8th Ed. Spell selection probability

[kroxigor01]

Lets try to calculate the chance of rolling a specific spell you want. Start with a level 2 with Plaque of Tepok:


P(gets a specific spell)=P(rolls the number) + P(gets a double) - P(rolls the number AND a double)

P(rolls the number) = 1/6 + 5/6*1/6 + (1- (1/6 + 5/6*1/6))*1/6 = 91/216

P(gets a double) = 1/6*5/6 + 5/6*1/6 + 1/6*1/6 + 5/6*1/6= 4/9

P(rolls the number AND a double) = 1/6^3 + (1/6*5/6 + 5/6*1/6 + 5/6*1/6)/3 = 11/72


P(gets a specific spell) = 91/216 + 4/9 - 11/72 = 77/108 = 71.30%


Hope I didn't screw up...

 

[Wolf]

In your calculations, what does it matter if you roll a double and the number? Haven't you already covered that under 'roll the number'?

Anyway, I wrote out the whole tree diagram ( ) and counted every time the number 1 (could be anything, I just went with 1) came up, added the amount of times a double was rolled (which also covers the triples) and subtracted the amount of cases they both appeared in. The result came to 189 (for all the doubles and triples (98) and amount of 1s rolled (91))) - 31 (every case where they both appeared) / 216 (total number of possibilities) = 73.15%

Basically what you got, but not quite I'm probably still doing something wrong...

 

[Aetylthor]

According to the program I wrote, these are the chances of getting a specific spell when rolling a certain number of dice:

Dice: ----- Chance to get the spell:
2 -------- 44.4 %
3 -------- 72.1 %
4 -------- 90.7 %
5 -------- 98.5 %

 

[kroxigor01]

If you go through all the cases there are:

24 if your first roll is any unwanted number

1 2 3 4 5 6
1 1 - - - - - -
1 2 - - - 0 0 0
1 3 - - - - - -
1 4 - 0 - - 0 0
1 5 - 0 - 0 - 0
1 6 - 0 - 0 0 -

(hmm, the format doesn't quite work...)

So that is 24x5 = 120 cases in total.

36 if your first roll is the spell you want.

So total successful cases is 156. 156/216 = 13/18 = 72.22222222

What is your program Aetylthor it seems pretty accurate.

 

[Aetylthor]

It simulates the launch of a specified number of dice and determines whether either a specific number was rolled or a double was rolled. It then repeats this process an arbitrary number of times (I used 10000000 for the calculations I posted) and divides the total of positive outcomes by the total number of tries, then multiplies the result by 100 to get the percentage.

With 10000000 rolls the first number after the dot in the results seems to be accurate.

It's a Java program. I could post the source code, kroxigor01, if you're interested.

 

[kroxigor01]

Monte Carlo method eh? Nice.

Well that settles it then. Approximately:

2 - 44.4 %
3 - 72.1 %
4 - 90.7 %
5 - 98.5 %

I have no idea what a source code entails or any programing though, so it probably wouldn't mean anything to me.

 

[JohnMavrick]

Jeez...my head hurts! Some of these calculations are out there. Are you guys using basic algebra for this or something like Calculus?

 

[Overhamsteren]

A friend of mine made this:

		   5 rolls  4 rolls  3 rolls  2 rolls  1 roll
1 spell	 98,5%	90,7%	72,2%	  44,5%	16,7%
2 spells	93,8%	72,3%	38,8%	  11,1%
3 spells	84,2%	46,6%	12,5%
4 spells	67,2%	19,8%
5 spells	40,3%

So the chance for a slann with 4 spells to get 3 specific spells is 46,6% -overall it's pretty easy to get the right spells now with switching on doubles.