[Math] How to calculate dice with addition and subtraction based on dice rolls

probabilityprobability distributions

I am trying to figure out how to calculate results on a group of dice where some results are positive and others are negative.

Example: I roll a group of dice that are fair and six-sided. Each roll of $5$ or $6$ is a $+1$. Each roll of $1$ is a $-1$. I need to know how to calculate the probability for different size groups of dice depending on the overall result being positive, zero, or negative.

Edit:

To clarify:
I roll $x$ dice with the possible results of $-1, 0, 0, 0, 1, 1$ for each die.

  1. What is the chance that the sum would be negative ($5\%, 20\%$, whatever)?
  2. What is the chance that the sum would be zero ($5\%, 20\%$, whatever)?
  3. What is the chance that the sum would be positive ($5\%, 20\%$, whatever)?

I am trying to find a formula that will let me input various numbers of dice and get the percentage chances for negative, zero, and positive. I am not sure what a multinomial distribution is and was thinking of just setting my computer for a brute force attack and count each possibility, but would rather not.

Further clarification:

Someone did a sample roll of 1000d6 and came up with a distribution result. that is nice but doesn't answer my question.
I did 4d6 by brute force since that is only 1296 possible combinations and came up with specific results.
There is a 19.68% chance of getting a negative result.
There is a 24.77% chance of getting a zero result.
There is a 55.56% chance of getting a positive result.

I truncated the results, figuring that was enough significant decimal places.
What I am trying to find is what is the formula to calculate these results.

Best Answer

Comment:

The Comment by @GiovanniResta gives results for getting negative, zero, or positive sums when two dice are rolled. I'm checking these results by simulating a million rolls of a pair of dice in R. I use dice as proposed by @Logophobic.

 die = c(-1, 0, 0, 0, 1, 1)
 B = 10^6;  n = 2
 faces = sample(die, B*n, repl=T)
 MAT = matrix(faces, nrow=B) # B x n matrix, each row an n die experiment
 x = rowSums(MAT);  table(x)/m  # totals on n dice
 ## x
 ##       -2       -1        0        1        2 
 ## 0.027880 0.166367 0.361880 0.332582 0.111291 

 mean(x < 0);  mean(x==0);  mean(x > 0)  # whether totals are neg, zero, or pos
 ## 0.194247  # compare:  7/36 =  0.1944444; (0.027880 + 0.166367 = 0.194247)
 ## 0.36188   # compare: 13/36 =  0.3611111
 ## 0.443873  # compare:  4/9  =  0.4444444

This is one possible interpretation of your question, and simulation results agree with the proposed exact ones to 2 or 3 places, as they should with a million rolls. I do not see how this method can be easily generalized to other values of $n$.

However, it occurs to me that you might want an answer using the $multinomial\,$ $distribution.$ It is not clear to me exactly what you $are$ asking. What do you mean by "the probability" in your Question. Please state your question more explicitly.

Addendum: Based on your clarification, here are simulation results for $n = 3$ and $n = 4$. Same code as for $n = 2$, just change the number.

 n = 3
 x
       -3       -2       -1        0        1        2        3 
 0.004721 0.041662 0.152111 0.291366 0.305879 0.167397 0.036864 
 mean(x < 0);  mean(x==0);  mean(x > 1)
## 0.198494
## 0.291366
## 0.204261

n = 4
x
      -4       -3       -2       -1        0        1        2 
0.000778 0.009277 0.048022 0.138436 0.247043 0.278741 0.191518 
        3        4 
 0.073865 0.012320 

 mean(x < 0);  mean(x==0);  mean(x > 1)
 ## 0.196513
 ## 0.247043
 ## 0.277703
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