Dice probabilities on opposed sets of 10-sided dice

[Sociopath]

Hi,

I was wondering if anyone might want to help me with some less common dice probability. The dice mechanic is similar to the board game Risk (two sets of dice being compared), but the dice pools are varying numbers of 10-sided dice on either side (2 pools compared of a varying number).

I'm having a little trouble wrapping my head around how to get statistical probabilities of who wins on either side of the dice roll.

My needs involve finding the probabilities of the chances of winning on either side of the equation. To win, a die must be higher than it's counterpart from the other dice pool, and dice are compared highest to highest down the line.

For instance, if I roll one pool of 3 10-sided dice against a pool of 4 10-sided dice, I receive 8, 7, and 6 from pool one and 8, 5, 6 in the second pool. The highest are compared to the highest (8 = 8, 7 > 6, 6 > 5). Equal pairs cancel each other out, so pool one wins with 2 hits (2 of his dice exceeded the remaining two of the opponent).

If I recall correctly, opposing pools of equal dice have about a 50% probability of winning and a 50% probability of losing (the chances for any result on either side are the same across the board). What I don't know how to calculate is the probability of unequal dice pools in win/lose terms.

Values for the following is what I'm trying to figure out:

Probability of 1 die pool vs. 5 dice pool, probability of the 1 die at least cancelling out a hit from 5 dice (the 1 die exceeding all 5 dice, at least negating one hit).

Probability of uneven pairs...
- 1d vs 2d
- 1d vs 3d
- 1d vs 4d
- 1d vs 5d
- 2d vs 3d
- 2d vs 4d
- 2d vs 5d
- 3d vs 4d
- 3d vs 5d
- 4d vs 5d
...with win chances vs. bigger and smaller dice pools.

Some special rules (if anyone takes this up), a roll of 1 in any pool is discarded automatically; they don't exceed any die, and, being the poorest result possible, is discarded even if it does not have a matching die. So, any roll of 1 is automatically thrown out and not considered - except for the purposes of losing a die in your pool if it shows up.

I was thinking of doing probability of a theoretical 9-sided die, but I quickly threw that idea out because it would skew the results; using a 9-sided die would result in probability for the dice always returning higher than 1, which would not be the case.

I'd really appreciate help on this matter. I'm decent at math, but probability and statistics is a bit beyond my normal day-to-day.

 

[Stephen Tashi]

I'm decent at math,

How are you at computer programming? This looks like the type of problem that can be solved that way.

When the results of N realizations of a random variable are sorted by their values from smallest to largest, this vector of outcomes is called the vector of "order statistics". If you roll five 10-sided dies then the probability of ( 9,8,5) being the "high three" of your roll is the probability of the event ( 10th order statistic = 9, 9th order statistic = 8, 8th order statistic = 5). If you look up "order statistics" on the web, you can find formulae for them.

The problem of computing who wins, attacked in a straightforward manner, involves looking at each possible vector of the top K order statistics from the smaller die pool versus each possible vector of the top K order statistics from the larger die pool and noting who wins and what the probability of this pair of vectors is.

I don't know if there is an elegant way to solve this problem with hand-calculations. If you want to try that approach, I suggest you look at some iteration scheme. For example, if we assume we know the answer for the case of M dies vs N dies, can we find it for M dies vs N+1 dies?

 

[Sociopath]

You bring up an interesting possibility with a computer program. I suppose I was searching for a formula and an elegant way to predict probability so hard that I overlooked the obvious possibility: simply generating large numbers of test rolls to provide data, making the amount of data so large that it could accurately predict outcomes in percentage chance values.

Would you have any recommendations for a number of rolls that could be considered enough to generate accurate results? I was thinking at least 10,000, but the more sets rolled, the more accurate the data would become.

Thank you for your reply, that really helped.

 

[Stephen Tashi]

I wasn't thinking of a Monte-Carlo program, although that's always an option in probability problems. I was thinking of a program that would generate all possible vectors of order statistics and apply the appropriate formula to calculate their probability. If we can't figure out the appropriate formula, that could also be found by exhaustive enumeration.

For example, take 3 dice agains two.

Generate the 1000 possible vectors representing 3 dice rolls, find the order statistics that give the "high two" of each vector:

Roll | High Two
10 10 10| 10 10
10 10 9 | 10 10
...
10 9 8 | 10 9
... etc.

Then generate each possible vector that reprsents the "high two" and count up how many times it appears in this list of 1000. Each vector in list of 1000 has a probabiltiy of ( 1/ 10^3) so this allows us to compute the probability of each vector representing the "high two".

Do a similar process to find the probability of each possible "high two" from rolling two dice.

To find the probability that 3 dice beat 2 dice, go through the list of the possible "high two" for 3 dice. Compute the probabilty that the 3 dice produce that "high two" and the two dice produce a "high two" that loses. For eample, the probability that the 3 dice produce (8,7) and the two dice produce ( (8,6) or (8,5) or ... or (7,6) or (7,5)... )

Add up all those probabilities to get the answer for the probability that the 3 dice win.

 

[blue_raver22]

Hello,

I understand your interest in calculating the probabilities for your dice game. Probability and statistics can be complex, but with some basic principles and formulas, we can calculate the probabilities for the scenarios you have mentioned.

First, let's define some variables to make our calculations easier. Let "n" represent the number of dice in the smaller pool and "m" represent the number of dice in the larger pool. We will also use "x" to represent the probability of a die roll exceeding the highest die in the opposing pool, and "y" to represent the probability of a die roll matching the highest die in the opposing pool.

Now, let's look at the scenario of 1 die vs. 5 dice. In this case, the probability of the 1 die exceeding all 5 dice is simply 1/10, or 0.1. The probability of the 1 die matching at least one of the 5 dice is 5/10, or 0.5. Therefore, the probability of the 1 die at least cancelling out a hit from the 5 dice is 0.5 - 0.1 = 0.4.

For the other scenarios, we can use a similar approach. For example, in the case of 1 die vs. 2 dice, the probability of the 1 die exceeding both dice is (1/10)^2 = 0.01. The probability of the 1 die matching at least one of the 2 dice is 2/10, or 0.2. Therefore, the probability of the 1 die at least cancelling out a hit from the 2 dice is 0.2 - 0.01 = 0.19.

For uneven pairs, we can use a combination of the above calculations. For example, in the case of 1 die vs. 3 dice, the probability of the 1 die exceeding all 3 dice is (1/10)^3 = 0.001. The probability of the 1 die matching at least one of the 3 dice is 3/10, or 0.3. Therefore, the probability of the 1 die at least cancelling out a hit from the 3 dice is 0.3 - 0.001 = 0.299.

I hope this helps you in calculating the probabilities for your game. Please note that these calculations assume fair dice and