Game design probability question: what is the most likely value of rolling these three dice?

DA2601

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Mar 29, 2019
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Hey so I’m part of a team working on a mechanic for a video game in essence it uses dice rolls to determine certain outcomes and I have been tasked with finding out the probability of some of these. In short here’s the deal I hope I’m explaining it clearly enough.

I have three six sided dice the values of each dice face are as follows:
1=0
2=0
3=0
4=1
5=1
6=2

First question: what is the most likely value of rolling these three dice? (values of all three dice in total)

I have it as 2 seeing as 6 faces on each die and the total value across those faces is 4 so 4 divided by 6 is 0.66667, and three dice so 0.66667 times 3 is 2. Is this correct?

Second question if I roll these three dice once am I just as likely to roll a value of 2 or higher as in the following scenario:

Rolling the same three dice twice, but on the first roll for every value under 2 I roll one less die in the second roll, however for every value above 2 on the first dice roll I roll an additional dice on the second dice roll.
Only the values rolled on the second dice roll are counted.

For this I honesty have no idea where to begin working this one out could really use some help.
 
Just to clarify on my second question when I say in the second roll for every value above 2 on the first dice roll I roll an additional dice on the second dice roll I am referring to the total value of the three dice.
 
Not too great at math, but I do like to code a bit, so though I can't give you a proof of how to get each probability, I can give you a pretty precise probability of your totals from the added dice to equal each number. Can't currently run it as I'm at work so I'll run it in about 2 hours, but here's the code (in BASIC) if you have an app to run it like AGK. Video game probabilities are very interesting to me so I hope it helps to see each individual percentage in a live-fire format. I'll get back to you in a few hours with the results, hope this helps if even just a little.
___________________________

dice1 = 0
dice2 = 0
dice3 = 0
rolls = 0
percent0 = 0
percent1 = 0
percent2 = 0
percent3 = 0
percent4 = 0
percent5 = 0
percent6 = 0
got0 = 0
got1 = 0
got2 = 0
got3 = 0
got4 = 0
got5 = 0
got6 = 0
i = 1

do

dice1 = random(1, 6)
if dice1 = 1 then roll1 = 0
if dice1 = 2 then roll1 = 0
if dice1 = 3 then roll1 = 0
if dice1 = 4 then roll1 = 1
if dice1 = 5 then roll1 = 1
if dice1 = 6 then roll1 = 2

dice2 = random(1, 6)
if dice2 = 1 then roll2 = 0
if dice2 = 2 then roll2 = 0
if dice2 = 3 then roll2 = 0
if dice2 = 4 then roll2 = 1
if dice2 = 5 then roll2 = 1
if dice2 = 6 then roll2 = 2

dice3 = random(1, 6)
if dice3 = 1 then roll3 = 0
if dice3 = 2 then roll3 = 0
if dice3 = 3 then roll3 = 0
if dice3 = 4 then roll3 = 1
if dice3 = 5 then roll3 = 1
if dice3 = 6 then roll3 = 2

total = roll1 + roll2 + roll3
rolls = rolls + 1

if total = 0 then got0 = got0 + 1
if total = 1 then got1 = got1 + 1
if total = 2 then got2 = got2 + 1
if total = 3 then got3 = got3 + 1
if total = 4 then got4 = got4 + 1
if total = 5 then got5 = got5 + 1
if total = 6 then got6 = got6 + 1

if got0 > 0 then percent0 = got0 / rolls
if got1 > 0 then percent1 = got1 / rolls
if got2 > 0 then percent2 = got2 / rolls
if got3 > 0 then percent3 = got3 / rolls
if got4 > 0 then percent4 = got4 / rolls
if got5 > 0 then percent5 = got5 / rolls
if got6 > 0 then percent6 = got6 / rolls

print("Total rolls: " + str(rolls))
print(" ")
print("Times rolls add to zero: " + str(got0) + " " + str(percent0))
print(" ")
print("Times rolls add to one: " + str(got1) + " " + str(percent1))
print(" ")
print("Times rolls add to two: " + str(got2) + " " + str(percent2))
print(" ")
print("Times rolls add to three: " + str(got3) + " " + str(percent3))
print(" ")
print("Times rolls add to four: " + str(got4) + " " + str(percent4))
print(" ")
print("Times rolls add to five: " + str(got5) + " " + str(percent5))
print(" ")
print("Times rolls add to six: " + str(got6) + " " + str(percent6))

if GetRawKeyPressed(27)
repeat i
print("Total rolls: " + str(rolls))
print(" ")
print("Times rolls add to zero: " + str(got0) + " " + str(percent0))
print(" ")
print("Times rolls add to one: " + str(got1) + " " + str(percent1))
print(" ")
print("Times rolls add to two: " + str(got2) + " " + str(percent2))
print(" ")
print("Times rolls add to three: " + str(got3) + " " + str(percent3))
print(" ")
print("Times rolls add to four: " + str(got4) + " " + str(percent4))
print(" ")
print("Times rolls add to five: " + str(got5) + " " + str(percent5))
print(" ")
print("Times rolls add to six: " + str(got6) + " " + str(percent6))
if GetRawKeyPressed(27) then i = 2
until i = 2
endif

sync()

loop
 
I have three six sided dice the values of each dice face are as follows:
1=0
2=0
3=0
4=1
5=1
6=2
First question: what is the most likely value of rolling these three dice? (values of all three dice in total)
Can you please help me understand the setup?
It seems as if there are three six-sided dice. Each die has three zero-faces, two one-faces and one two-face. Is that correct?
If not what does 6=2 mean?

Second question if I roll these three dice once am I just as likely to roll a value of 2 or higher as in the following scenario:
Rolling the same three dice twice, but on the first roll for every value under 2 I roll one less die in the second roll, however for every value above 2 on the first dice roll I roll an additional dice on the second dice roll.
Only the values rolled on the second dice roll are counted. ?????
 
Thanks for the replies!

So I figure I should probably explain in more detail, with my six sided dice:
The one face of a die has a value of 0
The two face of a die has a value of 0
The three face of a die has a value of 0
The four face of a die has a value of 1
The five face of a die has a value of 1
The six face of a die has a value of 2

I could be rolling any number of these dice, but for this example let’s say I roll six of these dice, maybe a better way to phrase my second question would be:

As we have established that the most likely result from rolling six of my dice is 4 is it equally likely that the result would be 4 if I do the following:

Roll six of my dice and then total the value of those dice, if my total value exceeds 4 I roll six dice plus an extra amount of dice equal to the amount that exceeds 4 in my second roll. Conversely if my first roll valued less than 4 in my second roll I would be rolling six minus the amount my fist roll was less than 4. If on the first roll the value rolled was 4 I simply roll six dice on the second roll.
In doing this would I be equally likely to end with a value of 4 on my second roll?

If this needs more explanation I’ll give an example: I’m going to roll three of my dice, my three dice land on the four face, two face and one face this equals a total value of 1, which is less than 2 by 1. Now when I make my second roll I deduct 1 from the number of dice I get to roll so I only roll two dice rather than three. The value I roll on my second roll is the value I keep.

But if I had rolled my three dice having my dice land on the six face, another six face and a five face my total value would be 5. Now as my value is 3 higher than 2 I now roll three dice in addition bringing the total dice rolled up to six for my second roll. I would now roll six dice to determine the value.

Basically would I be more likely to score a higher value rolling my dice just once or twice if following the rules I’ve set out above for rolling twice? Or does it not matter do I have the same probability rolling once as I do twice?

Hope that makes sense!
 
So I figure I should probably explain in more detail, with my six sided dice:
The one face of a die has a value of 0
The two face of a die has a value of 0
The three face of a die has a value of 0
The four face of a die has a value of 1
The five face of a die has a value of 1
The six face of a die has a value of 2
DA2601, did you read my question quoted below.
Can you please help me understand the setup?
It seems as if there are three six-sided dice. Each die has three zero-faces, two one-faces and one two-face. Is that correct?
If not what does 6=2 mean?
If you did why not model your question as I suggest. Look all probability questions turn upon the model used.
Can you see that the new die I propose models what you describe. Roll six of these dice.
Now look at this link. 11578
That expansion tell us all we want to know about the outcomes. Look at the term \(\displaystyle 90x^4\). That tells us that there are ninety ways that the sum of the faces equals four. Look at the image:
there is one way to get a twelve
there is one way to get all zeros
there are six ways to get a one
etc
Using that WolframApha site you can change the \(\displaystyle (1+x+x^2)^n\) to any integral value for \(\displaystyle n\).
The resulting expansion will tell you the results of rolling \(\displaystyle n\) of these dice.
I just changed \(\displaystyle n\text{ to }20\) the expansion contains \(\displaystyle 18062x^{11}\).
So now I know that there are \(\displaystyle 18062\) ways to get an \(\displaystyle 11\) if we roll \(\displaystyle 20\) of these dice.
 
Little later than I thought lol but here’s the outcome of my program, out of 10000 rolls these are what they added to each time
 

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