[Math] Dice Rolling Probability Space

measure-theoryprobabilityprobability theory

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

Best Answer

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.

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