Suppose A and B play game and use dice with 1...A and 1...B values, respectively - who tosses more, wins.
What is the chance that A wins, B wins or the game is tie, based on the values?
A few years ago, I have made a grid for exact values A/B and got the results, but I would appreciate more "advanced" solution, or at least hints.
Additionally: How would the solution change if I chose different minimum and maximum values (rather than one dice being a superset of the other), but the intersection of intervals would contain the same values (e.g. 1...6 and 3...10, but not 1,3,5 and 2,3,4)? Or, if the probability space was infinite (e.g. any rational number from given intervals can be rolled), what would change?
Edit: My attempt on solution (I hope succesful):
Let's say A has the smaller values (set A), where B has the bigger values (set B). Place the values on a line, and let be:
x - size of left area (|A-B|)
y - size of intersecting area (|A intersection B|)
z - size of right area (|B-A|)
n - total amount of different values (x+y+z, |A union B|)
I will also explicitly provide the solution if the value rolls into either exclusively left or right area, but will not sum the probabilities, just subtract "A wins" and "Tie" from 1, in order to get chance "B wins".
Left area: A has a chance x/n to get his value there. B always wins.
Right area: B has a chance z/n to get his value there, and wins regardless of where A gets the roll. But the left area was already counted, so the chance in this case is (y/n)*(z/n) for B.
Middle area:
Chance for A to get into the middle area is y/(x+y), and for B is y/(y+z), together:
y^2/((x+y)(y+z))
In this case, the probability for win/loss is the same. But first, let's have a look at tie, which has chance 1/y (A can get anywhere, B must match). That's exactly y/((x+y)(y+z)).
To get the win chance, one subtracts the tie chance and divides the remainder by 2. That gives:
y(y-1)/(2(x+y)(y+z)) chance for A to win (at all).
Chance to win for player A:
y(y-1)/(2(x+y)(y+z))
Chance to tie:
y/((x+y)(y+z))
Chance to win for player B:
1-(y(y+1))/(2(x+y)(y+z))
This ended up being harder than I first thought - it was easier when x=0 :)
Infinity solution:
If we used computer to evenly generate any number from given interval and increased the precision, the ratio between x,y and z would not change at all. Therefore, the only thing that would change (again, intuitively) is the tie - it's less probable that two same numbers on the same interval are generated, the lower the gap between numbers is.