Oddstuff - 4d6 probability question

[Ego Eram Reputo 2,559]

I'm designing a game. It uses custom d6 dice. I'm trying to figure out the probability of 1's outnumbering evens when you roll 4d6. Simple really.

 

I *think* I've got the solution.....but I'd like someone, anyone, to come to the same conclusion independently.

 

Anyone want to give it a crack?

 

Spoiler

1/54 or 0.01851851852

 

[BoltBait 3,037]

Give me a better definition of the problem and I can solve it for you.

 

You say the D6 are custom... what are on the 6 faces?

 

When you say "1's outnumbering evens"... give me examples of what that looks like.

[BoltBait 3,037]

Probability problems are very easy.

 

Probability is simply a fraction with the number of things you care about over the total number of all possibilities.

 

For example, the odds of rolling a number less than 3 of a standard D6 would be 2/6 = 1/3 = 33.33%

 

Broken down, the things we care about are 1 and 2, so 2 possibilities over 1,2,3,4,5 and 6... or 6 possibilities: 2/6.

 

Now, for unrelated events, (that is, events that do not rely on the outcome of the first event*) you simply multiply the probability of the individual events together.  For example, rolling a number less than 3 on 2D6 would be:

 

1/3 * 1/3 = 1/9 = 11.11%

 

That is, the probability of the first die roll multiplied by the probability of the second die roll.

 

 

*example of events that are related: drawing a card out of a deck since the first card would be 1/52 and the second card drawn would be 1/51 since the first card has been removed from the deck.  Or, pulling a marble out of a mixed bag of marbles.  The only difference here is that you have to adjust the bottom number of the fraction for each iteration.

 

The only time probability problems get difficult interesting is when you need to calculate the odds of something happening in a particular order, like what are the odds of flipping 3 heads in a row out of 6 total flips.  But, it doesn't sound like your problem includes that.

[Ego Eram Reputo 2,559]

The dice faces are customized this way:

1. Panic
2. Axe
3. Bow
4. Axe
5. Bow
6. Axe

I want to know the probability of rolling four of these die and having Panic outnumber Axes. The order is not important.

 

PPPP and PPP? obviously succeed.

PPBB and PPBA also succeeds.

PPAA fails.

Pxyz succeeds only if x, y and z are not Axes or if x, y & z make up one of the previous combinations.

 

4d6 = 6^4 = 1296 combinations ( many are duplicates).

[BoltBait 3,037]

Your answer is ... drum roll ...

 

P  would out number A ~11.5% of the time. 

and as a bonus:
 

P would out number B ~19.7% of the time. 
 

P would out number both A and B ~1.6% of the time.  As an explanation, to out number both A and B, there would need to be 4 P or 3 P and one other random die. That die has 3 A sides and 2 B sides. Therefore, these would be:

PPPP

PPPAx3

PPPBx2

PPAPx3

PPBPx2

PAPPx3

PBPPx2

APPPx3

BPPPx2

 

21/1296 * 100 = 1.62037037037%

 

You can test this out by rolling 4D6 about 10 times. You should group your 1’s together (call this grouping P), your 2’s and 3’s together (call this grouping B), and your 4’s, 5’s, and 6’s together (call this grouping A) each roll. The count of P group should out number A group around once per 10 rolls.

 

By the way, rolling 4 of your custom dice, you would roll a majority of P 7% of the time, A 48% of the time, B 22% of the time, and you’d roll a tie (no clear majority) 23% of the time. So, in your game, if a tie means indecision (do nothing) or defend (attacker is ineffective), you would run away 7%, attack 70%, and do nothing or defend 23% of the time. 
 

I hope all this math helps your game development. 
 

🙂👍

 

 

[Ego Eram Reputo 2,559]

Many thanks @BoltBait

 

I appreciate the time you've spent on this. The bonus figures are very welcome too!

 

Our figures differ. Either I've not explained it well or my calculations are amiss. I suspect the latter because every time I a worked it out I got a different result.

 

It feels like actually rolling the dice gives a result somewhere in between our figures. I'll do some more trials & report back.

[BoltBait 3,037]

I PM’d you my spreadsheet as a justification of my figures. Go argue with Excel if you want. 😜

 

If after reviewing the spreadsheet you still feel that my formulas are off, just let me know and I will explain why I'm right.  

[Ego Eram Reputo 2,559]

No, no no. Rest assured you are right. After all, you're on the Internet!

 

Manually rolling 4d6 for combat attacks with an Axe:

• first 50 rolls resulted in 7 Panic attacks (P outnumbered A)
• second 50 rolls = 6 Panic attacks
• third 50 = 5 Panic attacks

7+6+5=18    18/150 rolls = 0.12 or 12%

 

( BoltBait is always right)

[BoltBait 3,037]

You're just pandering for up votes now...  

[MJW 1,211]

I can also confirm BoltBait's result is correct. The exact odds are 149/1296. I did analytically, and confirmed it with a program:

using System;
// Solve the following problem:
// Roll four dice. What is the probability there will be more ones than even numbers?
// (There are, of course, 6^4 total combinations.)
// (Obviously written for simplicity, not efficiency!)

namespace Dice_Test
{
    class Program
    {
        static void Main(string[] args)
        {
            int count = 0;
            for (int d1 = 1; d1 <= 6; d1++)
                for (int d2 = 1; d2 <= 6; d2++)
                    for (int d3 = 1; d3 <= 6; d3++)
                        for (int d4 = 1; d4 <= 6; d4++)
                        {
                            int oneCnt = 0;

                            int evenCnt = 0;
                            if (d1 == 1)
                                oneCnt++;
                            else if ((d1 & 1) == 0)
                                evenCnt++;

                            if (d2 == 1)
                                oneCnt++;
                            else if ((d2 & 1) == 0)
                                evenCnt++;

                            if (d3 == 1)
                                oneCnt++;
                            else if ((d3 & 1) == 0)
                                evenCnt++;

                            if (d4 == 1)
                                oneCnt++;
                            else if ((d4 & 1) == 0)
                                evenCnt++;

                            if (oneCnt > evenCnt)
                                count++;
                        }
            Console.WriteLine("Total matching condition: " + count);
        }
    }
}

 

[BoltBait 3,037]

Math FTW!

 

I didn't roll a single die, yet I told you what your results would be for a completely random event. 😎

 

Props on you for doing the empirical study though.

 

[Ego Eram Reputo 2,559]

^^ Fabulous. Both of you have put me to shame 🤭

 

Final results for one to four dice

 

1d6

• Axes panic 16.7%
• Bows panic 16.7%

2d6

• Axes panic 13.9%
• Bows panic 19.4%

3d6

• Axes panic 13.0%
• Bows panic 19.9%

4d6

• Axes panic 11.5%
• Bows panic 19.7%

 

Despite being further from the battle, archers are cowards and more inclined to run!

Edited February 12 by Ego Eram Reputo
Edit: missed 1d6 decimal place.