# Dice Rolling Probability Space

I have a question in which the person asking has identified that the total sum of 11 comes up more often than a sum of 12 in the rolling of three dice and this is strange as they both have the same number of possible combinations of 3 numbers that make up 11 and 12.

Clearly 11 can be made up of combinations of : (6,4,1),(6,3,2),(5,5,1),(5,4,2),(5,3,3),(4,4,3). And I argued that when we have (a,b,c) with $a\ne b\ne c$, then we have 3! possible combinations, when two of a,b,c are equal then we have 3, and when we have $a=b=bc$ then we only have one combination.

I then argued that 12 has: (6,5,1), (5,5,2), (4,4,4), (5,4,3), (3,6,3), (6,2,4) and because of the (4,4,4) triplet which only has one possible combination, the pr of a 12 is slightly lower than the possibiity of an 11.

The question asks me 'introduce a probability space' to tackle this question and I'm a bit confused as to how to approach this. What I've put together:

$\Omega=\{1,2,3,4,5,6\}^3$

$\mathscr F=\mathscr P(\Omega)$

$\mathbb P: \mathscr F\to[0,1]:\mathscr F (x)=\frac{card(x)}{216}$

where card is cardinality of the set x. Or should the function be a piecewise that depends on different values of a,b,c in the triplets?

Can I define the function like this instead?:

let the triplet (a,b,c) correspond to the outcome on the toss of each die, such that a,b,c$\in${1,2,3,4,5,6}

Let F be a function mapping the power set to [0,1] and x=(a,b,c).

Case1 (a=b=c): f(x) = 1/216

case2 (a=b, a$n=$ c) f(x) = 3/216

case 3 ($a\ne b \ne c$)

f(x) = 3!/216 = 6/216

• For the sake of computing the number of different combinations that sum up to a certain value, the order might not matter (depending on how you define different combinations). But for the sake of computing probability, it does. The probability space is simply the set of all $216$ possible combinations. Mar 12, 2015 at 10:10
• @barakmanos I thought the probability space is $(\Omega,\mathscr F, \mathbb P)$, the set of all 216 possible outcomes is just describing the sample space $\Omega$ isn't it? Mar 12, 2015 at 10:16

The dice are distinct objects. By coloring the dice, it might be easier see how the numbers are counted. Here are the different ways to roll $11$ on $3$ dice:

RGB    RGB    RGB
641    461    335
632    452    326
623    443    263
614    434    254
551    425    245
542    416    236
533    362    164
524    353    155
515    344    146


Each of these has a $\frac1{216}$ chance of being rolled. Thus, an $11$ has a $\frac{27}{216}$ chance of being rolled.

• Thanks for the answer, I've actually solved for this, my challenge atm is finding the probability space, that is, I have omega, I want to find the sigma field and probability measure for this experiment... Mar 12, 2015 at 14:46
• So you have the sample space, which is the $216$ different dice rolls, each with a probability of $\frac1{216}$. Above I have listed the outcomes meeting the criteria that the sum of the dice is $11$. You can do the same for $12$, or think of some more clever way. I listed them all so that it would be clear that 641 was to be counted separately from 416, and 443 was to be counted separately from 434, etc.
– robjohn
Mar 12, 2015 at 14:52
• but how do you determine the probability measure and sigma field by knowing the sample space?... this is my question Mar 12, 2015 at 14:53
• The probability measure is discrete; $\frac1{216}$ given to each element of the probability space. The sigma algebra would be the power set of these $216$ different outcomes.
– robjohn
Mar 12, 2015 at 14:59