Dice question, 2 pair of competing dice

Quote: NowTheSerpent

Haven't reflected on this one yet. I'm not sure I understand the question.
This question sounds similar to a riddle I worked recently: What is the probability that a given dice pair will out-total another dice trio?

Just scroll up for the answer.

Edit: On Page 1 of course.

@1BB: thanks for the welcome :)

@ NowTheSerpent: Thanks for the expanded explanation of rolling against an opponent coming out to 50% for each side. I think I will stick with re rolling ties because it becomes too much to remember with tie breakers. The game would bog down at that point and its not really necessary since the odds are the same for each player on a re roll.

The thing I was trying to figure out is how much of an advantage each +1 gives an individual player. That way I can figure out how much it should cost to obtain the bonus.

What I ended up doing is filling out a spreadsheet that had lots of cells I input manually then added and multiplied as needed to get percentages. It turns out that the +1 stacking bonus runs at a diminishing return because, the more bonuses you have the less of a chance you have to be beat.

And if anyone is interested, here is the table of what a +1 through a +10 bonus will get you in expected win percentages.

Bonus	Win Percentage	Percent of Change
None	50%	None
+1	59.83%	9.83%
+2	69.15%	9.31%
+3	77.60%	8.45%
+4	84.86%	7.27%
+5	90.67%	5.80%
+6	94.75%	4.08%
+7	97.41%	2.66%
+8	98.97%	1.57%
+9	99.75%	0.77%
+10	100%	0.25%

Quote: NowTheSerpent

This question sounds similar to a riddle I worked recently: What is the probability that a given dice pair will out-total another dice trio?

Yeah, and mine has an added layer of complexity in that from that trio of dice you can only pick 2 dice for your total so that on one hand you have 36 ways to make 12 results and on the other hand you have 216 ways to make 12 results. AND what is the chances that one total will be more than the other.

My tables show that if you roll 2 dice and your opponent can roll 3 and use the best 2 then you will only win 41.56% of the time and he wins 58.44% of the time. The way I got this is I found out how many rolls you can beat for any given number from 2 to 12 then I divided that by the total number of possible rolls the opponent can make. Averaging all the win percentages gives me the 8.44% difference.

For instance: Player A has 2 dice and rolls a 5. Player B has 3 dice and if he rolls a 2, 3, or 4 he looses. Player B can roll a 2 only once, roll a 3 in 4 different ways, and roll a 4 in 7 different ways. This means Player A can beat Player B 11 times out of the possible 216 rolls. Player A will win 5.09% of the time.

Now lets look at Player B rolling a 5. Player A can roll a 2 only once, roll a 3 in 2 different ways, and roll a 4 only in 3 ways. Player B can win 6 times out of a possible 36 rolls. Player B will win 16.67% of the time.

Here is a table showing how many ways you can roll 2 through 12 with 3 dice

Roll	Possible Ways
2	1
3	3
4	7
5	12
6	19
7	28
8	34
9	36
10	33
11	27
12	16