Ye Olde Inn

I have a lot of trouble wrapping my mind around probabilities. For example, if you cast command on the Wizard versus the dwarf, I am doing the probability = 1- (5/6)^n where n= mind points. However, that is one event. What about the second turn? What is the cumulative probability of the Command spell STAYING on the wizard or dwarf?

I would assume it's the same probability as the former multiplied by 2. In essence all you're doing is rolling the dice a second time...

wallydubbs wrote:I would assume it's the same probability as the former multiplied by 2. In essence all you're doing is rolling the dice a second time...

It's not multiplied by two; you have to multiply the odds of it staying on for every turn. The more turns pass, the less likely it's going to be the spell will last. Of course, at any given turn during which the spell is still on, the past doesn't count anymore.

I'm too lazy to lookup the odds of breaking free from the spell, but let's assume the following: the Wizard had 75% chance to break free each turn, while the Dwarf only has 25% chance. That means that for the Wizard to have the spell actually stick is 25% (i.e. 1/4), whereas for the Dwarf it's 75% (i.e. 3/4). To have it stick for 2 turns, it first has to stick the first turn (1/4 vs. 3/4) and then through the second turn as well. For the Wizard, that's 1/4 * 1/4 = 1/16, or 6.25%. For the Dwarf, however, it's 3/4 * 3/4 = 9/16, or 56.25%. Do the same for 10 turns, and the chance that the Wizard is still under the spells' influence is 1/4 * 1/4 * 1/4 * 1/4 * 1/4 * 1/4 *1/4 * 1/4 * 1/4 * 1/4 = (1/4)^10 = 0.000095% while for the Dwarf, it's (3/4)^10 = 5,6%.

I think I have it.
The probability of them not breaking the spell is (5/6)^n where n is the number of mind points. On turn one it is the mind points as listed. On turn two it is two times the mind points listed (12 for the wizard, 6 for the dwarf).
So the formula per turn of it NOT breaking is (5/6)^(yn) where y is the number of turns (or chances they get to break it, including immediately) and n is the number of mind points.
So the probability of it breaking (not not breaking) is P = 1-(5/6)^(yn) where y is number of chances they have to break it so far and n is the number of mind points.

I haven't used this kind of witchcraft math since high school. Mr. Haich my calculus teacher would be so proud of me. I should send him this post.

I dont know how to exactly calculate it im not an ace in math but i think i got the logic behind it. What the previous guys said makes sense but its more than that. This is why:
When you roll a dice and lets say for example you get a 2. Then next time the probabilites of getting a 2 are getting less.
So the probability of breaking the spell gets higher but still only so little because just the chances of getting another 2 are getting less.
The probabilities of getting a 1,3,4,5 are still as high as getting a 6.
So each turn the probabliities of breaking a spell do increase but in a slower pace than what the previous guys suggested (if I understood correctly).

Drew wrote:When you roll a dice and lets say for example you get a 2. Then next time the probabilites of getting a 2 are getting less.

No, this is incorrect. Each die roll is independent, so rolling a 2 on any dice is at exactly the same odds. If you rolled a 2 before (at 1/6 chance), rolling another 2 is still at 1/6 chance. The reason is that when you make that second roll, the first one is already in history, you already know it's a 2.

When you look at it before rolling those two dice, you can combine the odds to see if you can roll two successive 2's - and that's at 1/6 * 1/6 = 1/36 chance.