I have no formal background in math, statistics, or anything. Just trying to figure out a fun problem with a game of dice.

Lets say you have 3 people sitting a table rolling dice (including yourself).

This is what the result set looks like:

You: [5, 4, 3, 2, 1] P1: [x, x, x, x] P2: [x, x, x, x, x]

So you know what dice you have but not what the others have. Now P1 guesses that there are 3 fives on the table. What is the probability that there are actually 3 fives on the table?

Since there are 9 unknown dice, I know there are 531,441 possibilities that could have been rolled (9**6). But past that I'm not sure what to calculate next.

EDIT: After thinking about it, 9**6 is probably invalid because although someone could get [5, 4, 3, 2, 1] and [1, 2, 3, 4, 5] that is the same thing for what I'm trying to do, which is find out how probable it is to have X amount of a specific face value (i.e 4 fours)

  • $\begingroup$ To be honest I'd want to say that because P1 is making this claim, he has at least one $5$ $\endgroup$ Jun 27, 2014 at 3:05
  • $\begingroup$ I agree, I think it is a good assumption $\endgroup$
    – sontek
    Jun 27, 2014 at 5:25
  • $\begingroup$ Did you intend P1 to throw four rather than five dice? $\endgroup$
    – Henry
    Jun 27, 2014 at 5:39

1 Answer 1


You need exactly $2$ of the remaining $9$ dice to be $5$'s for P1 to be correct.

There are $\dbinom{9}{2}$ ways to pick the $2$ other dice which need to be $5$'s.

The other $7$ dice need to show one of $\{1,2,3,4,6\}$, so there are $5^7$ ways for this to happen.

Now, you should be able to calculate the total number of ways for exactly $2$ of the remaining $9$ dice to be $5$'s, and thus, the probability that P1 is correct.

  • $\begingroup$ I'm not familiar with math notation is 9 over 2 meaning 9 divided by 2? $\endgroup$
    – sontek
    Jun 27, 2014 at 0:33
  • $\begingroup$ $$\binom{9}{2}=\frac{9!}{2!7!}=36$$ $\endgroup$ Jun 27, 2014 at 3:00
  • $\begingroup$ oh, and $\binom{n}{k}$ is always the number of ways to choose $k$ objects from $n$ objects $\endgroup$ Jun 27, 2014 at 3:02
  • $\begingroup$ The problem is that P1's statement reveals something about the unknown number of $5$s that P1 has, so you actually need a conditional probability, possibly using game theory which would depend on rules of the game we have not seen such as who wins. $\endgroup$
    – Henry
    Jun 27, 2014 at 5:39

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