Dee Famera
United Kingdom

Can anyone help with this:
Two players each roll a D6 and the higher total wins  simple. Each has a 50% chance of winning. However, what happens when you introduce a +1, +2, +3 and +4 modifier to one player?
I need to know the % odds of winning when one player has modifiers of +1, +2, +3 and +4 before casting the die. I feel I should be able to work it out but my brain hurts!



(I assume that ties are a reroll).
With a +4 mod, the favored side has a 29/30 chance of winning (lose only if the roll a 1 and the other side rolls a 6).
With a +3 mod, the favored side has a 27/30 chance of winning (lose only if they roll a 1 and the other side rolls a 5,6; or if they roll a 2 and the other side rolls a 6)
With a +2 mod, the favored side wins 24/30 (lose if 1:4,5,6; 2:5,6; 3:6)
With a +1 mod, the favored side wins 20/30 (lose if 1:3,4,5,6; 2:4,5,6; 3:5,6; 4:6)
With a +0 mod, the favored side wins 15/30 (lose if 1:2,3,4,5,6; 2:3,4,5,6; 3:4,5,6; 4:5,6; 5:6).

Andreas Kortegaard
Denmark Kbh N

higher total sounds odd? Do you mean highest roll wins? and what about rolling the same number then?
win/draw/lose in % from the perspective of the one receiving the bonus
1d6 vs. 1d6: 42/17/42
1d6+1 vs. 1d6: 58/14/28
1d6+2 vs. 1d6: 72/11/17
1d6+3 vs. 1d6: 83/9/8
16+4 vs. 1d6: 92/5/3
If you want to visualize it for another time. Take a piece of graph paper and do a 6 by 6 square (for d6's), do xaxis as one dies possible results and yaxis as the other die's, and then count the squares with a given result as the two intersect. Then you got a x/36 calculation you can convert to percentages.

Dee Famera
United Kingdom

RandomLetters wrote: (I assume that ties are a reroll).
With a +4 mod, the favored side has a 29/30 chance of winning (lose only if the roll a 1 and the other side rolls a 6).
With a +3 mod, the favored side has a 27/30 chance of winning (lose only if they roll a 1 and the other side rolls a 5,6; or if they roll a 2 and the other side rolls a 6)
With a +2 mod, the favored side wins 24/30 (lose if 1:4,5,6; 2:5,6; 3:6)
With a +1 mod, the favored side wins 20/30 (lose if 1:3,4,5,6; 2:4,5,6; 3:5,6; 4:6)
With a +0 mod, the favored side wins 15/30 (lose if 1:2,3,4,5,6; 2:3,4,5,6; 3:4,5,6; 4:5,6; 5:6).
Thanks  your answer is excellent. As you're an expert, in my given example what if one player by rolling a 6 wins if the other doesn't roll a 6 (irrespective of the sum totals)? What a conundrum!

Dee Famera
United Kingdom

robertflatt wrote: higher total sounds odd? Do you mean highest roll wins? and what about rolling the same number then?
win/draw/lose in % from the perspective of the one receiving the bonus
1d6 vs. 1d6: 42/17/42
1d6+1 vs. 1d6: 58/14/28
1d6+2 vs. 1d6: 72/11/17
1d6+3 vs. 1d6: 83/9/8
16+4 vs. 1d6: 92/5/3
If you want to visualize it for another time. Take a piece of graph paper and do a 6 by 6 square (for d6's), do xaxis as one dies possible results and yaxis as the other die's, and then count the squares with a given result as the two intersect. Then you got a x/36 calculation you can convert to percentages.
Your win/draw/lose % is great, thank you. I am now trying to figure how, if either player automatically wins with a roll of 6 if the other doesn't roll 6, how it alters things.

Andreas Kortegaard
Denmark Kbh N

Essentially what happens is that the person without modification will transform draws into wins (# out of 36, /36*100 each # to get percentages):
+4: 29/2/5 +3: 27/3/6 +2: 24/4/8 +1: 20/5/11 +0: 15/6/15
you can compare them to the numbers given by RandomLetters



>As you're an expert, in my given example what if one player by rolling a 6 wins if the other doesn't roll a 6 (irrespective of the sum totals)?
I would have to sit down and work through all of the possibilities to be certain.

Andreas Kortegaard
Denmark Kbh N

robertflatt wrote: Essentially what happens is that the person without modification will transform draws into wins (# out of 36, /36*100 each # to get percentages): +4: 29/2/5 +3: 27/3/6 +2: 24/4/8 +1: 20/5/11 +0: 15/6/15
you can compare them to the numbers given by RandomLetters
Actually forget the reference to RandomLetters. He forgot to take into account for the fewer draws, as the intervals change. e.x.: at +4 without "sixes wins" modless guy can roll 16 and modguy can roll 510, only leading to draws at 55 and 66. So the numbers, similar to the ones i made in the quote, without "sixes wins"rule are: +4: 33/2/1 +3: 30/3/3 +2: 26/4/6 +1: 21/5/10 +0: 15/6/15
again if you want percentages, divide each number by 36 and multiply by a 100.
edit: Actually what he did was count "losses for modguy", and deducted that from 30, but the premise that it was out of 30, instead of 36 was wrong unfortunately.



Ah, that's right, I did forget that there were fewer draws as the modifier increased. Work in haste, repent at leisure.

Nicholas Palmer
United States Athens Georgia

http://anydice.com/program/8f9d This will calculate the odds. You can change the +1 to whatever modifier you want.
Because the 6 on the die is counted as a 100, it will always win unless you put in an absurdly large modifier or the other person also rolls a "6".
You also never answered on who won ties.


