 on the
> first (and only) roll of the wild die. The second term corresponds to a
> 0 on the (nonexploding) wild die. And the third term corresponds to an
> exploding roll of 1 on the first roll of the wild die.
>
> I coded this formula in Mathematica to compute the probabilities.
>
> Then I doublechecked this formula by writing a Monte Carlo simulation
> in Perl, simulating one million random rolls. This simulation matched
> the theoretical values given above:
>
> sub dF {
> int(rand(3))  1;
> }
>
> $count = 1_000_000;
>
> %prob = ();
>
> sub getWild {
> my ($Wild) = @_;
> my $w = dF();
> if($w == 1){
> $Wild = getWild($Wild+1);
> } else {
> $Wild += $w;
> }
> }
>
> sub getSum {
> my ($x,$y,$z) = (dF(),dF(),dF());
> $x+$y+$z+getWild(0);
> }
>
> sub printDistrib {
> my ($prob) = @_;
> my ($x,$p,$sx,$sx2,$v,$cdf);
> for $x (sort {$a <=> $b} (keys %$prob)) {
> $p=$$prob{$x}/$count;
> $cdf+=$p;
> $sx+=$p*$x;
> $sx2+=$p*$x*$x;
> print "P[X=",$x,"]=",$p,", P[X<=",$x,"]=",$cdf,"\n";
> }
> $v=$sx2($sx*$sx);
> print "Sum P[X]=",$cdf,"\n";
> print "E[X]=",$sx,"\n";
> print "Var[X]=",$v,"\n";
> print "sd[X]=",sqrt($v),"\n";
> }
>
> for $_ (1..$count) {
> $prob{getSum()}++;
> }
>
> printDistrib(\%prob);
>
>
> [Nontext portions of this message have been removed]
>
>
>
[Nontext portions of this message have been removed]  Since I play with "the player always rolls" (so "being hit" is caused
by the PC rolling low, not the NPC rolling high ... becuase the NPC
never rolls against a PC) and the GM only rolls for NPC vs NPC events,
the currently presentation doesn't include the fear aspect you
mention. Only the PC can get the high result, so the PC has nothing
to fear.
The way to fix that is: let the wild die also "explode downward" (if
it rolls a , keep rerolling it as long as it continues to roll 's).
Anyone want to quickly work up the probability for a wild die that
can explode in both directions?
On Wed, Jul 11, 2012 at 4:27 AM, Mike Greene <mikedgreene@...> wrote:
> To your point, the Wild die adds very little skew to the outcome of the
> roll.
>
> On the other hand, it adds to the excitement of the test, and the fear of
> the opponent. There is ALWAYS a chance that anyone, can do anything
> at any time. That one point orc, could slay the invincible knight if it
> was lucky!!
>
> Geek content:
>
> I think your chart is a bit off.
>
> Rolls the yield 3 Rolls that yield +3
> 1 1 1 [0 ] 1 1 1 [0 ]
> 1 1 1 [1 1] 1 1 1 [1 1 ]
> 1 1 0 [1 ] 1 1 0 [1 1 1]
> 1 0 1 [1 ] 1 0 1 [1 1 1]
> 0 1 1 [1 ] 0 1 1 [1 1 1]
> 1 0 0 [1 1 0] *3
> 1 0 0 [1 1 1 1] *3
> 0 0 0 [1 1 1 0] *3
> 0 0 0 [1 1 1 1 1] *3
>
> So the chances of rolling a 3 should be less than the chances of
> rolling a +3.
> Your perl script consistently shows this, but the chart doesn't. I'm not
> that great at statistics, so I don't understand exactly why the math
> doesn't work. But as stated above, the differences are less than 1% almost
> negligible.
>
> Mike Greene
>
>
>
>
> On Wed, Jul 11, 2012 at 12:49 AM, Aher <elisha_ben_abuyah@...> wrote:
>
>> **
>>
>>
>> If you roll 3 normal fudge dice and one wild fudge die that keeps
>> exploding as long as you roll a plus, then the resulting probability
>> distribution isn't that much different from an ordinary 4dF roll. For
>> the purposes of making these calculations, I'm assuming the player
>> doesn't have a choice about when to stop rolling the wild die, but keeps
>> rolling until either a 1 or 0 appears. (I can't calculate when a player
>> would choose to stop rollingchoice isn't chance... [:)] )
>>
>> A normal 4dF roll has a mean of 0 and a standard deviation of
>> 2*sqrt(2/3) = 1.63299.
>>
>> The wild dice roll also has a mean of 0, but a standard deviation of
>> sqrt(3) = 1.73205.
>>
>> Here are the probabilities for each distribution from 4 to 10.
>>
>> x 4dF 3dF + Wild dF
>> 4 0.0123457 0.0123457
>> 3 0.0493827 0.0534979
>> 2 0.123457 0.128944
>> 1 0.197531 0.203475
>> 0 0.234568 0.228319
>> 1 0.197531 0.187217
>> 2 0.123457 0.111789
>> 3 0.0493827 0.0496085
>> 4 0.0123457 0.0165362
>> 5 0. 0.00551206
>> 6 0. 0.00183735
>> 7 0. 0.000612451
>> 8 0. 0.00020415
>> 9 0. 0.0000680501
>> 10 0. 0.0000226834
>>
>> If you're interested, there is an exact formula for the probability mass
>> function f of this fudge wild dice roll. Let w be the conditional
>> probability mass function of an exploding dF given that the first roll
>> is a plus, namely w(x) = 0 if x < 0, 1/3 if x = 0, and 4/(3^(x+1)) if x
>> > 0. The rationale for this is as follows: Given the first roll is 1,
>> there is no way for x to be less than 0. There is exactly one way for x
>> to be 0, namely roll a 1 on the second roll. This has probability 1/3.
>> There are two ways for x to be 1: roll a 0 on the second roll; or roll
>> another 1 followed by 1; these have probabilities 1/3 and 1/9,
>> respectively; and 1/3+1/9=4/9. This pattern continues for x larger than
>> 1, namely the probability is 4/(3^(x+1)).
>>
>> Let v be the probability mass function for one ordinary dF. Let * be the
>> convolution operator. Let u be the probability mass function for
>> ordinary 3dF. Then u(x) = (v*v*v)(x).
>>
>> Then the probability mass function f of the fudge wild dice roll is the
>> following mixture distribution:
>>
>> f(x) = (1/3) u(x+1) + (1/3) u(x) + (1/3) (w*u)(x)
>>
>> The intuition here is that on the first roll, the wild die can be either
>> 1, 0 or 1 with equal probabilities 1/3rd each. This gives rise to the
>> three terms in the mixture distribution with weights 1/3rd each. The
>> first term is 3dF shifted left, corresponding to a roll of 1 on the
>> first (and only) roll of the wild die. The second term corresponds to a
>> 0 on the (nonexploding) wild die. And the third term corresponds to an
>> exploding roll of 1 on the first roll of the wild die.
>>
>> I coded this formula in Mathematica to compute the probabilities.
>>
>> Then I doublechecked this formula by writing a Monte Carlo simulation
>> in Perl, simulating one million random rolls. This simulation matched
>> the theoretical values given above:
>>
>> sub dF {
>> int(rand(3))  1;
>> }
>>
>> $count = 1_000_000;
>>
>> %prob = ();
>>
>> sub getWild {
>> my ($Wild) = @_;
>> my $w = dF();
>> if($w == 1){
>> $Wild = getWild($Wild+1);
>> } else {
>> $Wild += $w;
>> }
>> }
>>
>> sub getSum {
>> my ($x,$y,$z) = (dF(),dF(),dF());
>> $x+$y+$z+getWild(0);
>> }
>>
>> sub printDistrib {
>> my ($prob) = @_;
>> my ($x,$p,$sx,$sx2,$v,$cdf);
>> for $x (sort {$a <=> $b} (keys %$prob)) {
>> $p=$$prob{$x}/$count;
>> $cdf+=$p;
>> $sx+=$p*$x;
>> $sx2+=$p*$x*$x;
>> print "P[X=",$x,"]=",$p,", P[X<=",$x,"]=",$cdf,"\n";
>> }
>> $v=$sx2($sx*$sx);
>> print "Sum P[X]=",$cdf,"\n";
>> print "E[X]=",$sx,"\n";
>> print "Var[X]=",$v,"\n";
>> print "sd[X]=",sqrt($v),"\n";
>> }
>>
>> for $_ (1..$count) {
>> $prob{getSum()}++;
>> }
>>
>> printDistrib(\%prob);
>>
>>
>> [Nontext portions of this message have been removed]
>>
>>
>>
>
>
> [Nontext portions of this message have been removed]
>
>
>
> 
>
> > Please visit the Fudge Community Yahoo Group home page at http://games.groups.yahoo.com/group/fudgecommunity to search the archives, upload or download files, and more! Note: No attachments are allowed on the fudgecommunity@yahoogroups.com email list, please upload any attachments to the Files section instead. Thank you! <Yahoo! Groups Links
>
>
>  I let the mathematicians do the math for me. Luckily I know of a
mathematician who is also a game designer (Torben Mogensen) he has created
a wonderful tool called troll:
http://www.diku.dk/~torbenm/Troll/
He also made a web interface for it: http://topps.diku.dk/torbenm/troll.msp
So go to the web interface and insert the following formula:
sum 2d34 +
sum (accumulate x:=1d32 while x=1) +
sum (accumulate y:=1d32 while y=1)
and look at the probabilities or make rolls.
Wolf
PS the exploding+ can also be described with this formula:
sum {3 # choose {1, 0, 1},
accumulate x := 1 # choose {1, 0, 1} while x = 1 }
On Wed, Jul 11, 2012 at 6:57 PM, John Rudd <john@...> wrote:
> Since I play with "the player always rolls" (so "being hit" is caused
> by the PC rolling low, not the NPC rolling high ... becuase the NPC
> never rolls against a PC) and the GM only rolls for NPC vs NPC events,
> the currently presentation doesn't include the fear aspect you
> mention. Only the PC can get the high result, so the PC has nothing
> to fear.
>
> The way to fix that is: let the wild die also "explode downward" (if
> it rolls a , keep rerolling it as long as it continues to roll 's).
> Anyone want to quickly work up the probability for a wild die that
> can explode in both directions?
>
>
> On Wed, Jul 11, 2012 at 4:27 AM, Mike Greene <mikedgreene@...>
> wrote:
> > To your point, the Wild die adds very little skew to the outcome of the
> > roll.
> >
> > On the other hand, it adds to the excitement of the test, and the fear of
> > the opponent. There is ALWAYS a chance that anyone, can do anything
> > at any time. That one point orc, could slay the invincible knight if it
> > was lucky!!
> >
> > Geek content:
> >
> > I think your chart is a bit off.
> >
> > Rolls the yield 3 Rolls that yield +3
> > 1 1 1 [0 ] 1 1 1 [0 ]
> > 1 1 1 [1 1] 1 1 1 [1 1 ]
> > 1 1 0 [1 ] 1 1 0 [1 1 1]
> > 1 0 1 [1 ] 1 0 1 [1 1 1]
> > 0 1 1 [1 ] 0 1 1 [1 1 1]
> > 1 0 0 [1 1 0] *3
> > 1 0 0 [1 1 1 1] *3
> > 0 0 0 [1 1 1 0] *3
> > 0 0 0 [1 1 1 1 1] *3
> >
> > So the chances of rolling a 3 should be less than the chances of
> > rolling a +3.
> > Your perl script consistently shows this, but the chart doesn't. I'm
> not
> > that great at statistics, so I don't understand exactly why the math
> > doesn't work. But as stated above, the differences are less than 1%
> almost
> > negligible.
> >
> > Mike Greene
> >
> >
> >
> >
> > On Wed, Jul 11, 2012 at 12:49 AM, Aher <elisha_ben_abuyah@...>
> wrote:
> >
> >> **
> >>
> >>
> >> If you roll 3 normal fudge dice and one wild fudge die that keeps
> >> exploding as long as you roll a plus, then the resulting probability
> >> distribution isn't that much different from an ordinary 4dF roll. For
> >> the purposes of making these calculations, I'm assuming the player
> >> doesn't have a choice about when to stop rolling the wild die, but keeps
> >> rolling until either a 1 or 0 appears. (I can't calculate when a player
> >> would choose to stop rollingchoice isn't chance... [:)] )
> >>
> >> A normal 4dF roll has a mean of 0 and a standard deviation of
> >> 2*sqrt(2/3) = 1.63299.
> >>
> >> The wild dice roll also has a mean of 0, but a standard deviation of
> >> sqrt(3) = 1.73205.
> >>
> >> Here are the probabilities for each distribution from 4 to 10.
> >>
> >> x 4dF 3dF + Wild dF
> >> 4 0.0123457 0.0123457
> >> 3 0.0493827 0.0534979
> >> 2 0.123457 0.128944
> >> 1 0.197531 0.203475
> >> 0 0.234568 0.228319
> >> 1 0.197531 0.187217
> >> 2 0.123457 0.111789
> >> 3 0.0493827 0.0496085
> >> 4 0.0123457 0.0165362
> >> 5 0. 0.00551206
> >> 6 0. 0.00183735
> >> 7 0. 0.000612451
> >> 8 0. 0.00020415
> >> 9 0. 0.0000680501
> >> 10 0. 0.0000226834
> >>
> >> If you're interested, there is an exact formula for the probability mass
> >> function f of this fudge wild dice roll. Let w be the conditional
> >> probability mass function of an exploding dF given that the first roll
> >> is a plus, namely w(x) = 0 if x < 0, 1/3 if x = 0, and 4/(3^(x+1)) if x
> >> > 0. The rationale for this is as follows: Given the first roll is 1,
> >> there is no way for x to be less than 0. There is exactly one way for x
> >> to be 0, namely roll a 1 on the second roll. This has probability 1/3.
> >> There are two ways for x to be 1: roll a 0 on the second roll; or roll
> >> another 1 followed by 1; these have probabilities 1/3 and 1/9,
> >> respectively; and 1/3+1/9=4/9. This pattern continues for x larger than
> >> 1, namely the probability is 4/(3^(x+1)).
> >>
> >> Let v be the probability mass function for one ordinary dF. Let * be the
> >> convolution operator. Let u be the probability mass function for
> >> ordinary 3dF. Then u(x) = (v*v*v)(x).
> >>
> >> Then the probability mass function f of the fudge wild dice roll is the
> >> following mixture distribution:
> >>
> >> f(x) = (1/3) u(x+1) + (1/3) u(x) + (1/3) (w*u)(x)
> >>
> >> The intuition here is that on the first roll, the wild die can be either
> >> 1, 0 or 1 with equal probabilities 1/3rd each. This gives rise to the
> >> three terms in the mixture distribution with weights 1/3rd each. The
> >> first term is 3dF shifted left, corresponding to a roll of 1 on the
> >> first (and only) roll of the wild die. The second term corresponds to a
> >> 0 on the (nonexploding) wild die. And the third term corresponds to an
> >> exploding roll of 1 on the first roll of the wild die.
> >>
> >> I coded this formula in Mathematica to compute the probabilities.
> >>
> >> Then I doublechecked this formula by writing a Monte Carlo simulation
> >> in Perl, simulating one million random rolls. This simulation matched
> >> the theoretical values given above:
> >>
> >> sub dF {
> >> int(rand(3))  1;
> >> }
> >>
> >> $count = 1_000_000;
> >>
> >> %prob = ();
> >>
> >> sub getWild {
> >> my ($Wild) = @_;
> >> my $w = dF();
> >> if($w == 1){
> >> $Wild = getWild($Wild+1);
> >> } else {
> >> $Wild += $w;
> >> }
> >> }
> >>
> >> sub getSum {
> >> my ($x,$y,$z) = (dF(),dF(),dF());
> >> $x+$y+$z+getWild(0);
> >> }
> >>
> >> sub printDistrib {
> >> my ($prob) = @_;
> >> my ($x,$p,$sx,$sx2,$v,$cdf);
> >> for $x (sort {$a <=> $b} (keys %$prob)) {
> >> $p=$$prob{$x}/$count;
> >> $cdf+=$p;
> >> $sx+=$p*$x;
> >> $sx2+=$p*$x*$x;
> >> print "P[X=",$x,"]=",$p,", P[X<=",$x,"]=",$cdf,"\n";
> >> }
> >> $v=$sx2($sx*$sx);
> >> print "Sum P[X]=",$cdf,"\n";
> >> print "E[X]=",$sx,"\n";
> >> print "Var[X]=",$v,"\n";
> >> print "sd[X]=",sqrt($v),"\n";
> >> }
> >>
> >> for $_ (1..$count) {
> >> $prob{getSum()}++;
> >> }
> >>
> >> printDistrib(\%prob);
> >>
> >>
> >> [Nontext portions of this message have been removed]
> >>
> >>
> >>
> >
> >
> > [Nontext portions of this message have been removed]
> >
> >
> >
> > 
> >
> > > Please visit the Fudge Community Yahoo Group home page at
> http://games.groups.yahoo.com/group/fudgecommunity to search the
> archives, upload or download files, and more! Note: No attachments are
> allowed on the fudgecommunity@yahoogroups.com email list, please upload
> any attachments to the Files section instead. Thank you! <Yahoo! Groups
> Links
> >
> >
> >
>
>
> 
>
> > Please visit the Fudge Community Yahoo Group home page at
> http://games.groups.yahoo.com/group/fudgecommunity to search the
> archives, upload or download files, and more! Note: No attachments are
> allowed on the fudgecommunity@yahoogroups.com email list, please upload
> any attachments to the Files section instead. Thank you! <Yahoo! Groups
> Links
>
>
>
>
[Nontext portions of this message have been removed]  @Mike Greene...As you say, there are more rolls that lead to a +3 than lead to a 3, but that doesn't imply that the probability of a +3 is greater than 3. The reason is that the probability of making most of the +3 rolls is way smaller than the probability of making any of the 3 rolls. Let's look at a concrete example from your list. The probability of rolling this plus three 0 0 0 [1 1 1 1 1] is a mere (1/3)^8=1/6561=0.0001524158, while the probability of rolling this minus three 0 1 1 [1 ] is (1/3)^4=0.01234568, which is about 2 orders of magnitude bigger. In short, the probability of any of the combinations you list is (1/3)^(the number of digits in the roll). In other words, each extra "1" on the wild die *decreases* the probability by 1/3rd. These reductions have a major impact on the final result. In any case, when you add up these probabilities for plus three, you get exactly the probability I put in the table.
Remember: There are an infinite number of possible rolls once you add the wild die. So you can't simply look at the number of combinations for a plus three and divide by the total number of rollsyou'd always get 0. That's why the computation isn't as straightforward as in the case of 4dF, where you can simply enumerate all the combinations for an outcome and divide by the total number of rolls, namely 3^4.
As another poster pointed out, you can double check the theoretical results I calculated using Troll. The website is
http://topps.diku.dk/torbenm/troll.msp
And the command is
sum {3 # choose {1, 0, 1}, accumulate x := 1 # choose {1, 0, 1} while x = 1 }
You can see there that Troll's probabilities are the same as mine.
Anyway, thanks for taking the time and effort to delve into my post, and I hope that my explanation here helped clear up your doubts about it.
Sincerely,
Elisha
 In fudgecommunity@yahoogroups.com, Mike Greene <mikedgreene@...> wrote:
> Geek content:
>
> I think your chart is a bit off.
>
> Rolls the yield 3 Rolls that yield +3
> 1 1 1 [0 ] 1 1 1 [0 ]
> 1 1 1 [1 1] 1 1 1 [1 1 ]
> 1 1 0 [1 ] 1 1 0 [1 1 1]
> 1 0 1 [1 ] 1 0 1 [1 1 1]
> 0 1 1 [1 ] 0 1 1 [1 1 1]
> 1 0 0 [1 1 0] *3
> 1 0 0 [1 1 1 1] *3
> 0 0 0 [1 1 1 0] *3
> 0 0 0 [1 1 1 1 1] *3
>
> So the chances of rolling a 3 should be less than the chances of
> rolling a +3.
> Your perl script consistently shows this, but the chart doesn't. I'm not
> that great at statistics, so I don't understand exactly why the math
> doesn't work. But as stated above, the differences are less than 1% almost
> negligible.
>
> ∩>  Interesting, a wild dice that explodes in either direction! I hadn't considered that before, but it certainly seems like it would add a lot more drama to the game. I love it!

Munkwunk
http://psipunk.blogspot.com  Psipunk, a Fudge Cyberpunk RPG
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