Probability one die shows a larger number than another

Question

What is the probability of a number (picked at random) from set $A= \{1,2...,6\},$ being larger than a number (picked at random) from set $B= \{1,2...,10\}.$ What would the probability be if sets $A$ and $B$ were generalized: Set $A=\{n,n+1,...,n+x\},$ and set $B = \{m,m+1,...,m+x\}?$ This is well beyond my realm of statistics knowledge so the more clear each step is would be appreciated.

BONUS: I am trying to find a general equation. If it could apply to multi-sided dice, sets that skipped numbers (e.g. Set $A=\{1,2,3,5,8,13\},$ or set $B=\{2,4,6,8\}),$ more than two sets with different numbers (e.g. Set $A=\{1,2,3\}, B=\{2,3,4\}, C=\{1,3,5\}),$ etc. that would be ideal. I am not sure if this is even possible in general form so please answer the first part, even if you cannot answer this portion.

This link has a similar question, but the answer is in matrix form. Is there a clean equation to solve this?

Answer 1

Extended Comment: As indicated in the Comment by @HagenvonEitzen, one way to work the initial problem (on the probability D6 shows a larger value than D10) is to enumerate cases. In particular, you might make a $10 \times 6$ array of possible pairs of outcomes and highlight the pairs that satisfy your condition. When I did that, it was pretty clear that there are 15 favorable outcomes out of 60, so the probability $P(\text{D6 > D10}) = 1/4.$

D10\D6   1    2    3    4    5    6
 1      11   12*  13*  14*  15*  16*
 2      21   22   23*  24*  25*  26*
 3      31   32   33   34*  35*  36*
 4      41   42   43   44   45*  45*
 5      51   52   53   54   55   56*
 6      61   62   63   64   65   66
...

A brief simulation can sometimes help to provide insurance against miscounting outcomes. In a simulation of a million pairs of dice rolls (D6 and D10), results ought to give two place accuracy, and that was the result. (The code is for R statistical software: event is a logical vector with a million TRUEs and FALSEs, and its mean is its proportion of TRUEs.)

m = 10^6
event = replicate(m, sample(1:6, 1) > sample(1:10, 1))
mean(event)
[1] 0.250383  # aprx 16/50 = 1/4

Using the array method might suggest a way to generalize to cases in which the two dice have consecutively numbered faces, such as in the 'generalization' of your first paragraph. I will leave it to you to figure that out. Or maybe you can see how to generalize Hagen von Eitzen's computation.

My guess is that it will be considerably messier to solve the 'bonus' problem.

Answer 2

Use the Law of Total Probability: for example $d6, d10$ the results of independen six and ten sided dice.

$$\begin{align}\mathsf P(d6>d10) &= \mathsf P(d10>6)\mathsf P(d6>d10\mid d10>6)+\mathsf P(d10\leq 6)\mathsf P(d6>d10\mid d10\leq 6) \\ &=\tfrac 4{10}\cdot 0+\tfrac 6{10}\cdot\mathsf P(d6>d10\mid d10\leq 6)\\ &=0+\tfrac 3 5\cdot\mathsf P(d6>d10\mid d10\leq 6)\end{align}$$

All that is left is to evaluate that last term

Hints: $1=\mathsf P(d6>d10\mid d10\leq 6)+\mathsf P(d6=d10\mid d10\leq 6)+\mathsf P(d6<d10\mid d10\leq 6)\\~~=\mathsf P(d6=d10\mid d10\leq 6)+\mathsf P(d6<d10\mid d10\leq 6)$

Also, when given the condition that it is at most 6, the distribution of a 10 sided die is identical to the distribution of a six sided die .

Extend this principle to account for selections from any two independent uniform discrete distributions from non-identical supports.