Given an N-sided die, what is the probability that the second roll of a greater than N-sided die will be greater than the first roll?

Question

First off, I am not a math guy - so please accept my apologies.

My 4th-grade daughter is making up a game to bring to school (dice wars), in the game two people roll a d6 and the higher number wins, ties "push" to the next round. Winners are rewarded with a piece of Easter candy.

To add spice to the game I suggested the "house" can use other larger dice (e.g., d8, d10, d12, and d20 - what else to do with those dnd dice).

I thought calculating the "odds" would be simple. I was wrong.

My internet sleuthing has lead me to the ((n-1)/2)n yields the probability for two equally-sided dice. What to do when the second die is of greater (or lesser) value than the first? Furthermore, I understand that probabilities do not equal "odds" per:

To convert from a probability to odds, divide the probability by one minus that probability. So if the probability is 10% or 0.10, then the odds are 0.1/0.9 or '1 to 9' or 0.111. To convert from odds to a probability, divide the odds by one plus the odds. (Source: [GraphPad] (https://www.graphpad.com/support/faq/probability-vs-odds/))

I've come up with:

Die Combinations	Odds
d6 vs. d6	1 : 1
d6 vs. d8	the answer didn't give a satisfying whole number so it was sh*t canned. At the end of the game, we are giving out Easter candy - no fractions
d6 vs. d12	1 : 2
d6 vs. d20	1 : 3 again, I rounded up to 3 vs. 2.3 for ease of use

I understand I have the odds wrong, but it is a children's game, so I'm okay with some amount of approximation.

What I am failing to grasp (admittedly, one of many things) is how to determine the probabilities in the first place. Initially, I would take the percentage of d6 from the larger die and divide it by the remainder of the larger number (e.g., 6 of 20 is 30%; 30% from 100% leaves 70%; 70%/30% is 2.3-ish; I rounded to 3 for ease of use).

I don't mean to offend with my sloppy math, but I was hoping something simple or at least an accurate way to generate the probabilities and then I can use the probability to odds conversion quoted above.

Thank you.

Answer 1

d6 vs. d20 calculation. $P(d20\gt 6)=.7$, When $d20\le 6$, the probability of a tie is $1/6$ so overall prob. $=.05$, otherwise d20 win = d6 win $=.125$.

Net probabilities d20 win $=.825$ and d6 win$=.125$ with tie $=.05$

Answer 2

If $m \le n$ in a d$m$ vs d$n$ game:

There are $m$ possible ways to tie ($1$ to $1$, or $2$ to $2$, or..., or $m$ to $m$).

There are $1 + 2 + 3 + \cdots + (m-1)$ ways for d$m$ to win ($2$ to $1$; or $3$ to $2$ or $1$; etc. And as you noted, this is equal to $\frac{m(m-1)}{2}$

The remaining possibilities are for d$n$ to win. So this will be $nm - m - \frac{m(m-1)}{2}$ ways for d$n$ to win. This simplifies to $nm-\frac{m(m+1)}{2}$ ways for d$n$ to win.

From there you should be able to calculate the odds of each die winning.