Let's take a classic opposed check: initiative.
There are two guys, one has a +2 bonus, the other a +6. I kind of understand that the one having the +4 is more likely to go first, but I'd like to know how to calculate the probability of that event. I understand that this is not "20% more likely", but I can't quite figure out what the probability should be.
I'd love to know a general formula, not necessarily one using a d20.
Thanks :)
You lost me after probability. :P
This is the best I could come up with. I am sure there is an easier way to do this, but this works. It is based on the idea that one of the participants has a higher mod or will win on a tie result. If they are so closely tied that it would just result in a roll-off, the odds of success for both players is 50%. It is also based on the RAW ruling that a 20 and 1 are not treated as critical. I know a lot of people house-rule that in, so I felt the need to mention it.
X = Die max (Usually 20)
A = Initiative that wins tie (Higher)
B = Initiative that loses tie (Lower)
((X-A+B-1)*(X-A+B))/8
The formula returns Player B's chance of success. To find Player A's chance, subtract the result from 100.
In the example above, the following is true: X=20, A=6, B=2
((20-6+2-1)*(20-6+2))/8
(15*16)/8
240/8
30
Player B has only a 30% chance of winning Initiative. Player A has 70%. This one came out even, but most other results will not return whole numbers.
Now, I think I see a way to make this much more simply, but I will have to look into that later.
In fact I think I should have posted in general discussion, or advice maybe. It's a math's question more than a rule's one. But rules are based on math, so here it is...
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Hardly the "easiest" way, but whenever you have a stats problem like this, and you can't figure out the math, just build a "truth table", and count.
Using your example:
There are 20 times 20 (400) possible roll combinations. That is, player 1 rolls 1, player two rolls 1; player 1 rolls 1, player two rolls 2; etc.
This sounds like a lot, but it's easy to build in an excel spreadsheet (there's a lot of stats you can do in excel, but never mind that now). For our purposes, we can eliminate anything where player 1 (+2 int) rolls less than 5 higher than player two (higher bonus wins ties).
So, that takes out 5 times 20 possible combinations (player one rolls 5 or less), or 100, leaving 300. Even if player 1 rolls a 6, that only wins if player two rolls a 1, so that's 19 more combinations gone. If Player 1 rolls a 7, he beats player 2 if they roll a 1 or 2, so 18 more combinations gone. Keep going and remove all those combinations, and remember, if player 2 rolls 18 or better, it doesn't matter what player 1 rolls, so that takes out 3 times 20 more combinations...
When you have all the possible pairs of rolls that player 1 can win, just divide 400 by that number to get the percentage chance.
And "voila!" Stats with just basic math!
I just realized something about my formula. By attempting to simplify it, I made it completely useless where X is anything but 20. Oops. That means the method I was going to use to simplify it even more will not work. The formula only works on a d20 and reads like this:
A = Initiative that wins tie (Higher)
B = Initiative that loses tie (Lower)
((19-A+B)*(20-A+B))/8
The formula returns Player B's chance of success. To find Player A's chance, subtract the result from 100.
Still works. Disappointed in my oversight though :(
Hardly the "easiest" way, but whenever you have a stats problem like this, and you can't figure out the math, just build a "truth table", and count.
Using your example:
There are 20 times 20 (400) possible roll combinations. That is, player 1 rolls 1, player two rolls 1; player 1 rolls 1, player two rolls 2; etc.
This sounds like a lot, but it's easy to build in an excel spreadsheet (there's a lot of stats you can do in excel, but never mind that now). For our purposes, we can eliminate anything where player 1 (+2 int) rolls less than 5 higher than player two (higher bonus wins ties).
So, that takes out 5 times 20 possible combinations (player one rolls 5 or less), or 100, leaving 300. Even if player 1 rolls a 6, that only wins if player two rolls a 1, so that's 19 more combinations gone. If Player 1 rolls a 7, he beats player 2 if they roll a 1 or 2, so 18 more combinations gone. Keep going and remove all those combinations, and remember, if player 2 rolls 18 or better, it doesn't matter what player 1 rolls, so that takes out 3 times 20 more combinations...
When you have all the possible pairs of rolls that player 1 can win, just divide 400 by that number to get the percentage chance.
And "voila!" Stats with just basic math!
I did it, thank you. A simple sheet with only if and count.if functions was enough.
Shalmdi, your formula is great, I missed it yesterday, thank you!