1

This is not a homework problem, I'm just trying to better my understanding of this concept because it's so interesting yet counterintuitive to me and. I'm hoping someone will confirm that this is either correct or provide me with a logical explanation for why it's wrong.

Say you have $3$ die. They're all indistinguishable, 6-sided and fair. Let's say you roll all $3$ die at the exact same time. $2$ of the die land on the table, and $1$ rolls off and hits the floor.

You see that the $2$ die on the table are both $3$'s. What is the probability that the die on the floor is also a $3$?

My understanding is that the correct probability that 3rd die is a 3 should be $1/16$.

$3$ die were rolled for a total of $6^3$ possible combinations = $216$ and by revealing that there are two 3's, we reduce our sample space to the following 16 possibilities:

enter image description here

of which only {3, 3, 3} satisfies our event, thus 1/16.

Because the die are unordered, the 3rd die would not be considered an independent event. If I said I have a red die, a blue die, and a green die and after rolling them I saw that the red and blue die landed on the table, then the green die that lands on the floor would be considered to be a 1/6 chance.

I'd also appreciate any additional information that might help me further wrap my head around why this is.

PDef
  • 37
  • 1
    See the boy-girl problem. Compare to the version of the question where "a girl answers the door." The correct answer is $\frac{1}{6}$. "At least two of the dice show a 3" would include the possibility of the die on the floor being a $3$ and exactly one of the two dice still on the table being a $3$. This is not the scenario you are describing. – JMoravitz Feb 17 '21 at 21:15
  • 2
    As an aside... brute force listing of all the possibilities like you have is generally a terrible idea. The whole point of a course in combinatorics or probability is learning how to count without needing to resort to such tactics. – JMoravitz Feb 17 '21 at 21:16
  • 2
    Indeed. The title of a book that served as my childhood introduction to this fascinating field said it perfectly: *Combinatorics: How to count without counting*. – David G. Stork Feb 17 '21 at 21:17
  • 1
    The very fact that you can say there are three dice means they are distinguishable. The probabilty that the die on the floor is a $3$ is $1/6$. – Rob Arthan Feb 17 '21 at 21:22

1 Answers1

1

Your fundamental mistake is that, contrary to your belief, the third die is an independent event. Even though the dice aren't numbered or painted and weren't initially distinguishable, the fact that the third die is the die on the floor and the two dice whose results you know are precisely the two dice that aren't on the floor makes the third die distinguishable. In the terms of your brute force listing, you are including combinations where one of the dice on the table shows something other than a $3$. The correct answer is therefore $\frac 16$.

As stated in the comments, if you knew that at least two dice showed a $3$ and exactly two dice ended up on the table, then the result of the "floor" die would not be independent from the information that you already have.

Robert Shore
  • 23,332
  • 1
    I think I understand now. I thought it was a mathematic concept when it seems more like it's just a language 'smoke and mirrors' type of situation. By stating that the dice on the table are 3's, I'm eliminating the chance of either of those 3's being on the floor – PDef Feb 17 '21 at 22:29